Morita Equivalence of Graph and Ultragraph Leavitt Path Algebras
aa r X i v : . [ m a t h . R A ] J un Morita Equivalence of Graph and UltragraphLeavitt Path Algebras
Michael Mekonen Firrisa Dartmouth College, Hanover, New Hampshire
May, 2020
Abstract
The primary purpose of this thesis is to show every ultragraph Leav-itt path algebra is Morita equivalent, as a ring, to a graph Leavitt pathalgebra. Takeshi Katsura, Paul Muhly, Aidan Sims, and Mark Tomfordeshowed every ultragraph C ∗ -algebra is Morita equivalent, in the C ∗ -sense,to a graph C ∗ -algebra; our result is an algebraic analog of this fact. Fur-ther, we will use our result to give an alternate proof for establishedconditions which guarantee the simplicity of an ultragraph Leavitt pathalgebra over a field. Before proceeding with the task at hand, we would like the reader to know thework presented here is the author’s Ph.D. thesis. The history of Leavitt pathalgebras, as with most areas of mathematics, is the confluence of ideas fromdifferent fields brought about by the work of numerous mathematicians. Theinterested reader can find a great description of the history in [2], but we willgive a comparatively brief treatment here.Our story starts with an investigation into the invariant basis number (IBN)property. A unital ring R satisfies the IBN property if the free modules R n and R m being isomorphic implies m = n . Some rings (e.g., non-trivial commutativerings, matrix rings over a field) always satisfy the IBN property, but not all do.In the early 60’s, the eponymous William G. Leavitt proved that given any twopositive integers n , m , with n = m , there exists a ring R such that R n ∼ = R m [22, Theorem 8]; for k = | m − n | , one can show R i ∼ = R j ⇐⇒ i ≡ j ( mod k ) . If m is the smallest integer such that there exists an integer l > m with R m ∼ = R l ,and n > m is the smallest such integer for which R m ∼ = R n , we say R has moduletype ( m, n ). Leavitt achieved his result in part by showing, given any field1 , one can construct a unital K -algebra (see Definition 3.1), L K ( m, n ), withmodule type ( m, n ). The algebra L K ( m, n ) is at times referred to as the Leavittalgebra of type ( m, n ). Leavitt goes on to prove interesting results regardingthese algebras. Of particular interest is the universal property they possess:if A is a unital K -algebra with module type ( m, n ), then there exists a unital K -algebra homomorphism φ : L K ( m, n ) → A. We will see later on that Leavitt path algebras have a similar universal property.While Leavitt path algebras do not live in the world of analysis, an importantcomponent of their story comes from C ∗ -algebras . To briefly describe what a C ∗ -algebra is, let A be a C -algebra. Suppose A is equipped with an involutionmap, ∗ : A → A , with a ∗ denoting the image of a ∈ A , such that( a + b ) ∗ = a ∗ + b ∗ , ( ab ) ∗ = b ∗ a ∗ , and ( ka ) ∗ = ka ∗ for all a, b ∈ A and k ∈ C . Lastly, suppose A is also equipped with a norm, k · k , such that k ab k≤k a kk b k , and k aa ∗ k = k a k = k a ∗ k ( C ∗ -identity) , for all a, b ∈ A . If A is a complete topological space with respect to the topologyinduced by the norm, A is a C ∗ -algebra. The study of these objects has its rootsin quantum mechanics. It has since grown into a vast area of functional analysisin its own right. Mathematicians have worked on classifying them, and givingexplicit constructions of C ∗ -algebras with desired properties, for decades since.In line with this endeavor, Joachim Cuntz gave an explicit construction forobtaining unital, simple, infinite, separable C ∗ -algebras ( Cuntz algebras ) in thelate 70’s [10]—a C ∗ -algebra A is separable if it contains a dense countable subset,it is infinite if there exits a ∈ A such that aa ∗ = 1 and a ∗ a = 1, and it is simpleif it doesn’t contain any non-trivial closed two-sided ideals (i.e., A and { } arethe only closed two-sided ideals). Mathematicians, including Cuntz himself,then took on the enterprise of generalizing Cuntz algebras, giving us Cuntz-Krieger algebras . By 1982, Yasuo Watatani had noticed each Cuntz-Kriegeralgebra has a generating set whose elements and relations can be encoded bya directed graph with a finite number of vertices, where each vertex receivesand emits a non-zero finite amount of edges. This laid the foundation for graphalgebras. Of course, as time went on, graph algebras came to include morearbitrary directed graphs than those introduced by Watatani. Although it didn’timmediately attract much attention, many had taken notice of the utility in thegraph approach before the 90’s were through. In showing one can deduce resultsregarding C ∗ -algebras by studying the properties of the much more tractabledirected graphs, papers such as [20], [9], and [21] helped to illumine the powerand elegance of graph algebras.Motivated by purely infinite simple C ∗ -algebras, the mathematicians PereAra, K. R. Goodearl, and Enrique Pardo introduced an algebraic analog to2hese objects in the early 2000’s: purely infinite simple rings [7]. The samegroup of mathematicians, along with M. A. Gonz´alez-Barroso, set out to findways to construct explicit examples of such rings. In [6], they introduced away to construct a class of rings called fractional skew monoid rings , which arethemselves purely infinite simple rings. In investigating the algebraic K -theoryof fractional skew monoid rings, the aforementioned authors introduced Leavittpath algebras in the same paper—having been inspired by the work of C ∗ -algebraists, the authors used the term “graph algebras” instead of “Leavitt pathalgebras.” As it turned out, another group of mathematicians had happenedupon Leavitt path algebras as well.During a conference held in 2004 on graph C ∗ -algebras, which some ring the-orists were invited to attend, one of the ring theorists present, Gene Abrams,noticed the algebraic data of graph C ∗ -algebras is quite similar to objects knownwell by algebraists as path algebras . Inspired by what he saw at the conference,Abrams, together with Aranda Pino, published [3]. In their paper, Abramsand Aranda Pino define Leavitt path algebras to stand as an algebraic analogto graph C ∗ -algebras. Much like the initial results established for graph C ∗ -algebras, the main result of [3] is a simplicity theorem for Leavitt path algebras;graph C ∗ -algebras are much older than Leavitt path algebras and have moreestablished results, a lot of the research in Leavitt path algebras is thus dedi-cated to proving analogs of these results (e.g., this thesis). Now, Abrams hadbeen studying Leavitt algebras prior to his joint work with Aranda Pino. Hisparticular focus had been the Leavitt algebra L K (1 , n ). Through the course oftheir work, Abrams and Aranda Pino realized the Leavitt algebra L K (1 , n ) isan example of a Leavitt path algebra. It is this realization which imparted the“Leavitt” to Leavitt path algebras. The name “Leavitt path algebra” becameubiquitous due to the fact [3] appeared in print before [6]. Regardless, both[3] and [6] are taken to be the foundation of Leavitt path algebras. It’s worthnoting that Leavitt path algebras were always taken to be over a field in theiroriginal definition. It was not until Mark Tomforde’s work in [31] that they weredefined for algebras over unital commutative rings.The main aim of this thesis is to establish the Morita equivalence of graphand ultragraph Leavitt path algebras as rings. As mentioned before, this en-deavor mirrors a result from the world of C ∗ -algebras as well; a result whichshowed how graph C ∗ -algebras are related to another class of C ∗ -algebras: Exel-Laca algebras. In their original formulation, Cuntz-Krieger algebras are C ∗ -algebras associated to finite matrices with entries in { , } . Watatani’s workled to their generalization as graph algebras. However, by dropping the finiterestriction on the associated matrices, with the only criteria being they don’tcontain a zero row, Exel-Laca algebras were introduced as another general-ization of Cuntz-Krieger algebras [11]. So, naturally, one is compelled to askexactly how these two class of C ∗ -algebras are related, seeing they both gener-alize Cuntz-Krieger algebras. It was worked out relatively quickly that they arenot the same—there are Exel-Laca algebras which cannot be realized as graph C ∗ -algebras [26, Example 4.2], and there are graph C ∗ -algebras which cannotbe realized as Exel-Laca algebras [28, Proposition A.16.2]. Then exactly how3re they related? In [19], the authors showed the two classes are the same up toMorita Equivalence. In [30], Tomforde introduced the notion of ultragraphs andultragarph C ∗ -algebras as a way of unifying graph algebras and Exel-Laca alge-bras. The class of ultragraph C ∗ -algebras properly contains graph algebras andExel-Laca algebras [29, Section 5]. Thus, to establish the Morita equivalenceof graph algebras and Exel-Laca algebras, it suffices to show any ultragraph C ∗ -algebra is Morita equivalent to a graph algebra, and vice versa. That isprecisely what was established in [19], an algebraic analog of which we wish toprove in this thesis. Definition 2.1. A graph , E = ( E , E , r E , s E ), consists of a set of countablevertices E , a set of countable edges E , and range and source maps r E , s E : E → E .A finite path in E , α = e e . . . e n , is a sequence of edges such that s E ( e i +1 ) = r E ( e i ) for 1 ≤ i ≤ n −
1. We set | α | := n to be the length of α ; the set of such paths of finite length is denoted by E ∗ ( E is included in E ∗ as paths of length 0). The range and source maps are extended to E ∗ by r E ( α ) = r E ( e n ) and s E ( α ) = s E ( e ); we set r E ( v ) = s E ( v ) = v for v ∈ E .Given α, β ∈ E ∗ , with r E ( α ) = s E ( β ), we get an element of E ∗ by concatenat-ing α with β which is denoted by αβ . A path α such that s E ( α ) = r E ( α ) iscalled a cycle . Finally, we call v ∈ E a singular vertex if | s − E ( v ) | = 0 or ∞ . Remark . The reader might find, when reading about graph C ∗ -algebras inparticular, a slightly different definition of a finite path: a sequence of edges α = e e . . . e n such that s E ( e i ) = r E ( e i +1 ) for all 1 ≤ i ≤ n −
1; further, r E ( α ) := r E ( e ) and s E ( α ) := s E ( e n ). This difference in convention doesn’tchange any of the established results so long as one accounts for the reversal inroles of the range and source maps. The author has yet to see this conventionused within the context of Leavitt path algebras.Pictorially, we depict graphs in terms of dots and arrows: a dot for eachvertex, an arrow for each edge, and the placement of the arrow indicates thesource and range vertices of the associated edge.4 xample . vve e n e e Figure 1: E = { v } , E = { e , e , . . . , e n } , and s E ( e i ) = r E ( e i ) = v for each i. Example . v v v v n − v n e e . . . e n − Figure 2: E = { v , v , . . . , v n } , E = { e , e , . . . , e n } , s E ( e i ) = v i − and r E ( e i ) = v i for each i. Definition 2.5. [30] An ultragraph G = ( G , G , r, s ) consists of a set of count-able vertices G , a set of countable edges G , a source map s : G → G , and arange map r : G → P ( G ) \ {∅} ( P ( G ) denotes the power set of G ). Further, G denotes the smallest subset of P ( G ) containing { v } for each v ∈ G and r ( e ) for each e ∈ G , and is closed under finite unions, finite intersections, andrelative complements.Similar to graphs, a finite path in G , α = e e . . . e n , is a sequence of edgessuch that s ( e i +1 ) ∈ r ( e i ) for 1 ≤ i ≤ n −
1; we define its length to be | α | := n .The set of finite paths in G is denoted by G ∗ . We extend the range and sourcemaps to r : G ∗ → G and s : G ∗ → G by s ( α ) = s ( e ) and r ( α ) = r ( e n ); further,since G sits in G ∗ as paths of length 0, for A ∈ G , we set r ( A ) = s ( A ) = A .Given α, β ∈ G ∗ , with s ( β ) ∈ r ( α ), we get an element of G ∗ by concatenating α with β ; we denote this element by αβ . If α is a path such that s ( α ) ∈ r ( α ), itis called a cycle . Also, we call v ∈ G a singular vertex if | s − ( v ) | = 0 or ∞ .Again, similar to graphs, we depict ultragraphs in terms of dots and arrows:a dot for each vertex, an arrow for each edge, and the placement of the arrowindicates the source vertex and the range “generalized” vertex of the associatededge. The following is a simple example but one which will be useful later.5 xample . v v v v n . . .e e e . . .. . . . . . Figure 3: G = { v , v , v , . . . , v n , . . . } , G = { e } , G = {S ∈ P ( G ) : S , or G \ S , is finite } , s ( e ) = v , r ( e ) = { v , v , . . . , v n , . . . } . The following construction, from [19], plays a fundamental role in our work. For n ∈ N , let { , } n denote the set of words of length n whose entry are 0’s and 1’s.Given ω ∈ { , } n , and 1 ≤ i ≤ n , ω i := i -th entry of ω . We will at times write ω = ( ω ω ...ω n ) and set its length to be | ω | := n . Further, we define elements( ω, , ( ω, ∈ { , } n +1 by ( ω,
0) := ( ω ω ...ω n
0) and ( ω,
1) := ( ω ω ...ω n m ≤ n , we define ω | m ∈ { , } m by ω | m := ( ω ω ...ω m ) . Finally, we let0 n (similarly, 1 n ) denote the sequence of n n n − ,
1) to denote the sequence of ( n −
1) 0’s followed by a 1.With this in mind, let G be an ultragraph and { e , e , ... } an enumeration of G . For ω ∈ { , } n \ { n } , set r ( ω ) := \ { i : ω i =1 } r ( e i ) ! \ [ { j : ω j =0 } r ( e j ) ! ⊆ G , also set ∆ n := { ω ∈ { , } n \ { n } : | r ( ω ) | = ∞} , ∆ := ∞ [ i =1 ∆ n , Γ := { (0 n ,
1) : n ≥ , | r ((0 n , | = ∞} , Γ + := ∆ \ Γ ,W + := [ ω ∈ ∆ r ( ω ) ⊆ G , and W := G \ W + . Lemma 2.7.
There exists a function σ : W + → ∆ such that v ∈ r ( σ ( v )) foreach v ∈ W + , and σ − ( ω ) is finite (possibly empty) for each ω ∈ ∆ .Proof. [19, Lemma 3.7]. 6s for σ in Lemma 2.7, it can be extended to a function σ : G → ∆ ∪ {∅} by setting σ | W + := σ and σ | W := ∅ . Abusing notation, we will write “ σ ” toalso denote σ . Now, for each n ∈ N , set X ( e n ) := { v ∈ r ( e n ) : | σ ( v ) | < n } ⊔ { ω ∈ ∆ n : ω n = 1 } ⊆ G ⊔ ∆;we have the following useful lemma. Lemma 2.8.
For each n ∈ N , X ( e n ) is nonempty and finite.Proof. [19, Lemma 3.11]. Definition 2.9.
Let G = ( G , G , r, s ) be an ultragraph. Fix an enumerationof G and let σ be the function described following Lemma 2.7, we can thenconstruct a graph E G = ( E G , E G , r E , s E ) associated to G . First, set E G := { v ι } ι ∈ G ⊔ ∆ , E G := { e κ } κ ∈{ W + ⊔ Γ + }⊔{ ( e n ,x ): e n ∈G , x ∈ X ( e n ) } ;then, define the range and source maps as follows:if κ = v ∈ W + , then r E ( e v ) = v v and s E ( e v ) = v σ ( v ) , if κ = ω ∈ Γ + , then r E ( e ω ) = v ω and s E ( e ω ) = v ω | | ω |− , and finally, r E ( e ( e n ,x ) ) = v x and s E ( e ( e n ,x ) ) = v s ( e n ) . For the sake of convenient notation, we will write “( e n , x )” for e ( e n ,x ) . Wewill also usually write the other elements of E G as “ v ”, or “ ω .” It’s importantto note E G depends on our choice of ordering of G as well as our choice of σ (please see Figures 4 and 5 in Example 2.12 for an illustration). Definition 2.10.
Let E = ( E , E , r E , s E ) be a graph. A subgraph F =( F , F , r F , s F ) of E is a graph such that: F ⊆ E , F ⊆ E , r F = r E | F , and s F = s E | F , where, for each e ∈ F , s E ( e ) , r E ( e ) ∈ F . Remark . There is an important subgraph F of E G which we will call uponin later chapters; it is the subgraph where F = E G and F = { e κ } κ ∈ W + ⊔ Γ + . For more details, see Section 4 of [19]. 7 xample . v v v v v v k w w w w w w w w w w w w w . . . . . . ... ... ... ... ... e e e e e e e e e e e e e e e e e e . . . . . . Figure 4: Ultragraph G : The dashed gray arrows illustrate the ordering mech-anism for G . 8 v v v v v v v v v v v k v (1) v (10) v (100) v (1000) v (1 , k − ) v w v w v w v w v w v w v w v w v w v w v w v w v w . . . . . . ... ... ... ... ... ( e , w e , w e , w e , w
11) ( e , w e , w e , w e , v
12) ( e , w e , w e , w
13) ( e , w e , w . . . . . . ( e , (1)) e (10) e (100) e (1000) e (1 , k − ev ev ev ev evk Figure 5: Graph E G : σ : W + → ∆ is given by v i (1 , i − ).9 Leavitt Path Algebras
Before we begin with the meat of this section, we would like to let our readerknow algebras are taken to be over unital commutative rings for the remainderof this thesis.
Definition 3.1.
Let R be a commutative unital ring. An R -algebra A is a ring(not necessarily unital) equipped with a map · : R × A → A such that:1) r · ( x + y ) = r · x + r · y , and r · ( xy ) = ( r · x ) y = x ( r · y ), for all r ∈ R andall x, y ∈ A ,2) ( r + r ) · x = r · x + r · x for all r , r ∈ R and for all x ∈ A ,3) ( r r ) · x = r · ( r · x ) for all r , r ∈ R and for all x ∈ A ,4) 1 · x = x for all x ∈ A .Moving forward, we will simply write “ rx ” to denote r · x . Also, if A happensto be a unital ring, we say A is a unital R -algebra . Definition 3.2.
Let A and B be R -algebras. A map f : A → B is an R -algebra homomorphism if f ( r x + r y ) = r f ( x ) + r f ( y ) and f ( xy ) = f ( x ) f ( y )for all r , r ∈ R and x, y ∈ A . If A and B are unital R -algebras, with units1 A and 1 B respectively, we say f is a unital R -algebra homomorphism if it alsosatisfies the condition f (1 A ) = 1 B . Given a graph E = ( E , E , r E , s E ), we define a set of “ghost edges” by associ-ating an element e ∗ to each edge e ,( E ) ∗ := { e ∗ } e ∈ E . And, for each α = e e . . . e n ∈ E ∗ , we define an associated “ghost path” by α ∗ := e n ∗ e ( n − ∗ . . . e ∗ . Definition 3.3.
Let E be a graph. A Leavitt E -family in an R -algebra A is aset { Q v , T e , T e ∗ } v ∈ E ,e ∈ E ⊆ A such that:( LP1 ) Q v Q w = δ v,w Q v for all v, w ∈ E ,( LP2 ) Q s E ( e ) T e = T e Q r E ( e ) = T e , Q r E ( e ) T e ∗ = T e ∗ Q s E ( e ) = T e ∗ for all e ∈ E ,( LP3 ) T e ∗ T f = δ e,f Q r E ( e ) for all e, f ∈ E ,( LP4 ) Q v = P e ∈ s − E ( v ) T e T e ∗ whenever 0 < | s − E ( v ) | < ∞ .For the sake of avoiding cluttered notation, we will write “ { Q v , T e , T e ∗ } ” todenote { Q v , T e , T e ∗ } v ∈ E ,e ∈ E . 10 efinition 3.4. [3, 6, 31] A Leavitt path algebra of a graph E over R , L R ( E ),is an R algebra generated by a Leavitt E - family { q v , t e , t e ∗ } ⊆ L R ( E ) havingthe following universal mapping property: given an R -algebra A and a Leavitt E -family { Q v , T e , T e ∗ } ⊆ A , there exists an R -algebra homomorphism φ : L R ( E ) → A with φ ( q v ) = Q v , φ ( t e ) = T e , and φ ( t e ∗ ) = T e ∗ . We will write “ L R ( { q, t } ),” or“ L R ( E ),” to denote such an algebra.We will briefly mention how to construct such an algebra later on in this sec-tion. It’s worth noting that, due to its universal mapping property, it’s uniqueup to isomorphism. For this reason, once the existence of such an algebra isestablished, we will say “the” Leavitt path algebra of a graph. The followingconstruction (see proof of [31, Proposition 3.4]) is of important utility in deter-mining some properties of L R ( E ). For the reader unfamiliar with R -modules,please see Definitions 5.1 and 5.2 before proceeding. Let R be a unital commu-tative ring and E a graph. Set Z := L n ∈ N R ; then, for each e ∈ E , let Z e := Z. For each v ∈ E , let Z v := L e ∈ s − E ( v ) Z e if 0 < | s − E ( v ) | < ∞ ,Z ⊕ (cid:18) L e ∈ s − E ( v ) Z e (cid:19) if | s − E ( v ) | = ∞ ,Z if | s − E ( v ) | = 0 . Lastly, let X := L v ∈ E Z v .Now, for each v ∈ E , take the identity map Id Z v : Z v → Z v . We extendId Z v to an R -module map Q v : X → X , where Q v is such that, given theinclusion map M v = w ∈ E Z w i ֒ −→ X ,Q v ◦ i = 0 (and Q v ◦ Id Z v : Z v → X is the inclusion map of Z v into X ). Foreach e ∈ E , note that Z e and Z r E ( e ) are both free R -modules on a countableset. Thus, by the universal mapping property of free R -modules, we can definean isomorphism T ′ e : Z r E ( e ) → Z e —note that Z e is a summand of Z s E ( e ) , andso a summand of X as well. We then extend T ′ e to a map T e : X → X , where,identifying Z e with its inclusion in X , im T e ⊆ Z e , and given the inclusion map M r E ( e ) = w ∈ E Z w i ֒ −→ X ,T e ◦ i = 0, and for the inclusion map i e : Z r E ( e ) → X and the projection map p e : X → Z e , we have i e ◦ T e ◦ p e = T ′ e . We can similarly extend ( T ′ e ) − to a map11 e ∗ : X → X . It is easy to check the set { Q v , T e , T e ∗ } forms a Leavitt E -familyin End R ( X ) (the ring of R -module endomorphisms of X ).What is particularly useful about the above construction is the fact rQ v =0 ⇐⇒ r = 0, and that T e ◦ T f (similarly, T f ∗ ◦ T e ∗ ) is non-zero if and only if ef ∈ E ∗ . Thus, by the universal property of L R ( E ), we have rq v = 0 ⇐⇒ r = 0 , and t e t f = 0 (similarly, t f ∗ t e ∗ = 0) ⇐⇒ ef ∈ E ∗ , in L R ( E ). These facts are useful when figuring out how multiplication worksin L R ( E ). Even more important, they play a crucial role in establishing someisomorphism theorems for L R ( E ).Given α = e e . . . e n ∈ E ∗ , we set t α := t e t e · · · t e n ; similarly, t α ∗ := t e ∗ n t e ∗ ( n − · · · t e ∗ . With this in mind, we will now give a description of multipli-cation in L R ( E ). For α, β ∈ E ∗ , we have t α t β = ( t αβ if αβ ∈ E ∗ , , t β ∗ t α ∗ = ( t ( αβ ) ∗ if αβ ∈ E ∗ , , and t α t β ∗ = 0 ⇐⇒ r E ( α ) = r E ( β )—this follows from LP2 and, as witnessedby the construction above, the universal property of L R ( E ); further, by [31,Equation 2.1], we have t β ∗ t α = t γ if α = βγ for γ ∈ E ∗ ,t γ ∗ if β = αγ for γ ∈ E ∗ ,q r E ( α ) if β = α, . To put all of this more concisely, we have the following lemma.
Lemma 3.5.
For a given graph E , L R ( E ) = span R { t α t β ∗ : α, β ∈ E ∗ and r E ( α ) = r E ( β ) . Proof. [31, Proposition 3.4]Before we proceed to give examples of Leavitt path algebras, and continueour discussion of them, we will need the following remark regarding free R -algebras. Remark . Let R be a unital ring. For a set X , let w ( X ) denote the set ofnonempty words in X , i.e., w ( X ) := { w = x x . . . x n : n ∈ N , x i ∈ X } . F R ( w ( X )) denote the free R -module over w ( X ); we may think of the ele-ments of F R ( w ( X )) as formal sums of the form n X i =1 r i · w i , with r i ∈ R, w i ∈ w ( X ) . Note that since R is unital, we can embed X in F R ( w ( X )) by the map i : X → F R ( w ( X )) where i ( x ) = 1 · x . Further, we can define multiplication on F R ( w ( X )) by (cid:0) n X i =1 r i · w i (cid:1)(cid:0) m X j =1 r j · w j (cid:1) := n X i =1 m X j =1 r i r j · w i w j , where w i w j is given by concatenation. Then, with the multiplication definedabove, F R ( w ( X )) is the associative free R -algebra on X . Further, it has thefollowing universal property (see [25, Proposition 2.6]): let A be any R -algebraand f : X → A a set map, then, there exists a unique R -algebra homomorphism φ : F R ( w ( X )) → A making the following diagram commute: F R ( w ( X )) X A fi φ Using the same construction as above, if we allow for the empty word ∅ in w ( X )(with ∅ w := w for all w ∈ w ( X )), we get the unital associative free R -algebra ,where 1 · ∅ is the unit, which we will denote by F R ( w ( X )). So, if A in the abovediagram is a unital R -algebra, F R ( w ( X )) has the universal property that φ isthe unique unital R -algebra homomorphism making the diagram commute.Recall from Chapter 1 the Leavitt algebra L K (1 , n ), with n ≥
2. We willgive its construction and show, as our first example, it is a Leavitt path alge-bra. Given the set X := { x , . . . , x n , y , . . . , y n } , let F K ( w ( X )) be the unitalassociative free K -algebra as in Remark 3.6. Further, let I ⊳ F K ( w ( X )) be theideal generated by the set { − n X i =1 x i y i , δ i,j − y i x j : i, j ∈ { , . . . , n }} . Then, L K (1 , n ) := F K ( w ( X )) /I. xample . Let K be field, and E the graph in Example 2.3. Then, L K ( E ) ∼ = L K (1 , n ) . To see this, first note that q v is the identity of L K ( E ); this fact followsfrom Lemma 3.5. Now, take the set { Q v , T e , T e ∗ } ⊆ L K (1 , n ) to be Q v = 1 , T e i = x i , and T e ∗ i = y i . One can quickly check { Q v , T e , T e ∗ } is a Leavitt E -family in L K (1 , n ). And so,by the universal mapping property of L K ( E ), we have a K -algebra homomor-phism φ : L K ( E ) → L K (1 , n ) such that φ ( q v ) = 1 , φ ( t e i ) = x i , and φ ( t e ∗ i ) = y i .On the other hand, since L K ( E ) is a unital K -algebra, by the universal map-ping property of F K ( w ( X )), there exists a unital K -algebra homomorphism ϕ : F K ( w ( X )) → L K ( E ) such that ϕ (1) = q v , ϕ ( x i ) = t e i , and ϕ ( y i ) = t e ∗ i .Moreover, by the properties of a Leavitt E -family, we have I ⊆ ker ϕ . Thus,there exists a unital K -algebra homomorphism ϕ : L K (1 , n ) → L K ( E ) where ϕ = ϕ ◦ π , meaning ϕ (1) = q v , ϕ ( x i ) = t e i , and ϕ ( y i ) = t e ∗ i . It is clear φ and ϕ are inverses of each other, hence L K ( E ) ∼ = L K (1 , n ). In the case where E consists of a single vertex and a single looped edge, note a similar argument asabove shows L K ( E ) ∼ = K [ x, x − ]—the Laurent polynomials over K . Example . Let R be a unital commutative ring, and E the graph in Example2.4. We will show L R ( E ) ∼ = M n ( R ) . Let E i,j denote the n × n matrix whose( ij )-th entry is 1, and all other entries are 0. Let { Q v , T e , T e ∗ } ⊆ M n ( R ) begiven by Q v i = E i,i , T e i = E i,i +1 , and T e ∗ i = E i +1 ,i . One can quickly verify { Q v , T e , T e ∗ } is a Leavitt E -family in M n ( R ); thus, by theuniversal mapping property of L R ( E ), there exists an R -algebra homomorphism φ : L R ( E ) → M n ( R ) . Given E i,j , suppose w.l.o.g. that i < j . Then, we caneasily work out that E i,j = E i,i +1 E i +1 ,i +2 · · · E j − ,j . A similar argument shows E i,j can be written as the product of elements in { Q v , T e , T e ∗ } when i > j ; since M n ( R ) = span R { E i,j : i, j ∈ { , . . . , n }} , φ mustbe surjective. The fact each E i,j can be written as the product of elementsin { Q v , T e , T e ∗ } means that E i,j = φ ( t α i t β ∗ j ) for some α i , β j ∈ E ∗ such that r E ( α i ) = r E ( β j ) . And so, M n ( R ) = span R { φ ( t α i t β ∗ j ) : i, j ∈ { , . . . , n } , α i , β j ∈ E ∗ such that r E ( α i ) = r E ( β j ) } , which means φ is injective as well since0 = φ (cid:18) X i,j r i,j t α i t β ∗ j (cid:19) ⇐⇒ r i,j = 0 for all i, j. Thus, L R ( E ) ∼ = M n ( R ).One of the main endeavors within the study of Leavitt path algebras is tosee how certain properties of E translate to properties of L R ( E ). While this14ursuit has resulted in some important theorems, it’s worth noting that verydifferent looking graphs can produce the same Leavitt path algebra, as Example3.8 and the following example show. Example . Let R be a unital commutative ring, and E the graph illustratedbelow. Then, L R ( E ) ∼ = M n ( R ) .v we e e n − Figure 6Let E i,j ∈ M n ( R ) be as in Example 3.8, let Id n − ∈ M n ( R ) denote thematrix whose first n − n − × ( n −
1) sub-matrix is the identity on M n − ( R ) andthe n -th row and column are 0’s). Let { Q v , T e , T e ∗ } ⊆ M n ( R ) be given by Q v = Id n − , Q w = E n,n , T e i = E i,n , and T e ∗ i = E n,i . Then, { Q v , T e , T e ∗ } is a Leavitt E -family in M n ( R ), and so there exists an R -algebra homomorphism φ : L R ( E ) → M n ( R ) . Again, an argument similar to the one used in Example 3.8 shows φ is anisomorphism; thus, L R ( E ) ∼ = M n ( R ).While we will not go into great detail here, since we will see this more explic-itly when constructing Exel-Laca algebras, but the usual method of constructing L R ( E ) is to take F R ( w ( X )) (see Remark 3.6), where X = { v } v ∈ E ∪{ e, e ∗ } e ∈ E ,and quotient by I ⊳ F R ( w ( X )), where I is the ideal generated by the union ofsets of the form: • { vw − δ v,w v : v, w ∈ E } , • { s E ( e ) e − e, er E ( e ) − e, e ∗ s E ( e ) − e ∗ , r E ( e ) e ∗ − e ∗ : e ∈ E } , • { e ∗ f − δ e,f r E ( e ) : e, f ∈ E } , • { P e ∈ s − E ( v ) ee ∗ − v : v ∈ E such that 0 < | s − E ( v ) | < ∞} .The reason we take F R ( w ( X )), and not F R ( w ( X )), is due to the fact L R ( E ) neednot be unital. Lastly, the Leavitt E -family { q v , t e , t e ∗ } is given by q v := π ( v ), t e := π ( e ), and t e ∗ := π ( e ∗ ), where π : F R ( w ( X )) → F R ( w ( X )) /I
15s the projection map. The following lemma is mentioned in the paragraph after[31, Definition 4.2].
Lemma 3.10.
For a given graph E , L R ( E ) is unital if and only if E is finite.Proof. Suppose E is finite. Then, using the properties of a Leavitt E -familyand Lemma 3.5, we can show P v ∈ E q v is a unit for L R ( K ). On the other hand,suppose E is infinite and L R ( E ) is unital. Let { v , . . . , v n , . . . } be an enumer-ation of E . Set u k := k X i =1 q v i . By applying Lemma 3.5 and
LP2 , we can show L R ( E ) = [ k ∈ N u k L R ( E ) u k . Further, it is easy to check u k is an idempotent element, which means u k is the identity for u k L R ( E ) u k ; and, u k u k +1 = u k +1 u k = u k , which means u k L R ( E ) u k ⊆ u k +1 L R ( E ) u k +1 . And so for 1 ∈ L R ( E ), there exists k ∈ N such that 1 ∈ u k L R ( E ) u k ; but, since u k is the identity for u k L R ( E ) u k ,it must mean 1 = u k . However, taking v i such that i > k , LP1 implies q v i = q v i q v i u k = 0 . ⇒⇐ —to see why, refer to the paragraph after Defi-nition 3.4. Thus, E infinite = ⇒ L R ( E ) is nonunital; all together, we have L R ( E ) is unital ⇐⇒ E is finite.While there is a lot more which can be said regarding Leavitt path algebras(there’s an entire book on the topic, see [2]), we will close our discussion ofLeavitt path algebras by stating two important results. But before we do so,we will quickly discuss Z -graded rings. Definition 3.11.
A ring R is said to be a Z -graded ring if there are subgroups R i ⊆ R such that R = M i ∈ Z R i as an internal direct sum, and R n R m ⊆ R n + m for each n, m ∈ Z . The elementsof R i are called homogeneous elements of degree i . If we want to say an elementbelongs to one of the factor subgroups, we will call it a homogeneous element . Definition 3.12.
Let R = L i ∈ Z R i and S = L i ∈ Z S i be two Z -graded rings. A ringhomomorphism f : R → S is called a graded homomorphism if f ( R i ) ⊆ S i for each i . Definition 3.13.
Given a Z -graded R , an ideal I ⊆ R is a homogeneous (orgraded) ideal if it is generated by its homogeneous elements.16he following remark will be of use to us in later sections. Remark . An ideal I being homogeneous is equivalent to saying I = L i ∈ Z ( I ∩ R i ) . Further, if I is a homogeneous ideal, then R/I is a Z -graded ring with R/I = M i ∈ Z ( R i + I ) /I. For specific details, see [15, Section 1.1.5.].
Proposition 3.14.1.
For any graph E , L R ( E ) is a Z -graded ring with L R ( E ) i = span R { t α t β ∗ : α, β ∈ E ∗ , r E ( α ) = r E ( β ) , and | α | − | β | = i } . Proof. [31, Proposition 4.7]
Theorem 3.15.
Let E be a graph, and S a Z -graded ring. Suppose there existsa graded ring homomorphism π : L R ( E ) → S such that π ( rq v ) = 0 for all v ∈ E and r ∈ R \ { } . Then, π is injective.Proof. [31, Theorem 5.3] Similar to graphs, given an ultragraph G = ( G , G , r, s ), we define a set of“ghost edges” by associating an element e ∗ to each edge e ,( G ) ∗ := { e ∗ } e ∈G . And, for each α = e e . . . e n ∈ G ∗ , we define an associated “ghost path” by α ∗ := e n ∗ e ( n − ∗ . . . e ∗ . Definition 3.16.
Let G be an ultragraph. A Leavitt G -family in an R -algebra A is a set { P A , S e , S e ∗ } A ∈G ,e ∈G ⊆ A such that:( uLP1 ) P ∅ = 0 , P A ∩ B = P A P B , P A ∪ B = P A + P B − P A ∩ B for all A, B ∈ G ,( uLP2 ) P s ( e ) S e = S e P r ( e ) = S e and P r ( e ) S e ∗ = S e ∗ P s ( e ) = S e ∗ for all e ∈ G ,( uLP3 ) S e ∗ S f = δ e,f P r ( e ) for all e, f ∈ G ,( uLP4 ) P v = P e ∈ s − ( v ) S e S e ∗ for v ∈ G such that 0 < | s − ( v ) | < ∞ .For v ∈ G , we take P v := P { v } .Notice that, for each A ∈ G , uLP1 implies P A is idempotent: P A = P A ∩ A = P A P A . Also, the following lemma expands on uLP2 . Its statement and proofis exactly that of [30, Lemma 2.8]. Lemma 3.17.
Given a Leavitt G -family { P A , S e , S e ∗ } ⊆ A , for A ∈ G and e ∈ G : A S e = ( S e if s ( e ) ∈ A, otherwise , and S e ∗ P A = ( S e ∗ if s ( e ) ∈ A, otherwise . Proof.
Let A ∈ G and e ∈ G . By uLP2 , we have P s ( e ) S e = S e , and uLP1 implies P A P s ( e ) = P A ∩ s ( e ) . And so P A S e = P A P s ( e ) S e = P A ∩ s ( e ) S e = ( P s ( e ) S e = S e if s ( e ) ∈ A,P ∅ = 0 otherwise . A similar argument shows S e ∗ P A = ( S e ∗ if s ( e ) ∈ A, . Definition 3.18. A Leavitt path algebra of an ultragraph G over R , L R ( G ), isan R algebra generated by a Leavitt G - family { p A , s e , s e ∗ } ⊆ L R ( G ) havingthe following universal property: given an R -algebra A and a Leavitt G -family { P A , S e , S e ∗ } ⊆ A , there exists an R -algebra homomorphism φ : L R ( G ) → A with φ ( p A ) = P A , φ ( s e ) = S e , and φ ( s e ∗ ) = S e ∗ . We will sometimes denote theLeavitt path algebra of G by L R ( { p, s } ).As with the Leavitt path algebra of a graph, the universal mapping propertyguarantees L R ( G ) is unique up to isomorphism. We construct L R ( G ) by takingthe associative free algebra F R ( w ( X )), where X = { A } A ∈G ∪ { e, e ∗ } e ∈G , andtaking its quotient by the ideal I ⊳ F R ( w ( X )), where I is the ideal generatedby the union of sets of the form: • { A ∩ B − AB, A ∪ B − ( A + B ) + A ∩ B : A, B ∈ G such that A = ∅ , B = ∅ , A ∩ B = ∅} , • { s ( e ) e − e, er ( e ) − e, r ( e ) e ∗ − e ∗ , e ∗ s ( e ) − e ∗ : e ∈ G } , • { e ∗ f − δ e,f r ( e ) : e, f ∈ G } , • { P e ∈ s − ( v ) ee ∗ − v : v ∈ G such that 0 < | s − ( v ) | < ∞} ;given the projection map π : F R ( w ( X )) → F R ( w ( X )) /I, the Leavitt G -family { p A , s e , s e ∗ } is given by s e := π ( e ), s e ∗ := π ( e ∗ ), p A := π ( A ) (when A = ∅ ), and p ∅ := 0. 18uppose A is an R -algebra generated by a Leavitt G and satisfies the hy-pothesis of Definition 3.18. We can use the universal mapping properties of L R ( G ) and A to define maps φ : L R ( G ) → A and ϕ : A → L R ( G )which are inverses of each other. Thus, as with L R ( E ), L R ( G ) is unique upto isomorphism. Much like the construction described in the paragraph afterDefinition 3.4, there is an analogous construction (see [16, Theorem 2.6]) whichshows rp A = 0 for all A ∈ G \ {∅} and r ∈ R \ { } ;and s α s β = 0 , s β ∗ s α ∗ = 0 ⇐⇒ αβ ∈ G ∗ . We will now give a description of how multiplication works in L R ( G ). Un-surprisingly, it is similar to that of L R ( E ). Let α, β ∈ G , then: if | α | , | β | > s α s β = ( s αβ if αβ ∈ E ∗ , , and s β ∗ s α ∗ = ( s ( αβ ) ∗ if αβ ∈ E ∗ , | α | > , | β | = 0, then for β = A ∈ G , uLP1 and uLP2 imply s α p A = s α p r ( α ) ∩ A , which is 0 if r ( α ) ∩ A = ∅ , similarly, p A s α ∗ = p r ( α ) ∩ A s α ∗ ; the casewhere | α | = | β | = 0 is addressed in uLP1 , and the case where | α | = 0 is Lemma3.17; note, s α s β ∗ = s α ( p r ( α ) ∩ r ( β ) ) s β ∗ = 0 ⇐⇒ r ( α ) ∩ r ( β ) = ∅ ;finally, by [16, Lemma 2.4], we have s β ∗ s α = s γ if α = βγ for γ ∈ G ∗ ,s γ ∗ if β = αγ for γ ∈ G ∗ ,p r ( α ) if β = α, . Putting it all together, we have the following theorem which is analogous to thecombination of Lemma 3.5 and Proposition 3.14.1.
Theorem 3.19.
For any ultragraph G :1) L R ( G ) = span R { s α p A s β ∗ : α, β ∈ G ∗ , A ∈ G , and r ( α ) ∩ A ∩ r ( β ) = ∅} ,2) L R ( G ) is a Z -graded ring with L R ( G ) i = span R { s α p A s β ∗ : α, β ∈ G ∗ , A ∈ G , r ( α ) ∩ A ∩ r ( β ) = ∅ and | α | −| β | = i } . Proof. [16, Theorem 2.5]We also have the following theorem which is analogous to Theorem 3.15.19 heorem 3.20.
For a given ultragraph G , and a Z -graded ring S , suppose π : L R ( G ) → S is a graded ring homomorphism such that π ( rp A ) = 0 for all A ∈ G \ {∅} and r ∈ R \ { } . Then, π is injective.Proof. [16, Corollary 3.4]Just like the Leavitt path algebras of graphs, the Leavitt path algebras ofultragraphs need not be unital. The following result is a restatement of [16,Lemma 6.11]. Lemma 3.21.
For a given ultragraph G , L R ( G ) is unital ⇐⇒ G ∈ G .Proof. Suppose G ∈ G . Then, by part 1) of Theorem 3.19, p G is a unit for L R ( G ). On the other hand, suppose L R ( G ) is unital. Again, Theorem 3.19 1)implies 1 = n X i =1 r i s α i p A i s β ∗ i . Let A = n S i =1 s ( α i ). Since A is the finite union of elements in G , we have A ∈ G .If G / ∈ G , it must mean there exists a v ∈ G \ A . But then, by uLP1, p v = p v p v n X i =1 r i s α i p A i s β ∗ i ! = n X i =1 r i p v s α i p A i s β ∗ i = 0 ⇒⇐ . Thus, L R ( G ) is unital ⇐⇒ G ∈ G . Example . Every graph Leavitt path algebra is an ultragraph Leavitt pathalgebra. This follows from the fact that a graph is an ultragraph. To See this,for E = G , note that { p v , s e , s e ∗ } is a Leavitt E -family in L R ( G ). Thus, thereexists an R -algebra homomorphism φ : L R ( E ) → L R ( G ) . On the other hand, for each A ∈ G = G , define: P A := X v ∈ A q v , S e := t e , and S e ∗ := t e ∗ . One can quickly show { P A , S e , S e ∗ } is a Leavitt G -family in L R ( E ). And sothere exists an R -algebra homomorphism ϕ : L R ( G ) → L R ( E );it is easily verified φ and ϕ are inverses of each other, hence, L R ( G ) ∼ = L R ( E ).20 emark . Let G be the ultragraph from Example 2.6. It turns out L Z ( G )is not isomorphic to the Leavitt path algebra of any graph. So, just like the C ∗ -algebra case, the class of ultragraph Leavitt path algebras properly containsthe class of graph Leavitt path algebras. Showing this fact is a bit involvedand would require further discussion of Leavitt path algebras; in particular, wewould need to know a lot more about their ideal structure. See [16, Example6.12] for further details.While the class of ultragraph Leavitt path algebras is strictly larger, we dohave the following: Proposition 3.23.1.
For an ultragraph G , let E G be the graph from Definition2.9. Suppose | E G | < ∞ , then, L R ( G ) ∼ = L R ( E G ) . Proof.
Note that | E G | < ∞ = ⇒ | G | < ∞ . But, by the construction of E G , | G | < ∞ = ⇒ ∆ = ∅ . And so we can identify E G with G , and E G with theset { ( e n , v ) : e n ∈ G , v ∈ r ( e n ) } . Define a set { Q v , T e , T e ∗ } ⊆ L R ( G ) by Q v := p v , T ( e n ,v ) := s e n p v and T ( e n ,v ) ∗ := p v s e ∗ n . It can quickly be verified that { Q v , T e , T e ∗ } is a Leavitt E G -family in L R ( G );thus, there exists an R -algebra homomorphism φ : L R ( E G ) → L R ( G )such that φ ( q v ) = Q v , φ ( t ( e n ,v ) ) = T ( e n ,v ) , and φ ( t ( e n ,v ) ∗ ) = T ( e n ,v ) ∗ . ByProposition 3.14.1, Theorem 3.19, and Theorem 3.20, φ is an injective, Z -graded,homomorphism. Note that | G | < ∞ implies | A | < ∞ for each A ∈ G and | r ( e n ) | < ∞ for each e n ∈ G . Further, for each A ∈ G and each e n ∈ G , p A = X v ∈ A Q v , s e n = X v ∈ r ( e n ) T ( e n ,v ) , and s e ∗ n = X v ∈ r ( e n ) T ( e n ,v ) ∗ ;since L R ( G ) is generated by { p A , s e , s e ∗ } , this means φ is surjective as well.Thus, L R ( G ) ∼ = L R ( E G ).Because Morita equivalence is strictly weaker than being isomorphic, Propo-sition 3.23.1 implies L R ( G ) is Morita equivalent to L R ( E G ) when | E G | < ∞ . Andso it only remains to establish Morita equivalence when | E G | = ∞ ; notice that,by Lemma 3.10, this necessarily means L R ( E G ) is nonunital.While Morita equivalence of rings has many formulations, it is at heart anequivalence of categories. To clearly understand what this means and how thedifferent formulations of Morita equivalence arise, it is useful to review somefoundational category theory. 21 Category Theory
Definition 4.1. A category , C , consists of:1) a class of objects , ob C ;2) for each pair of objects, ( A, B ) ∈ ob C× ob C , a set of morphisms , Hom C ( A, B ),we can visualize f ∈ Hom C ( A, B ) as an arrow A f −→ B (indeed, morphisms aresometimes called “arrows”), A is the domain of f , and B is the codomain ;3) further, for each ( A, B ) , ( B, C ) ∈ ob C × ob C , there is a set map (the com-position map), Hom C ( A, B ) × Hom C ( B, C ) → Hom C ( A, C );the image of ( f, g ) ∈ Hom C ( A, B ) × Hom C ( B, C ) is denoted by “ g ◦ f ;”finally, the objects and morphisms must satisfy the conditions that,( C1 ) for ( A, B ) , ( C, D ) ∈ ob C × ob C ,( A, B ) = ( C, D ) = ⇒ Hom C ( A, B ) ∩ Hom C ( C, D ) = ∅ , ( C2 ) given f ∈ Hom C ( A, B ), g ∈ Hom C ( B, C ), and h ∈ Hom C ( C, D ), h ◦ ( g ◦ f ) = ( h ◦ g ) ◦ f, ( C3 ) for each A ∈ ob C , there exists a unique element 1 A ∈ Hom C ( A, A ) suchthat,i) for each B ∈ ob C , and each f ∈ Hom C ( A, B ) , f ◦ A = f, ii) for each B ∈ ob C , and each g ∈ Hom C ( B, A ) , A ◦ g = g . Example
Grp ) . Here, ob
Grp is the class of groups.For (
A, B ) ∈ ob Grp × ob Grp , Hom
Grp ( A, B ) is the set of group homomor-phisms from A to B . For ( f, g ) ∈ Hom
Grp ( A, B ) × Hom
Grp ( B, C ), g ◦ f ∈ Hom
Grp ( A, C ) is given by the usual composition of group homomorphisms.
Example
Rng ) . In this case, ob
Rng is the class ofrings. Hom
Rng ( A, B ) is the set of ring homomorphisms from A to B .For ( f, g ) ∈ Hom
Rng ( A, B ) × Hom
Rng ( B, C ), g ◦ f ∈ Hom
Rng ( A, C ) is givenby composition of ring homomorphisms.
Example
Top ∗ ) . Here, ob
Top ∗ consists of pointed topological spaces. Given (cid:0) ( X, x ) , ( Y, y ) (cid:1) ∈ ob Top ∗ × ob Top ∗ , Hom Top ∗ (cid:0) ( X, x ) , ( Y, y ) (cid:1) consists of continuous maps f : X → Y such that f ( x ) = y .For ( f, g ) ∈ Hom
Top ∗ (cid:0) ( X, x ) , ( Y, y ) (cid:1) × Hom
Top ∗ (cid:0) ( Y, y ) , ( Z, z ) (cid:1) , g ◦ f ∈ Hom
Top ∗ (cid:0) ( X, x ) , ( Z, z ) (cid:1) is given by the usual composition of continuousmaps.To get some idea of how general categories can be, consider the followingexample. Example . Let X be a topological space. We can define a category C asfollows: 22et ob C consist of the points of X ; that is, ob C := { x : x ∈ X } . For x, y ∈ ob C , let Hom C ( x, y ) be the set of equivalence classes of paths from x to y .For ([ f ] , [ g ]) ∈ Hom C ( x, y ) × Hom C ( y, z ), [ g ] ◦ [ f ] ∈ Hom C ( x, z ) is given by[ g ] ◦ [ f ] := [ f g ], where [ f g ] is the equivalence class of the concatenation of f by g . Finally, for x ∈ ob C , 1 x is the equivalence class of the constant path at x . Definition 4.6.
Let C be a category. A subcategory of C is a category, D , suchthat:1) ob D is a subclass of ob C .2) For each ( A, B ) ∈ ob D × ob D , Hom D ( A, B ) ⊆ Hom C ( A, B ).3) For each A ∈ ob D , 1 A ∈ Hom D ( A, A ) is the same as 1 A ∈ Hom C ( A, A ); and,the composition mapHom D ( A, B ) × Hom D ( B, C ) → Hom D ( A, C )is the restriction of the composition mapHom C ( A, B ) × Hom C ( B, C ) → Hom C ( A, C ) . Example Ab ) . ob Ab is the class of abeliangroups, and the morphisms are group homomorphisms. It is straightforward tocheck Ab is a subcategory of Grp .Recall that every ring has an abelian group structure. Moreover, every ringhomomorphism is a group homomorphism as well. Thus,
Rng is a subcategoryof Ab . As a result, Rng is subcategory of
Grp as well. However, since not everygroup homomorphism is a ring homomorphism, Hom
Rng ( A, B ) is a strict subsetof Hom Ab ( A, B ). If D is a subcategory of C such that, for every A, B ∈ ob D ,Hom D ( A, B ) = Hom C ( A, B ), we say D is a full subcategory. For example, Ab is a full subcategory of Grp .Much like maps between sets, we have a notion of mapping between cate-gories.
Definition 4.8.
Let C and D be categories. A (covariant) functor from C to D , F : C → D , consists of:1) a mapping, A F ( A ), from ob C to ob D ,2) a mapping, f F ( f ), from Hom C ( A, B ) to Hom D ( F ( A ) , F ( B )).Further, the following conditions must be satisfied. F1:
For every A ∈ ob C , F (1 A ) = 1 F ( A ) ∈ Hom D ( F ( A ) , F ( A )) . F2:
For every ( f, g ) ∈ Hom C ( A, B ) × Hom C ( B, C ), and g ◦ f ∈ Hom C ( A, C ), F ( g ◦ f ) = F ( g ) ◦ F ( f ) ∈ Hom D ( F ( A ) , F ( C )) .
23e can also compose functors. Let C , D , and E be categories with functors F : C → D and G : D → E . Then, we get a functor, G ◦ F : C → E , where G ◦ F ( A ) = G ( F ( A )) for each A ∈ ob C , and G ◦ F ( f ) = G ( F ( f )) for each f ∈ Hom C ( A, B ). Example . For any category C , the indentity functor , 1 C : C → C , consists of:the identity mapping on ob C ; the identity mapping on Hom C ( A, B ) for each
A, B ∈ ob C . Example . Given a pointed topological space (
X, x ), let π ( X, x ) de-note the fundamental group of X with base point x . There is a functor F : Top ∗ → Grp consisting of a map from ob
Top ∗ to ob Grp , given by (
X, x ) π ( X, x ), and a map from Hom Top ∗ (cid:0) ( X, x ) , ( Y, y ) (cid:1) to Hom Grp (cid:0) π ( X, x ) , π ( Y, y ) (cid:1) ,given by f F ( f ) where F ( f ) : π ( X, x ) → π ( Y, y ) is the group homomor-phism given by [ γ ] [ f ◦ γ ].As we stated earlier, Morita equivalence is a statement regarding the equiv-alence of categories. We need the following definition in order to precisely saywhat an equivalence of categories is. Definition 4.11.
Let C and D be categories, and F, G : C → D functors. A natural transformation , η , from F to G is a map which assigns a morphism, η A ∈ Hom D ( F ( A ) , G ( A )), for each A ∈ ob C such that, given any two A, B ∈ ob C and f ∈ Hom C ( A, B ), the following diagram commutes: F ( A ) G ( A ) F ( B ) G ( B ) η A η B F ( f ) G ( f )Given a morphism f ∈ Hom C ( A, B ), f is an isomorphism if there exists amorphism g ∈ Hom C ( B, A ) such that g ◦ f = 1 A and f ◦ g = 1 B ; note this means g is an isomorphism as well. If η A in Definition 4.11 is an isomorphism for each A ∈ ob C , we say η is a natural isomorphism . If F, G : C → D are functorsfor which there is a natural isomorphism, we say they are naturally isomorphic ,which we will denote this by writing “ F ≃ G .” Definition 4.12.
Two categories, C and D , are equivalent if there exist functors F : C → D and G : D → C such that G ◦ F ≃ C and F ◦ G ≃ D . Morita equivalence of rings is the equivalence of module categories. The theorywas first developed within the context of unital rings. If the reader wishes to24earn more about Morita equivalence for unital rings, section 3.12 of NathanJacobson’s
Basic Algebra II , [17], is an excellent source. The theory was thenextended to certain nonunital rings by numerous mathematicians over severaldecades. In our discussion of Morita equivalence, we will solely rely on Jacobson,[1], [5], and [12]. We will start our brief foray into this vast topic by laying thebasic foundation.
Definition 5.1.
Let R be a ring. An abelian group, M , is a left R -module ifthere exists a map · : R × M → M such that:1) r · ( x + y ) = r · x + r · y for all r ∈ R and all x, y ∈ M ,2) ( r + r ) · x = r · x + r · x for all r , r ∈ R and for all x ∈ M ,3) ( r r ) · x = r · ( r · x ) for all r , r ∈ R and for all x ∈ M .If R is unital, then we also require4) 1 · x = x for all x ∈ M . Definition 5.2.
Let M and N be left R -modules. A map, f : M → N , is a left R -module homomorphism if f ( r · x + r · y ) = r · f ( x ) + r · f ( y )for all r , r ∈ R and for all x, y ∈ M. We denote the set of R -module homomorphisms from M to N by “Hom R ( M, N ),”and “End R ( M )” denotes the set of R -modules endomorphisms of M . One cancheck End R ( M ) is a ring under pointwise addition and with multiplication givenby composition. Given a ring R , the category of left R -modules, R-Mod , con-sists of left R -modules as objects and R -module homomorphisms as morphisms.We similarly define a right R -module to be an abelian group M equippedwith a map · : M × R → M satisfying similar axioms to those in Definition 5.1, of course, the difference beingelements of R are now acting on the right. A right R -module homomorphism isthen a map, f : M → N , where M and N are right R -modules such that f ( x · r + y · r ) = f ( x ) · r + f ( y ) · r for all r , r ∈ R and for all x, y ∈ M. Much like
R-Mod , the category of right R -modules, Mod-R , consists of right R -modules as objects and right R -module homomorphisms as morphisms.Throughout this paper, however, we will take R -modules to be left R -modules unless specified otherwise. Further, given an R -module M , we willwrite “ rx ” to denote r · x . Example . Let R be a ring. Then, R is a left, and right, R -module over itselfwhere · : R × R → R is given by the ring multiplication. Example . A vector space, V , over a field, K , is just a K -module. In fact,modules are a generalization of vector spaces.25 xample . Let R = M n ( Z ), the ring of n × n matrices over Z , and let M = M n × ( Z ), the group of n × Z . Then, M is a left R -module with · : R × M → M given by matrix multiplication. Note that M isnot a right R -module under matrix multiplication. In general, having a left R -module structure doesn’t automatically give a right R -module structure. If R isa commutative ring, a left R -module M can be made into a right R -module butsetting m · r := rm ; this fails when R is noncommutative since 3) of Definition5.1 won’t hold. However, taking m · r := rm does give M a right R op -modulestructure (see Definition 5.25 for an explanation of “ R op ”).If R is a unital ring, then 4) of Definition 5.1 implies RM := { rx : r ∈ R, x ∈ M } = M. If R is a nonunital ring, then it isn’t always the case RM = M . The followingsimple example illustrates this point. Example . Let R = 2 Z . Viewed as a 2 Z -module over itself,(2 Z )(2 Z ) = 4 Z = 2 Z . Moreover, 4) of Definition 5.1 again implies that, in the unital case, Rx := { rx : r ∈ R } = { } = ⇒ x = 0;this also need not hold when R is nonunital (e.g., take R = 2 Z , M = Z / Z , and x = 1).For a nonunital ring R , an R -module M is said to be unital if RM = M ; it issaid to be nondegenerate if for any x ∈ M , Rx = { } = ⇒ x = 0. Throughoutthis paper, R − M OD denotes the full subcategory of R -Mod whose objectsare unital nondegenrate left R -modules. Note if R is unital, then R − M OD isthe same category as R -Mod . We define M OD − R in the same way. Now,finally, Definition 5.7.
Two rings, R and S , are Morita equivalent if R − M OD isequivalent to S − M OD . It can be shown, certainly in the cases we are interestedin, R − M OD is equivalent to S − M OD if and only if
M OD − R is equivalentto M OD − S , [5, Corollary 2.3]There are several ways Morita equivalence can be characterized for certaintypes of rings. We are particularly interested in the case where R is a ring withlocal units. Definition 5.8. R is a ring with local units if, for each finite subset S ⊆ R ,there exists an idempotent e ∈ R such that S ⊆ eRe . Note that eRe is a subringof R .It’s worth clarifying that the results in [12] are proved for idempotent rings;a ring is idempotent if R := { rs : r, s ∈ R } = R.
26t is straightforward to see that a ring with local units is also an idempotentring. And so the results in [12] apply to rings with local units as well. Withthat, we work toward the different characterizations of Morita equivalence.
Definition 5.9.
Let R be a ring, M a right R -module, and N a left R -module.An R -balanced product of M and N is a pair, ( P, f ), where P is an abeliangroup, and f : M × N → P a map such that:1) f ( x + x , y ) = f ( x , y ) + f ( x , y ) for all x , x ∈ M and y ∈ N ,2) f ( x, y + y ) = f ( x, y ) + f ( x, y ) for all x ∈ M and y , y ∈ N ,3) f ( xr, y ) = f ( x, ry ) for all x ∈ M , y ∈ N , and r ∈ R . Definition 5.10.
Let R be a ring, M a right R -module, and N a left R -module.The tensor product of M and N over R is an R -balanced product of M and N ,( M ⊗ R N, ⊗ ) , such that, given any other R -balanced product, ( P, f ), there exists a uniquegroup homomorphism, φ : M ⊗ R N → P , making the following diagram com-mute: M ⊗ R NM × N Pf ⊗ φ To construct M ⊗ R N , we start with the free abelian group on M × N , whichwe will denote by “ F ab ( M × N ).” We can think of the elements of F ab ( M × N )as commuting formal sums of the form n X i =1 ( x i , y i )where ( x i , y i ) ∈ M × N for each i . Further, F ab ( M × N ) has the universalproperty that, given any abelian group, P , and a set map, f : M × N → P ,there exists a unique group homomorphism e φ : F ab ( M × N ) → P making thefollowing diagram commute: 27 ab ( M × N ) M × N Pfi e φ In the diagram above, i : M × N → F ab ( M × N ) is the inclusion of M × N into F ab ( M × N ). Taking G to be the subgroup of F ab ( M × N ) generated byelements of the form ( x + x , y ) − ( x , y ) − ( x , y ), ( x, y + y ) − ( x, y ) − ( x, y ),and ( xr, y ) − ( x, ry ), where x , x ∈ M , y , y ∈ N , and r ∈ R , we set M ⊗ R N := F ab ( M × N ) /G, and, given the projection p : F ab ( M × N ) → M ⊗ R N , we set ⊗ : M × N → M ⊗ R N to be the group homomorphism p ◦ i . For ( x, y ) ∈ M × N ,we write “ x ⊗ y ” to denote ⊗ ( x, y ). Elements of the form x ⊗ y are referred to as simple tensors . Each element of M ⊗ R N can be expressed as the sum of simpletensors; inconveniently, this expression need not be unique. From here on out,we will simply write “ M ⊗ R N ” instead of “( M ⊗ R N, ⊗ ).” If it’s clear whichring we are taking the tensor product over, we will simply write “ M ⊗ N .”While M ⊗ R N is an abelian group, it can be endowed with a module struc-ture under certain circumstances. Definition 5.11.
Let R and S be rings. An R - S -bimodule is an abelian group M such that:1) M is a left R -module,2) M is a right S -module,3) r ( xs ) = ( rx ) s for all r ∈ R , s ∈ S , and x ∈ M . M is unital if RM = M S = M . Further, given two R - S -bimodules M and N , a map f : M → N is an R - S -bimodule homomorphism if it is simultaneouslya left R -module homomorphism and a right S -module homomorphism. Wehave the following standard proposition relating tensor products, modules, andbimodules: Proposition 5.11.1.
Let R , S , and T be rings. Then: 1) M ⊗ R N is a left S -module, with s · ( x ⊗ y ) = ( sx ) ⊗ y on simple tensors, if M is an S - R -bimodule,2) M ⊗ R N is a right T -module, with ( x ⊗ y ) · t = x ⊗ ( yt ) on simple tensors,if N is an R - T -bimodule. The first different formulation of Morita equivalence we will give relies onthe existence of a
Morita context . Definition 5.12. A Morita context is a sextuple, (
R, S, M, N, τ, µ ), where R and S are rings, M is a unital S - R -bimodule, N is a unital R - S -bimodule, τ : N ⊗ S M → R is an R - R -bimodule homomorphism, and µ : M ⊗ R N → S isan S - S -bimodule homomorphism such that:1) µ ( x ⊗ y ) x ′ = xτ ( y ⊗ x ′ ) for all x, x ′ ∈ M and y ∈ N ,28) yµ ( x ⊗ y ′ ) = τ ( y ⊗ x ) y ′ for all x ∈ M and y, y ′ ∈ N .A Morita context ( R, S, M, N, τ, µ ) is of particular interest when τ and µ aresurjective; in fact, it isn’t uncommon to find the surjectivity condition includedin the definition of a Morita context. Theorem 5.13.
Let R and S be idempotent rings. Then, R is Morita equivalentto S if and only if there exists a Morita context, ( R, S, M, N, τ, µ ) , with τ and µ surjective.Proof. [12, Propositions 2.5 and 2.7]Consider the following example, [17]. Example . Consider a unital commutative ring R and M n ( R ). Under theusual rules of matrix multiplication, and multiplying matrices by scalars, wecan see M n × ( R ) is an M n ( R )- R -bimodule; likewise, M × n ( R ) is an R - M n ( R )-bimodule. Now, define maps µ : M n × ( R ) × M × n ( R ) → M n ( R ) and τ : M × n ( R ) × M n × ( R ) → R by( x ... x n , (cid:2) y . . . y n (cid:3) ) x y . . . x y n ... . . . ... x n y . . . x n y n , and ( (cid:2) y . . . y n (cid:3) , x ... x n ) n X i =1 x i y i . One can easily check ( M n ( R ) , µ ) and ( R, τ ) are R -balanced products witheach map being surjective. So, by exploiting the universal mapping property oftensor products, we have surjective group homomorphisms µ : M n × ( R ) ⊗ R M × n ( R ) → M n ( R ) and τ : M × n ( R ) ⊗ M n ( R ) M n × ( R ) → R. In fact, µ and τ are, respectively, M n ( R )- M n ( R ) and R - R -bimodule homomor-phisms satisfying the conditions listed in Definition 5.12. Thus, (cid:0) R, M n ( R ) , M n × ( R ) , M × n ( R ) , τ, µ (cid:1) is a Morita context with µ and τ surjec-tive; by Theorem 5.13, this means R and M n ( R ) are Morita equivalent.It may be tedious but it isn’t difficult to show being isomorphic impliesMorita equivalence—this can be done working directly from definitions, with-out having to establish a Morita context. On the other hand, Example 5.14illustrates Morita equivalence is strictly weaker than being isomorphic. Still,Morita equivalence is of interest in part because it preserves certain ideal struc-tures. 29 efinition 5.15. A partially ordered set , ( P, ≤ ), is a set, P , together with abinary relation, ≤ , such that:1) x ≤ x for all x ∈ P ,2) for all x, y ∈ P , x ≤ y and y ≤ x = ⇒ x = y ,3) for all x, y, z ∈ P , x ≤ y and y ≤ z = ⇒ x ≤ z .Given two partially ordered sets, ( P, ≤ p ) and ( Q, ≤ q ), a map f : P → Q is order preserving if x ≤ p y = ⇒ f ( x ) ≤ q f ( y ) for all x, y ∈ P . Let ( P, ≤ ) be apartially ordered set and S ⊆ P . Then, x ∈ P is a lower bound of S if x ≤ y for all y ∈ S ; x is the greatest lower bound if, for any other lower bound z of S , z ≤ x . Similarly, x ∈ P is an upper bound of S if y ≤ x for all y ∈ S ; it isthe least upper bound if, for any other upper bound z of S , x ≤ z . Note that2) of Definition 5.15 implies least upper bounds, and greatest lower bounds, areunique should they exist. Definition 5.16. A lattice is a partially ordered set, ( P, ≤ ), such that every two-element subset, { x, y } ⊆ P , has a greatest lower bound and a least upper bound.A lattice homomorphism is an order preserving map between two lattices.Let R be a ring and I := { I : I is an ideal of R } . Under inclusion, ( I , ⊆ ) isa partially order set. Moreover, for any two I, J ∈ I , I ∩ J is the greatest lowerbound, and I + J is the least upper bound, of { I, J } ; I is the lattice of ideals of R . Proposition 5.16.1.
Let R and S be two Morita equivalent rings, and let ( R, S, M, N, τ, µ ) be a Morita context. Set I R := { I ⊳ R : RIR = I } , and I S := { I ⊳ S : SIS = I } . Then, the map I µ ( M I ⊗ R N ) defines a latticeisomorphism from I R to I S with the inverse given by I τ ( N I ⊗ S M ) .Proof. [31, Proposition 4.12]In Proposition 5.16.1, M I ⊗ R N := span S { xr ⊗ R y : r ∈ I ⊳ R, x ∈ M, and y ∈ N } ; similarly N I ⊗ S M := span R { ys ⊗ S x : s ∈ I ⊳ S, x ∈ M, and y ∈ N } . Also, note that if R is a ring with local units, every ideal I in R satisfies the condition RIR = I . While the study of Morita equiva-lence is extensive in its own right, we will conclude our discussion of Moritaequivalence by setting up and stating one last characterization of it. A crucialcomponent of our last characterization is the categorical notion of direct limits. Definition 5.17.
Let ( I, ≤ ) be a partially ordered set, and let C be a category.A direct system in C over I is a set of objects, { A α } α ∈ I ⊆ ob C , and a set ofmorphisms, { ϕ α,β : A α → A β } α,β ∈ I : α ≤ β , such that:1) ϕ α,γ = ϕ β,γ ◦ ϕ α,β for all α ≤ β ≤ γ ,2) ϕ α,α = 1 A α for each α .We will write “ (cid:10) A α , ϕ α,β (cid:11) ” to denote direct systems.30 efinition 5.18. Let ( I, ≤ ) be a partially ordered set, and let C be a category.Given a direct system, (cid:10) A α , ϕ α,β (cid:11) , its direct limit , should it exist, is an object,lim −→ α ∈ I A α ∈ ob C , together with a set of morphisms, { η α : A α → lim −→ α ∈ I A α } α ∈ I , such that:1) η α = η β ◦ ϕ α,β for all α ≤ β ,2) Given any object B ∈ ob C and a set of morphisms { ζ α : A α → B } α ∈ I satisfying ζ α = ζ β ◦ ϕ α,β for all α ≤ β , there exists a unique morphism θ : lim −→ α ∈ I A α → B such that ζ α = θ ◦ η α for all α .Part 2) of Definition 5.18 ensures that, when it exists, lim −→ α ∈ I A α is unique up toisomorphism. Fortunately for us, the direct limit always exists in the categorieswe will consider. Definition 5.19.
Let ( I, ≤ ) be a partially ordered set, and let C be a category.A compatible set in C over I , { A α , ϕ α,β , ψ β,α , I } , consists of a set of objects, { A α } α ∈ I , and a set of morphisms, { ϕ α,β : A α → A β , ψ β,α : A β → A α } α,β ∈ I , such that:1) ϕ α,α = ψ α,α = 1 A α for all α ,2) ϕ β,γ ◦ ϕ α,β = ϕ α,γ and ψ β,α ◦ ψ γ,β = ψ γ,α for all α ≤ β ≤ γ ,3) ψ β,α ◦ ϕ α,β = 1 A α for all α ≤ β ,4) For all α, β, γ ∈ I such that α, β ≤ γ , ϕ β,γ ◦ ψ γ,β ◦ ϕ α,γ ◦ ψ γ,α = ϕ α,γ ◦ ψ γ,α ◦ ϕ β,γ ◦ ψ γ,β . Note that, given a compatible set { A α , ϕ α,β , ψ β,α , I } , (cid:10) A α , ϕ α,β (cid:11) forms adirect system. Our next, and last, formulation of Morita equivalence relies onthe existence of certain direct limits and compatible sets. Definition 5.20.
Let R be a ring. An R -module, M , is a finitely generated module if there exists a finite set, X ⊆ M , such that M = span { rx : r ∈ R, x ∈ X } . Definition 5.21. An R -module, P , is projective if for every surjective R -modulehomomorphism, f : N → M , and any R -module homomorphism, g : P → M ,there exists an R -module homomorphism, h : P → N making the diagram31 P Mgh f commute.
Definition 5.22. An R -module, M , is a generator of R − M OD , if for any R -module, N , there exists a set I and a surjective R -module homomorphism ρ : M i ∈ I M → N. The notion of a generator is actually categorical. Given a category C , wesay G ∈ ob C is a generator of C if, for any f, g ∈ Hom C ( A, B ) with f = g ,there exists h ∈ Hom C ( G, A ) such that f ◦ h = g ◦ h . Generators may not existwithin a category in general. However, for abelian categories (see Definition 6.2in Jacobson’s Basic Algebra II ), which admit products and coproducts , not onlydo generators exist, but one finds the following equivalent definition: let C bean abelian category, then G ∈ ob C is a generator of C if, given any A ∈ ob C ,there exists a set I such that the coproduct, L i ∈ I G , admits an epimorphism ρ : M i ∈ I G → A ;which is to say, the morphism ρ ∈ Hom C (cid:16) L i ∈ I G, A (cid:17) is such that, given any f, g ∈ Hom C ( A, B ), f ◦ ρ = g ◦ ρ = ⇒ f = g. Since, as a
Grothendieck category, R − M OD is abelian [12], Definition 5.22 isappropriate. Because we will make use of the equivalence of the two definitionsonly once in this thesis, we will not go through the trouble of giving the ratherlaborious definition of an abelian category. We will, however, give a standardproof showing the equivalence of the two definitions. The following proof isfrom a lecture note posted online which the author cannot seem to relocate forproper citation.
Claim . Let C be an abelian category. For G ∈ ob C , the following defini-tions are equivalent:1) for any f, g ∈ Hom C ( A, B ) with f = g , there exists h ∈ Hom C ( G, A ) suchthat f ◦ h = g ◦ h ;2) given any A ∈ ob C , there exists a set I with the coproduct, L i ∈ I G , admittingan epimorphism ρ : L i ∈ I G → A . 32 roof. ⇒
1) Suppose G ∈ ob C is such that, for any A ∈ ob C , there exists aset I and an epimorphism ρ ∈ Hom C (cid:16) L i ∈ I G, A (cid:17) . Let f, g ∈ Hom C ( A, B ) be suchthat f = g . Suppose f ◦ ρ = g ◦ ρ . Since we are in an abelian category, we cancarry out the following calculations: f ◦ ρ = g ◦ ρ = ⇒ ( f − g ) ◦ ρ = 0 ◦ ρ ; but,because ρ is an epimorphism, ( f − g ) ◦ ρ = 0 ◦ ρ = ⇒ f − g = 0 = ⇒ f = g. ⇒⇐ Thus, f ◦ ρ = g ◦ ρ .1) ⇒
2) On the other hand, suppose G ∈ ob C is such that, for any f, g ∈ Hom C ( A, B ), with f = g , there exists h ∈ Hom C ( G, A ) satisfying f ◦ h = g ◦ h .Let I := Hom C ( G, A ) . Since coproducts exist in abelian categories, we have afamily of morphisms, { φ j : G → L i ∈ I G } j ∈ I , and a morphism, ρ : L i ∈ I G → A, such that ρ ◦ φ i = i for each i ∈ I . Now, suppose f, g ∈ Hom C ( A, B ) such that f ◦ ρ = g ◦ ρ but f = g . By our assumption, there exists h ∈ Hom C ( G, A )with f ◦ h = g ◦ h . But, h = ρ ◦ φ h , and so f ◦ ρ = g ◦ ρ = ⇒ f ◦ ρ ◦ φ h = g ◦ ρ ◦ φ h = ⇒ f ◦ h = g ◦ h. ⇒⇐ Thus, f ◦ ρ = g ◦ ρ = ⇒ f = g , meaning ρ isan epimorphism. Definition 5.24. An R -module, M , is locally projective if there exists a compat-ible set { M α , ϕ α,β , ψ β,α , I } in R − M OD such that each M α is a finitely gener-ated, projective, R -module, and, for the direct system (cid:10) M α , ϕ α,β (cid:11) , M ∼ = lim −→ α ∈ I M α . Proposition 5.24.1.
Let M be a locally projective R -module. Define ϕ α,β : End R M α → End R M β by ϕ α,β ( φ ) = ϕ α,β ◦ φ ◦ ψ β,α . Then, D End R M α , ϕ α,β E is a direct system over I in the category of rings.Proof. [14, Proposition 2.5]. Definition 5.25.
Let R be a ring. Its opposite ring , R op , is defined to be thering where R and R op are the same as abelian groups, with multiplication, · ,in R op is given by r · r := r r for all r , r ∈ R .In both [1] and [5], “End R ( M )” should be interpreted as “ (cid:16) End R ( M ) (cid:17) op .”The reason for this is originally, in the case of two unital rings R and S , Moritaequivalence is characterized by the existence of an “anti-isomorphism” from S onto End R P , where P is a projective generator of R − M OD ; but this is pre-cisely equivalent to saying there is an isomorpshim from S onto (End R P ) op .One of the main goals of both [1] and [5] is to generalize this characterization.For reasons unclear to the author, instead of defining the opposite ring, ele-ments of End R ( M ) are thought of as “right operators” in the aforementionedpapers. Further, it’s worth noting Proposition 5.24.1 still holds if we replace D End R M α , ϕ α,β E with D(cid:16)
End R M α (cid:17) op , ϕ α,β E .33 heorem 5.26. Two rings with local units, R , S , are Morita equivalent if andonly if there exists a locally projective R -module, M , such that M is a generatorof R − M OD and S ∼ = lim −→ α ∈ I (cid:16) End R M α (cid:17) op . Proof. [5, Theorem 2.5].
Recall, given an ultragraph, G , E G denotes the graph constructed from G asin Definition 2.9. In the case where G doesn’t have any singular vertices, thecontents of this section are essential to establishing the Morita equivalence of L R ( G ) and L R ( E G ). The result we establish here need not hold in general, butit does hold for rings which share a certain property with Leavitt path algebras.As we have shown, Leavitt path algebras need not be unital. However, they dohave “the next best thing.” Definition 6.1.
A ring, R , is σ -unital if there exists a collections of elements { u k } k ∈ N ⊆ R such that R = S k ∈ N u k Ru k and u k = u k u k +1 = u k +1 u k for all k ≥ { u k } k ∈ N is called a σ -unit .For any graph E with | E | = ∞ , the set { u k } k ∈ N from the proof of Lemma3.10 is a σ -unit for L R ( E ). Although it requires a bit more work, one can showthat, for G with | G | = ∞ , L R ( G ) is σ -unital; we will prove it later on. In[19], the authors make use of multiplier algebras in order to show ultragraph C ∗ -algebras are Morita Equivalent to graph C ∗ -algebras. We will need to dosomething similar. But first, we need an algebraic analogue to the multiplieralgebra found in [19]. Definition 6.2.
For a ring, R , let M ( R ) denote the set of pairs ( T, S ) where
T, S : R → R are additive homomorphisms such that: (i) xT ( y ) = S ( x ) y , (ii) T ( xy ) = T ( x ) y , (iii) S ( xy ) = xS ( y )for all x, y ∈ R . M ( R ) is a ring with respect to the operations( T , S ) + ( T , S ) := ( T + T , S + S ) , and ( T , S )( T , S ) := ( T ◦ T , S ◦ S );it is called the multiplier ring of R .Given a ring R , for each x ∈ R , we can define additive homomorphisms, L x , R x , by left and right multiplication by x , respectively; one can quickly verify( L x , R x ) ∈ M ( R ). Further, it is straightforward to check the map given by x ( L x , R x ) defines a ring homomorphism from R into M ( R ), the kernel ofwhich is ann( R ) := { x ∈ R : xR = Rx = 0 } . efinition 6.3. Let R be a ring. I ⊳ R is an essential ideal if, for all x ∈ R , xI = Ix = 0 = ⇒ x = 0 .R is an essential ring if R is an essential ideal of itself.If R is an essential ring, the ring homomorphism x ( L x , R x ) gives anembedding of R into M ( R ); in this case, we identify R with its image in M ( R ).How R behaves in M ( R ) is of interest to us. Proposition 6.3.1.
Let R be a σ -unital ring, and let i : R → M ( R ) be thering homomorphism given by x ( L x , R x ) . Then,1) i ( R ) is an essential ideal in M ( R ) ,2) given a ring, S , and an injective ring homomorphism, φ : R → S , such that φ ( R ) is as an ideal in S , there exists a ring homomorphism, θ : S → M ( R ) ,such that the following diagram commutes: M ( R ) R S ; i φθ further, if φ ( R ) is an essential ideal in S , θ is injective.Proof. [23, Theorem 7.1.4]By 1) of Proposition 6.3.1, since L R ( E G ) is σ -unital, we can think of L R ( E G )as an essential ideal in M ( L R ( E G )). Following the blueprint set in [19], our re-sult rests on showing the existence of a full idempotent element, Q ∈ M ( L R ( E G )),such that QL R ( E G ) Q is Morita Equivalent to L R ( E G ). From there, in latersections, we will show L R ( G ) ∼ = QL R ( E G ) Q , thereby establishing the Moritaequivalence of L R ( G ) and L R ( E G ). To that end: Definition 6.4.
Let R be a ring and Q ∈ M ( R ) an idempotent element. Q isa full idempotent if RQR = R .The technique used in proving the following theorem is modeled on [1, Corollary4.3]. Theorem 6.5.
Let { u k } k ∈ N be a σ -unit for R , and let Q ∈ M ( R ) be a fullidempotent element such that u k Q = Qu k for each k . Then, QRQ and R areMorita equivalent. Before proving Theorem 6.5, we would like to first prove the following minor,but important, claim.
Claim . Let R be a ring and e ∈ R an idempotent. Then, Re := { re : r ∈ R } is a projective R -module. 35 roof. Suppose f : Re → M and g : N ։ M are R -module homomorphisms.Since g is surjective, we can fix an element n ∈ N be such that g ( n ) = f ( e ). Now,define h : Re → N by h ( re ) = re · n . It’s easy enough to check h is an R -modulemap; moreover, g ( h ( re )) = g ( re · n ) = re · g ( n ) = re · f ( e ) = f ( re ) = f ( re );meaning the diagram below commutes: NRe Mfh g
Hence, Re is projective. Proof of Theorem 6.5.
Set M := RQ , Q k := u k Q (= Qu k ), and M k := RQ k .We will prove our result by showing M is a locally projective generator of R − M OD such that lim −→ End R ( M k ) ∼ = QRQ . To that end, recall from Proposition6.3.1 that R a σ -unital ring = ⇒ R sits in M ( R ) as an essential ideal;since Q k is an idempotent element in R for each k , by Claim 6.6, M k is a finitelygenerated, projective, R -module for each k . By the property of a σ -unit, for i ≤ k , we have u i u k = u i ; this, along with the fact the elements of the σ -unitof R commute with Q , implies M i ⊆ M k . Define ϕ i,k : M i → M k to be theinclusion map. One can quickly check h M i , ϕ i,k i is a direct system of R -modulesover N , let lim −→ M i be its direct limit.Consider now the R -module M together with the inclusion maps { Φ i : M i → M } i ∈ N . Since Φ i = Φ k ◦ ϕ i,k , by the universal property of lim −→ M i , there existsa unique R -module homomorphism θ : lim −→ M i → M such that Φ i = θ ◦ η i for each i . For { u k } k ∈ N a σ -unit of R , we have R = S Ru k , which in turnmeans M = S M k . From this, along with the fact each Φ k is an inclusion map,we can conclude θ is surjective, by [32, 24.3, 1) and 2)], θ is injective as well.Thus, lim −→ M i ∼ = M. Finally, for each i ≤ k , define ψ k,i : M k → M i to be rightmultiplication by u i . One can easily check { M i , ϕ i,k , ψ k,i , N } is a compatibleset, allowing us to conclude M is a locally projective R -module.To see M is a generator of R − M OD , let N by any R -module. For each n ∈ N , define ρ n : M → N to be right multiplication by n ; then, extend ρ n tohomomorphism from L n ∈ N M to N in the usual way. After which, set ρ := M n ∈ N ρ n : M n ∈ N M → N, where the restriction of ρ to the n -coordinate is ρ n . Since M = RQ , noteim ρ = RQN . Moreover, because RN = N , recall we are assuming all modules36o be unital, and Q is full, we haveim ρ = RQN = RQRN = RN = N. Thus, ρ is surjective, implying M is a generator of R − M od .Now, for a fixed i , consider the R -module M i and let φ ∈ End R M opi . For rQ i ∈ M i , note φ ( rQ i ) = φ ( rQ i ) = rQ i φ ( Q i ) , and so φ is just right mul-tiplication by φ ( Q i ); moreover, φ ( Q i ) = φ ( Q i ) = Q i φ ( Q i ) implies φ ( Q i ) ∈ Q i M i := Q i RQ i ⊆ M i . Let ev Q i : End R M opi → Q i RQ i ; be the evalua-tion map at Q i (i.e., φ φ ( Q i )), it’s straight forward to work out ev Q i isa ring homomorphism. Since right multiplication by any element of Q i RQ i defines an element of End R M i , ev Q i is surjective. Further, if φ ( Q i ) = 0,im φ = φ ( RQ i ) = Rφ ( Q i ) = { } ; this means φ = 0, and so ev Q i is injec-tive as well. Hence, we have End R M opi ∼ = Q i RQ i . Lastly, the family of maps { ev Q i } i ∈ N are such that the following diagram commutes for each i ≤ k ∈ N :End R M opi Q i RQ i End R M opk Q k RQ k ev Q i ev Q k ϕ i,k ϕ i,k where the set { ϕ i,k : Q i RQ i → Q k RQ k } i,k ∈ N consists of inclusion maps, andthe set { ϕ i,k : End R M opi → End R M opk } i,k ∈ N consists of maps as in Proposition 5.24.1. By [32, 24.4], the family of maps { ev Q i } i ∈ N induce a ring isomorphism ev Q : lim −→ End R M opi → lim −→ Q i RQ i , thus,lim −→ End R M opi ∼ = lim −→ Q i RQ i . Using the same argument as the one we used toshow lim −→ M i ∼ = M , we can show lim −→ Q i RQ i ∼ = QRQ, meaning lim −→ End R M opi ∼ = QRQ, and so, byTheorem 5.26, R and QRQ are Morita equivalent.At this juncture, we encourage the reader to see the paragraph precedingLemma 2.7 for the definitions of W and Γ . With that, we will proceed withusing the machinery we have built up to establish an important result regard-ing graph Leavitt path algebras—the existence of an algebraic analog to theprojection constructed just before [19, Proposition 5.20]. Lemma 6.7.
There exists a full idempotent Q ∈ M ( L R ( E G )) such that, forany t α t β ∗ ∈ L R ( E G ) , Qt α t β ∗ = (cid:26) t α t β ∗ if s ( α ) ∈ W ⊔ Γ , otherwise. roof. Let X = W ⊔ Γ . From [31, Proposition 3.4], we can deduce L R ( E G ) ∼ = (cid:16) M v ∈ X L R ( E G ) q v (cid:17) M (cid:16) M v ∈ E G \ X L R ( E G ) q v (cid:17) ∼ = (cid:16) M v ∈ X q v L R ( E G ) (cid:17) M (cid:16) M v ∈ E G \ X q v L R ( E G ) (cid:17) as abelian groups. Let X P be the projection of L R ( E G ) onto L v ∈ X q v L R ( E G ),and let P X be the projection onto L v ∈ X L R ( E G ) q v , then set Q := ( X P, P X ).Fix t α t β ∗ ∈ L R ( E G ). Since L t α t β ∗ is left multiplication by t α t β ∗ , and X P isthe projection of L R ( E G ) onto L v ∈ X q v L R ( E G ) , we have X P ◦ L t α t β ∗ = (cid:26) L t α t β ∗ if s ( α ) ∈ X, LP1 , R t α t β ∗ = 0 on M { v ∈ E G : v = s ( α ) } L R ( E G ) q v , and since P X is the projection of L R ( E G ) onto L v ∈ X L R ( E G ) q v , R t α t β ∗ ◦ P X = (cid:26) R t α t β ∗ if s ( α ) ∈ X, L R ( E G ) with i ( L R ( E G )), “ t α t β ∗ ” should be understood to mean( L t α t β ∗ , R t α t β ∗ ) ∈ M ( L R ( E G )) . In which case, “ Qt α t β ∗ ” should be taken to mean( X P, P X )( L t α t β ∗ , R t α t β ∗ ) := ( X P ◦ L t α t β ∗ , R t α t β ∗ ◦ P X ) , which we have just shown is equal to ( L t α t β ∗ , R t α t β ∗ ). And so, by notationalabuse and all, we have Qt α t β ∗ = (cid:26) t α t β ∗ if s ( α ) ∈ X, Q is idempotent follows directly from the fact X P and P X are projec-tions. To see Q is full, fix t α t β ∗ ∈ L R ( E G ). By [19, Lemma 4.6], there exists apath, α ′ , such that s ( α ′ ) ∈ X and r ( α ′ ) = s ( α ). Note then t α ′∗ t α ′ α t β = t α t β ∗ ,moreover, since s ( α ′ ) ∈ X , Qt α ′ α t β = t α ′ α t β . And so we have t α t β ∗ = t α ′∗ t α ′ α t β = t α ′∗ Qt α ′ α t β . Since L R ( E G ) is the span of elements of the form t α t β ∗ , we can conclude Q is full. Corollary 6.8.
Let G be an ultragraph, E G the graph as in Definition 2.9, and Q ∈ M ( L R ( E G )) as in Lemma 6.7. Then, L R ( E G ) is Morita equivalent to QL R ( E G ) Q . roof. We have { u k } k ∈ N , as defined in Lemma 3.10, is a σ -unit for L R ( E G ). Let Q ∈ M ( L R ( E G )) be as in Lemma 6.7. For a given v ∈ E G , we have, by 6.7, Qq v = (cid:26) q v if v ∈ X, L q v , R q v )( X P, P X ) := ( L q v ◦ X P, P X ◦ R q v ) = (cid:26) ( L q v , R q v ) if v ∈ X, q v Q = (cid:26) q v if v ∈ X, u k Q = Qu k for each k ∈ N ; and so, by Theorem 6.5, L R ( E G )is Morita equivalent to QL R ( E G ) Q .Let G be an ultragraph without any singular vertices and E G its associatedgraph. The first major result of this chapter will be showing L R ( G ) and L R ( E G )are Morita equivalent. We will prove this fact by showing L R ( G ) ∼ = QL R ( E G ) Q, where Q is as in Lemma 6.7. We will later show that even in the case where G contains singular veritces, L R ( G ) is still Morita equivalent to a graph Leavittpath algebra; in particular, L R ( G ) is Morita equivalent to L R ( E F ), where F isthe desingularization of G .An important component of our endeavor is to define an algebraic analog tothe Exel-Laca algebras defined in [18, Definition 3.3]. For λ, µ , finite subsets of G , let r ( λ, µ ) := \ e ∈ λ r ( e ) \ [ f ∈ µ r ( f ) Definition 7.1.
Let G be an ultragraph and A an R -algebra. A collection ofidempotent elements { P v , Q e : v ∈ G , e ∈ G } in A satisfy condition (EL) if:( EL1 ) the elements of { P v } are mutually orthogonal,( EL2 ) the elements of { Q e } pairwise commute,( EL3 ) P v Q e = Q e P v = ( P v if v ∈ r ( e )0 if v / ∈ r ( e )( EL4 ) Q e ∈ λ Q e Q f ∈ µ (1 − Q f ) = P v ∈ r ( λ,µ ) P v for all λ, µ , finite subsets of G suchthat λ ∩ µ = ∅ , λ = ∅ , and r ( λ, µ ) is finite.39f A is nonunital, we can take the identity in EL4 to be the identity in A ∗ . Here, A ∗ is the unitization of A . As an abelian group, A ∗ = R ⊕ A ;multiplication is given by ( r , a )( r , a ) = ( r r , r a + r a + a a ); scalarmultiplication is given by r ( r , a ) = ( rr , ra ). It’s easy enough to check A ∗ isan R -algebra. More importantly, since R is unital, A ∗ is unital with unit (1 , a (0 , a ) embeds A into A ∗ as an ideal. Definition 7.2.
Let G be an ultragraph and A an R -algebra. An Exel-Laca G -family in A is a collection of idempotent elements { P v } v ∈ G , and elements { S e , S e ∗ } e ∈G such that:( ExL1 ) the collection of elements { P v } v ∈ G ∪ { S e ∗ S e } e ∈G satisfy condition(EL),( ExL2 ) P s ( e ) S e = S e S e ∗ S e = S e and S e ∗ S e S e ∗ = S e ∗ P s ( e ) = S e ∗ for all e ∈ G ,( ExL3 ) S f ∗ S e = 0 when e = f ,( ExL4 ) P v = P e ∈ s − ( v ) S e S e ∗ for each v ∈ G such that 0 < | s − ( v ) | < ∞ . Lemma 7.3.
Let A be an R -algebra. Given a Leavitt G -family { P A , S e , S ∗ e } in A , the set { P v , S e , S ∗ e } forms an Exel-Laca G -family in A .Proof. The proof given here is the same as the first part of the proof of[18, Lemma 2.10]. Let { P A , S e , S ∗ e } be a Leavitt G -family in A . The fact { P v , S e , S e ∗ } satisfies ExL2 , ExL3 , and
ExL4 , as well as the fact { P v , S e ∗ S e } = { P v , P r ( e ) } satisfies EL1 , EL2 , and
EL3 , follows directly from the propertiesof a Leavitt G -family. By uLP4 , we have Y e ∈ r ( λ ) P r ( e ) Y f ∈ µ (1 − P r ( f ) ) = P T e ∈ r ( λ ) r ( e ) − P T e ∈ r ( λ ) r ( e ) P S f ∈ r ( µ ) r ( f ) = P T e ∈ r ( λ ) \ S f ∈ r ( µ ) = X v ∈ r ( λ,µ ) P v ;meaning EL4 is satisfied, which in turn means
ExL1 is satisfied. Thus, the set { P v , S e , S ∗ e } forms an Exel-Laca G -family in A . Definition 7.4.
Given an ultragraph G , we define the Exel-Laca algebra of G , EL R ( G ), to be the R -algebra generated by an Exel-Laca G -family { p v , s e , s e ∗ } ⊆EL R ( G ) such that, for any R -algebra A and an Exel-Laca G -family { P v , S e , S e ∗ } ⊆A , there exists an R -algebra homomorphism φ : EL R ( G ) → A with φ ( p v ) = P v , φ s e = S e , φ ( s e ∗ ) = S e ∗ . E L R ( G ) For X = G ∪ G ∪ ( G ) ∗ , let F R ( w ( X )) be the nonunital, associative, free R -algebra as in Remark 3.6, and let I ⊳ F R ( w ( X )) be the ideal generated by theunion of the sets: 40 { vw − δ v,w v : v, w ∈ G } , { e ∗ ef ∗ f − f ∗ f e ∗ e : e, f ∈ G } , • { ve ∗ e − v, e ∗ ev − v : v ∈ G , e ∈ G such that v ∈ r ( e ) } , { ve ∗ e, e ∗ ev : v ∈ G ,e ∈ G such that v / ∈ r ( e ) } , • { Q e ∈ λ e ∗ e Q f ∈ µ (1 − f ∗ f ) − P v ∈ r ( λ,µ ) v : λ, µ finite subsets of G with λ ∩ µ = ∅ , λ = ∅ , and r ( λ, µ ) is finite } . • { e ∗ f : e, f ∈ G such that e = f } , { e − s ( e ) e, ee ∗ e − e, e ∗ − e ∗ s ( e ) , e ∗ ee ∗ − e ∗ : e ∈ G } , • { v − P e ∈ s − ( v ) ee ∗ : v ∈ G such that 0 < | s − ( v ) | < ∞} . Then, EL R ( G ) := F R ( w ( X )) /I ;given the projection π : F R ( w ( X )) → F R ( w ( X )) /I, the Exel-Laca G -family, { s e , s e ∗ , p v } , is taken to be the family { π ( e ) , π ( e ∗ ) , π ( v ) } .To see that EL R ( G ) and { s e , s e ∗ , p v } have the desired univeral propery, let A bean R -algebra and { P v , S e , S e ∗ } an Exel-Laca G -family in A . By the univerisalmapping property of F R ( w ( X )), we have an R -algebra homomorphism φ : F R ( w ( X )) → A such that φ ( v ) = P v , φ ( e ) = S e , φ ( e ∗ ) = S e ∗ . Further, since { P v , S e , S e ∗ } is anExel-Laca G -family, I ⊂ ker φ . Thus, there exists an R -algebra homomorphism φ : F R ( w ( X )) /I → A such that φ = φ ◦ π . Put more desirably, there exists an R -algebra homomor-phism φ : EL R ( G ) → A such that φ ( p v ) = P v , φ ( s e ) = S e , φ ( s e ∗ ) = S e ∗ .Let A be an R -algebra with an Exel-Laca G -family satisfying the propertiesof Definition 7.4. Then, there exist R -homomorphims ϕ : A → EL R ( G ) and φ : EL R ( G ) → A which are inverses of each other. Thus, EL R ( G ) is unique up to isomorphism. E L R ( G ) Lemma 8.1.
There exists a surjective R -algebra homomorphism ϕ : EL R ( G ) → L R ( G ) such that ϕ ( p v ) = p v , ϕ ( s e ) = s e , ϕ ( s e ∗ ) = s e ∗ . roof. By Lemma 7.3, the set { p v , s e , s e ∗ } forms an Exel-Laca G -family in L R ( G ). And so by the universal property of EL R ( G ), there exists an R -algebrahomomorphism ϕ such that ϕ ( p v ) = p v , ϕ ( s e ) = s e , ϕ ( s e ∗ ) = s e ∗ . Further, foreach e ∈ G , p r ( e ) ∈ im ϕ since p r ( e ) = s e ∗ s e = ϕ ( s e ∗ s e ); which, by Theorem3.19, means p A ∈ im ϕ for each A ∈ G . Since L R ( G ) is the R -linear span ofelements of the form s α p A s ∗ β , with r ( α ) ∩ A ∩ r ( β ) = ∅ [16, Theorem 2.5], itfollows ϕ is surjective.We now need to establish the injectivity of ϕ . This endeavor is a bit moreinvolved. We will start by giving a useful characterization of EL R ( G ). Firstly,for e, f ∈ G , Exl2 implies s e s f = ( s e s e ∗ s e )( p s ( f ) s f ). But, by EL3 , this means s e s f = s e ( s e ∗ s e p s ( f ) ) s f = 0 if s ( f ) / ∈ r ( e ). Which is to say s e s f = 0 unless ef isa path in G . A similar argument also shows s f ∗ s e ∗ = 0 unless ef is a path. Onthe other hand, from the construction of EL R ( G ), we can see ef ∈ G ∗ = ⇒ s e s f = 0 , s f ∗ s e ∗ = 0 . Thus, just as with L R ( G ), s e · · · s e n = 0 ⇐⇒ e · · · e n ∈ G ∗ . In which case, for α := e · · · e n , we take s α := s e · · · s e n and s α ∗ := s e ∗ n · · · s e ∗ .Recall that each vertex v is considered a path of length 0; and so for α = v , wetake s α = s α ∗ = p v .The following useful lemma shows results of the form Theorem 3.19 andProposition 3.5 hold for EL R ( G ); its method of proof is also similar to how oneobtains the aforementioned results. Lemma 8.2.
Let G be an ultragraph. Then, EL R ( G ) = span R n { s α , s β ∗ , s α p v , p v s β ∗ : v ∈ r ( α ) , v ∈ r ( β ) } ∪{ s α p v s β ∗ : v ∈ r ( α ) ∩ r ( β ) } ∪{ Y e ∈ S s e ∗ s e , , s α (cid:0) Y e ∈ S s e ∗ s e (cid:1) s β ∗ : S is a finite subset of G ,r ( α ) ∩ r ( β ) ∩ (cid:0) \ e ∈ S r ( e ) (cid:1) = ∅} o . Proof.
We can see from its construction that EL R ( G ) is generated as an R -algebra by the set { s α , s β ∗ } α,β ∈G ∗ ; and so to prove our lemma, we only needto argue a non-zero product of elements in { s α , s β ∗ } α,β ∈G ∗ can be reduced toone of the prescribed forms. To that end, suppose 0 = x ∈ EL R ( G ) is such anelement.1) Suppose there is a term of the form s α s α ′ (similarly s β ∗ s β ′∗ ) in the expres-sion of x . If both α and α ′ have lengths larger than 0, then by our assumption x = 0, it must be αα ′ ∈ G ∗ and s α s α ′ = s αα ′ . If | α | = 0, then it must be that α = s ( α ′ ) and s α s α ′ = p s ( α ′ ) s α ′ = s α ′ . Lastly, if | α ′ | = 0, then we can conclude42 ′ = v ∈ r ( α ) and s α s α ′ = s α p v . We can similarly deduce that s β ∗ s β ′∗ will bein one of the following forms: s ( β ′ β ) ∗ , s β ∗ , or p v s β ′∗ with v ∈ r ( β ′ ). Note thatthe result stated here comes as a direct consequence of EL3 and
ExL2 .2) Suppose there is a term of the form s α s β ∗ in the expression of x . If | α | = 0, EL3 and
ExL2 give us s α s β ∗ = p v s β ∗ , where α = v ∈ r ( β ); similarly, if | β | = 0,we get s α s β ∗ = s α p v where β = v ∈ r ( α ). Alternatively, if α = e · · · e n and β = f · · · f m , with n, m > ExL2 implies s α s β ∗ = s α (cid:0) ( s e ∗ n s e n )( s f ∗ m s f m ) (cid:1) s β ∗ . The fact x = 0, along with EL3 and
ExL2 , implies r ( e n ) ∩ r ( f m ) = ∅ .3) Finally, suppose α = e · · · e n , β = f · · · f m ∈ G ∗ . Then, s β ∗ s α := ( s f ∗ m · · · s f ∗ )( s e · · · s e n ) = s f ∗ m · · · s f ∗ ( s f ∗ s e ) s e · · · s e n . By ExL3 , s f ∗ s e = 0 if f = e ; otherwise, s f ∗ s e = s e ∗ s e . Consequently, EL3 and
Exl2 imply s e ∗ s e s e = s e ∗ s e p s ( e ) s e = s e . This means s f ∗ m · · · s f ∗ ( s f ∗ s e ) s e · · · s e n =0 if f = e , otherwise, s f ∗ m · · · s f ∗ ( s f ∗ s e ) s e · · · s e n = s f ∗ m · · · s f ∗ s e · · · s e n . Continuing in this manner, we can see that: s β ∗ s α = s e ∗ n s e n if α = β, s β ′∗ if β = αβ ′ , where β ′ = f n +1 · · · f m , s α ′ if α = βα ′ , where α ′ = e m +1 · · · e n , . Should | β | = 0, or | α | = 0, note that Exl2 implies the above formula still holds.By using 1), 2), and 3), as needed, we can reduce x to one of the desiredforms.An important component to showing the injectivity of ϕ from Lemma 8.1rests on the fact EL R ( G ) is a Z -graded ring. Lemma 8.3. EL R ( G ) is a Z -graded ring with EL R ( G ) ∼ = L i ∈ Z EL R ( G ) i , where EL R ( G ) i = span R n { s α , s β ∗ , s α p v , p v s β ∗ : v ∈ r ( α ) , v ∈ r ( β ) , | α | = | β | = i } ∪{ s α p v s β ∗ : v ∈ r ( α ) ∩ r ( β ) , | α | − | β | = i } ∪{ Y e ∈ S s e ∗ s e , , s α (cid:0) Y e ∈ S s e ∗ s e (cid:1) s β ∗ : S is a finite subset of G , \ e ∈ S r ( e ) = ∅ , r ( α ) ∩ r ( β ) ∩ (cid:0) \ e ∈ S r ( e ) (cid:1) = ∅ , | α | − | β | = i } o . roof. Let X = G ∪ G ∪ ( G ) ∗ and let F R ( w ( X )) be the free R -algebra on X .For each w ∈ w ( X ), take | w | := { the number of elements in G which appear in w }− { the number of elements in ( G ) ∗ which appear in w } . Since F R ( w ( X )) is the free R -module on w ( X ), we have F R ( w ( X )) = M i ∈ Z F R ( w ( X )) i , where F R ( w ( X )) i := span R { w ∈ w ( X ) : | w | = i } . Moreover, for w , w ∈ w ( X ),we have | w w | = | w | + | w | ; which means F R ( w ( X )) n · F R ( w ( X )) m ⊆ F R ( w ( X )) n + m . Thus, F R ( w ( X )) is a Z -graded ring with respect to the grading given above.Now, recall EL R ( G ) := F R ( w ( X )) /I , where I is generated by the union ofthe sets: • { vw, f ∗ e : v = w, e = f } , { e ∗ ef ∗ f − f ∗ f e ∗ e } , • { ve ∗ e − v, e ∗ ev − v : v ∈ r ( e ) } , { ve ∗ e, e ∗ ev : v / ∈ r ( e ) } , • { Q e ∈ λ e ∗ e Q f ∈ µ (1 − f ∗ f ) − P v ∈ r ( λ,µ ) v : λ, µ finite subsets of G with λ ∩ µ = ∅ , λ = ∅ , and r ( λ, µ ) is finite } , • { f ∗ e : e = f } , { e − s ( e ) e, ee ∗ e − e, e ∗ − e ∗ s ( e ) , e ∗ ee ∗ − e ∗ } , • { v − P e ∈ s − ( v ) ee ∗ : v ∈ G with 0 < | s − ( v ) | < ∞} .Note that I is generated by homogeneous elements of degree 0; which means EL R ( G ) := F R ( w ( X )) /I is a Z -graded ring (see Remark 3.14). In particular, F R ( w ( X )) /I = M i ∈ Z ( F R ( w ( X )) i + I ) /I as an internal direct sum. Since( F R ( w ( X )) i + I ) /I = span R { w : w ∈ ( F R ( w ( X )) i } , Lemma 8.2 implies EL R ( G ) i := ( F R ( w ( X )) i + I ) /I = span R n { s α , s β ∗ , s α p v , p v s β ∗ : v ∈ r ( α ) , v ∈ r ( β ) , | α | = | β | = i } ∪{ s α p v s β ∗ : v ∈ r ( α ) ∩ r ( β ) , | α | − | β | = i } ∪{ Y e ∈ S s e ∗ s e , , s α (cid:0) Y e ∈ S s e ∗ s e (cid:1) s β ∗ : S is a finite subset of G , \ e ∈ S r ( e ) = ∅ , r ( α ) ∩ r ( β ) ∩ (cid:0) \ e ∈ S r ( e ) (cid:1) = ∅ , | α | − | β | = i } o . Theorem 8.4.
Suppose G is an ultragraph with no singular vertices and S a Z -graded ring. Let π : EL R ( G ) → S be a graded ring homomorphism such that π ( rp v ) = 0 and π ( rs e ∗ s e ) = 0 for all v ∈ G , e ∈ G , and r ∈ R \ { } ; then, π is injective. We will prove Theorem 8.4 using similar techniques employed by the afore-mentioned papers. Namely, we will “approximate” EL R ( G ) by the Leavitt pathalgebras of finite graphs. To that end, we will show how one defines a finitegraph G F from an ultragraph G .Let G be an ultragraph with no singular vertices, and let F ⊆ G be finite.Recall that for λ, µ , finite subsets of G , r ( λ, µ ) := T e ∈ λ r ( e ) \ S f ∈ µ r ( f ). Withthat, we will define the finite graph G F as follows: G F := F ∪ { X : ∅ 6 = X ⊆ F for which { e ∈ G : s ( e ) ∈ r ( X, F \ X ) } * F } ,G F := { ( e, f ) ∈ F × F s ( f ) ∈ r ( e ) } ∪ { ( e, X ) : e ∈ X } ;with the range and source maps given by s F (( e, f )) = e, s F (( e, X )) = e,r F (( e, f )) = f, r F (( e, X )) = X. In order to prove Theorem 8.4, we will need the following lemmas. Theyestablish an analogous result to [16, Lemma 3.1] for EL R ( G ). We will also takethis time to state that, in general, Leavitt path algebras and Exel-Laca algebraswill not be unital, but we will still find ourselves referring to a unit. We caninterpret “1” to be the unit in their respective unitizations (see the note inDefinition 7.1). Lemma 8.5.
Let { P i } ni =1 be a set of commuting idempotent elements in a ring R . Then, X ∅6 = Y ⊆{ , ··· ,n } (cid:16) Y i ∈ Y P i Y i/ ∈ Y (1 − P i ) (cid:17) = 1 − Y i ∈{ , ··· ,n } (1 − P i ) . Note: If R is not unital, we again take 1 to be the identity in its unitization.Proof. [30, Lemma 5.2] states X Y ⊆{ , ··· ,n } (cid:16) Y i ∈ Y P i Y i/ ∈ Y (1 − P i ) (cid:17) = 1 . Since Q i ∈ Y P i Q i/ ∈ Y (1 − P i ) = Q i ∈{ , ··· ,n } (1 − P i ) for Y = ∅ , we have X ∅6 = Y ⊆{ , ··· ,n } (cid:16) Y i ∈ Y P i Y i/ ∈ Y (1 − P i ) (cid:17) = 1 − Y i ∈{ , ··· ,n } (1 − P i ) . emma 8.6. Let G be an ultragraph with no singular vertices, F ⊆ G finite,and G F the associated finite graph as constructed above. Then, P e := s e s e ∗ , P X := (cid:16) Y e ∈ X s e ∗ s e Y e ′ ∈ F \ X (1 − s e ′∗ s e ′ ) (cid:17)(cid:16) − X f ∈ F s f s f ∗ (cid:17) ,S ( e,f ) := s e P f , S ( e,X ) := s e P X , S ( e,f ) ∗ := P f s e ∗ , S ( e,X ) ∗ := P X s e ∗ forms a Leavitt G F -family in EL R ( G ) .Proof. ( LP1 ). We will first show that { P e , P X } forms a set of pairwise orthog-onal idempotents. For e, f ∈ F such that e = f , we have P e P f = s e s e ∗ s f s f ∗ = s e ( s e ∗ s f ) s f ∗ = 0since s e ∗ s f = 0 by ExL3 ; further, by
ExL2 , P e is an idempotent for each e . By EL4 , we have Y e ∈ X s e ∗ s e Y e ′ ∈ F \ X (1 − s e ′∗ s e ′ ) = X v ∈ r ( X,F \ X ) p v . In particular, by applying
ExL2 , we can see that P X = (cid:16) X v ∈ r ( X,F \ X ) p v (cid:17)(cid:16) − X f ∈ F s f s f ∗ (cid:17) = (cid:16) X v ∈ r ( X,F \ X ) p v − X f ∈ F : s ( f ) ∈ r ( X,F \ X ) s f s f ∗ (cid:17) ;using this fact, and exploiting ExL2 again, we can work out P X is an idempotentfor each X .Of particular utility in showing the pairwise orthogonality of the set { P e , P X } is the fact s e ∗ p v = s e ∗ if v = s ( e ), and s e ∗ p v = 0 otherwise. With this in mind,for e, X ∈ G F , note that s e s e ∗ (cid:16) P v ∈ r ( X,F \ X ) p v (cid:17) = 0 if s ( e ) / ∈ r ( X, F \ X ),meaning P e P X = 0; if s ( e ) ∈ r ( X, F \ X ), then by ExL2 and
ExL3 we have s e s e ∗ (cid:16) X v ∈ r ( X,F \ X ) p v (cid:17) = s e s e ∗ and s e s e ∗ (cid:16) X f ∈ F : s ( f ) ∈ r ( X,F \ X ) s f s f ∗ (cid:17) = s e s e ∗ , which again gives us P e P X = s e s e ∗ (cid:16) X v ∈ r ( X,F \ X ) p v − X f ∈ F : s ( f ) ∈ r ( X,F \ X ) s f s f ∗ (cid:17) = 0 . A similar argument shows P X P e = 0 for each e, X ∈ G F ; thus, P e P X = P X P e =0 for each e, X ∈ G F .Now suppose X, Y ∈ G F such that X = Y . WLOG, let e ′ ∈ X \ Y . Thismeans r ( X, F \ X ) := \ e ∈ X r ( e ) \ [ f ∈ F \ X r ( f ) ⊆ r ( e ′ )46ince e ′ ∈ X ; and r ( Y, F \ Y ) ∩ r ( e ′ ) = ∅ since e ′ ∈ F \ Y . And so we canconclude r ( X, F \ X ) ∩ r ( Y, F \ Y ) = ∅ . As a result, for P X = (cid:16) X v ∈ r ( X,F \ X ) p v − X f ∈ F : s ( f ) ∈ r ( X,F \ X ) s f s f ∗ (cid:17) and P Y = (cid:16) X v ∈ r ( Y,F \ Y ) p v − X f ∈ F : s ( f ) ∈ r ( Y,F \ Y ) s f s f ∗ (cid:17) , we have P X P Y = P Y P X = 0. Thus, { P e , P X } forms a set of pairwise orthogonalidempotents in EL R ( G ).( LP2 ). Recall G F := { ( e, f ) ∈ F × F s ( f ) ∈ r ( e ) } ∪ { ( e, X ) : e ∈ X } . Witha straight forward application of ExL2 , we can deduce P s F (( e,f )) S ( e,f ) = S ( e,f ) P r F (( e,f )) = S ( e,f ) , P s F (( e,X )) S ( e,X ) = S ( e,X ) P r F (( e,X )) = S ( e,X ) P r F (( e,f )) S ( e,f ) ∗ = S ( e,f ) ∗ P s F (( e,f )) = S ( e,f ) ∗ , and P r F (( e,X )) S ( e,X ) ∗ = S ( e,X ) ∗ P s F (( e,X )) = S ( e,X ) ∗ ( LP3 ). By definition, S ( e,f ) ∗ S ( e ′ ,f ′ ) = P f s e ∗ s e ′ P f ′ . If e = e ′ , then s e ∗ s e ′ = 0by ExL2 , which in turn means S ( e,f ) ∗ S ( e ′ ,f ′ ) = 0. If e = e ′ , then P f s e ∗ s e ′ P f ′ = P f P f ′ by ExL2 and
EL3 . Moreover, by
LP1 , P f P f ′ = 0 if f = f ′ . All inall, this means S ( e,f ) ∗ S ( e ′ ,f ′ ) = 0 if ( e, f ) = ( e ′ f ′ ), and S ( e,f ) ∗ S ( e ′ ,f ′ ) = P f = P r F (( e,f )) if ( e, f ) = ( e ′ , f ′ ). Similarly, S ( e,X ) ∗ S ( e ′ ,X ′ ) = P X s e ∗ s e ′ P X ′ . As before,if e = e ′ , then S ( e,X ) ∗ S ( e ′ ,X ′ ) = 0. If e = e ′ , then for e ∈ X ′ , EL2 implies s e ∗ s e (cid:16) Y g ∈ X ′ s g ∗ s g (cid:17) = Y g ∈ X ′ s g ∗ s g . And so s e ∗ s e P X ′ = s e ∗ s e (cid:16) Y g ∈ X ′ s g ∗ s g Y g ′ ∈ F \ X ′ (1 − s g ′∗ s g ′ ) (cid:17)(cid:16) − X f ∈ F s f s f ∗ (cid:17) = P X ′ . Therefore, for e = e ′ , we have S ( e,X ) ∗ S ( e ′ ,X ′ ) = P X P X ′ . Exploiting LP1 again,we have S ( e,X ) ∗ S ( e ′ ,X ′ ) = 0 if ( e, X ) = ( e ′ X ′ ), and S ( e,X ) ∗ S ( e ′ ,X ′ ) = P X = P r F (( e,X )) if ( e, X ) = ( e ′ , X ′ ). Finally, for ( e, f ) , ( e ′ , X ′ ) ∈ G F , we can deduce S ( e,f ) ∗ S ( e ′ ,X ′ ) = S ( e ′ ,X ′ ) ∗ S ( e,f ) = 0 due to the fact P f P X ′ = P X ′ P f = 0.( LP4 ). Since X ∈ G F is a sink, we only need to consider e ∈ F . To thatend, suppose X ⊆ F such that X / ∈ G F ; which is to say X ⊆ F for which { e ∈ G : s ( e ) ∈ r ( X, F \ X ) } ⊆ F. Then, by
ExL4 , we have (cid:16) Y e ∈ X s e ∗ s e Y e ′ ∈ F \ X (1 − s e ′∗ s e ′ ) (cid:17)(cid:16) − X f ∈ F s f s f ∗ (cid:17) = (cid:16) X v ∈ r ( X,F \ X ) p v − X { f ∈ F : s ( f ) ∈ r ( X,F \ X ) } s f s f ∗ (cid:17) = X v ∈ r ( X,F \ X ) (cid:16) p v − X { f ∈ F : s ( f )= v } s f s f ∗ (cid:17) = 0 .
47n the other hand, for a fixed e ∈ F , and X ⊆ F such that e / ∈ X (and so e ∈ F \ X ), EL2 implies s e ∗ s e (cid:16) Y g ∈ X s g ∗ s g Y g ′ ∈ F \ X (1 − s g ′∗ s g ′ ) (cid:17)(cid:16) − X f ∈ F s f s f ∗ (cid:17) = s e ∗ s e (1 − s e ∗ s e ) (cid:16) Y g ∈ X s g ∗ s g Y g ′ ∈ F \ X, g ′ = e (1 − s g ′∗ s g ′ ) (cid:17)(cid:16) − X f ∈ F s f s f ∗ (cid:17) = 0 . Combining the previous two calculations, we can deduce s e ∗ s e (cid:16) X X ∈ G F : e ∈ X P X (cid:17) = s e ∗ s e (cid:18) X ∅6 = X ⊆ F (cid:16) Y g ∈ X s g ∗ s g Y g ′ ∈ F \ X (1 − s g ′∗ s g ′ ) (cid:17)(cid:19)(cid:16) − X f ∈ F s f s f ∗ (cid:17)! , which by Lemma 8.5,= s e ∗ s e (cid:18) − Y g ′ ∈ F (1 − s g ′∗ s g ′ ) (cid:19)(cid:16) − X f ∈ F s f s f ∗ (cid:17)! = (cid:16) s e ∗ s e − s e ∗ s e (1 − s e ∗ s e ) Y g ′ ∈ F,g ′ = e (1 − s g ′∗ s g ′ ) (cid:17)(cid:16) − X f ∈ F s f s f ∗ (cid:17) = s e ∗ s e (cid:16) − X f ∈ F s f s f ∗ (cid:17) , and since s e ∗ s e s f s f ∗ = 0 if s ( f ) / ∈ r ( e ) , = s e ∗ s e (cid:16) − X { f : s ( f ) ∈ r ( e ) } s f s f ∗ (cid:17) = s e ∗ s e (cid:16) − X { f : s ( f ) ∈ r ( e ) } P f (cid:17) . Note that, by
ExL2 , s e (cid:16) X { X : e ∈ X } P X (cid:17) = s e (cid:18) s e ∗ s e (cid:16) X { X : e ∈ X } P X (cid:17)(cid:19) = s e (cid:18) s e ∗ s e (cid:16) − X { f : s ( f ) ∈ r ( e ) } P f (cid:17)(cid:19) = s e (cid:16) − X { f : s ( f ) ∈ r ( e ) } P f (cid:17) Thus, X { f : s ( f ) ∈ r ( e ) } S ( e,f ) S ( e,f ) ∗ + X { X : e ∈ X } S ( e,X ) S ( e,X ) ∗ = X { f : s ( f ) ∈ r ( e ) } S ( e,f ) S ( e,f ) ∗ + X { X : e ∈ X } S ( e,X ) S ( e,X ) ∗ = s e (cid:16) X { f : s ( f ) ∈ r ( e ) } P f (cid:17) s e ∗ + s e (cid:16) X { X : e ∈ X } P X (cid:17) s e ∗ = s e (cid:16) X { f : s ( f ) ∈ r ( e ) } P f (cid:17) s e ∗ + s e (cid:16) − X { f : s ( f ) ∈ r ( e ) } P f (cid:17) s e ∗ = s e s e ∗ = P e . emma 8.7. Let G be an ultragraph with no singular vertices, F ⊆ G finite,and G F the associated finite graph. Then, there exists a Z -graded injective R -algebra homomorphism π F : L R ( G F ) → EL R ( G ) such that π F ( p e ) = P e , π F ( p X ) = P X , π F ( s ( e,f ) ) = S ( e,f ) , and , π F ( s ( e,X ) ) = S ( e,X ) .Proof. By [31, Proposition 3.4], we have L R ( G F ) := L R ( s, p ) = span R { s α s β ∗ : r ( α ) = r ( β ) } ;and is a Z -graded ring. Specifically, L R ( G F ) = L i ∈ Z L R ( G F ) i , where L R ( G F ) i = span R { s α s β ∗ : r ( α ) = r ( β ) and | α | − | β | = i } . By Lemma 8.6, we have a Leavitt G F -family { P, S } ⊆ EL R ( G ); and so by theuniversal mapping property of L R ( G F ), we have an R -algebra homomorphism π F : L R ( G F ) → EL R ( G )such that π F ( p e ) = P e , π F ( p X ) = P X , π F ( s ( e,f ) ) = S ( e,f ) , and , π F ( s ( e,X ) ) = S ( e,X ) . Further, using Lemma 8.3 and Tomforde’s aforementioned result on the Z -grading of L R ( G F ), we can deduce π F ( L R ( G F ) i ) ⊆ EL R ( G ) i ;thus, π F is a graded homomorphism. Finally, Lemma 8.1 and [16, Theorem 2.6]imply that, for r = 0: r p v = 0, r s e = 0, and r s e ∗ = 0. This in turn implies that,for r = 0, rP e = 0 , rP X = 0 , rS ( e,f ) = 0 , and rS ( e,X ) = 0 . And so, by Theorem 3.15, π F is injective.We are now ready to prove Theorem 8.4. Proof of Theorem 8.4.
Let F ⊆ F ⊆ · · · ⊆ F n · · · be an increasing sequenceof finite subsets of G such that S n ∈ N F n = G . For each F n , let G F n be theassociated finite graph. By Lemma 8.7, for each n , there exists an injective R -algebra homomorphism π F n : L R ( G F n ) → EL R ( G ) . This in turn gives us a Z -graded ring homomorphism π ◦ π F n : L R ( G F n ) → S for each n . By our assumption on π , Theorem 3.15, and Lemma 8.7, π ◦ π F n isinjective for each n . 49inally, let V ⊆ G and F ⊆ G be finite, and s − ( v ) ⊆ F for each v ∈ V .Fix n large enough such that F ⊆ F n . Since we are assuming G doesn’t haveany singular vertices, we have p v = X { e : s ( e )= v } s e s e ∗ = X { e : s ( e )= v } π F n ( p e )for each v ∈ V ; and so, for each v ∈ V , p v ∈ im π F n . For e ∈ F ⊆ F n , we have X { f : s ( f ) ∈ r ( e ) } π F n ( s ( e,f ) ) + X { X : e ∈ X } π F n ( s ( e,X ) ) = s e (cid:16) X { f : s ( f ) ∈ r ( e ) } P f + (1 − X { f : s ( f ) ∈ r ( e ) } P f ) (cid:17) = s e ;we can similarly show X { f : s ( f ) ∈ r ( e ) } π F n ( s ( e,f ) ∗ ) + X { X : e ∈ X } π F n ( s ( e,X ) ∗ ) = s e ∗ . Thus, for each v ∈ V and e ∈ F , we have that p v , s e , s e ∗ ∈ im π F n . That is, the R -algebra generated by { p v , s e , s e ∗ } is a subset of im π F n . In turn, this implies EL R ( G ) = [ n ∈ N im π F n . Combining this with the fact π ◦ π F n is injective for each n , we can conclude π is injective as well. Theorem 8.8. If G is an ultragraph with no singular vertices, then EL R ( G ) ∼ = L R ( G ) . Proof.
We have, by Theorem 3.19, L R ( G ) = span R { s α p A s β ∗ : r ( α ) ∩ A ∩ r ( β ) = ∅} and is a Z -graded ring. Particularly, L R ( G ) = L i ∈ Z L R ( G ) i where L R ( G ) i = span R { s α p A s β ∗ : r ( α ) ∩ A ∩ r ( β ) = ∅ , | α | − | β | = i } . By Lemma 8.1, there exists a surjective map ϕ : EL R ( G ) → L R ( G ) such that ϕ ( p v ) = p v , ϕ ( s e ) = s e , ϕ ( s e ∗ ) = s e ∗ . It is straightforward to see ϕ ( EL R ( G ) i ) ⊆ L R ( G ) i , meaning ϕ is a Z -graded R -algebra homomorphism. By Theorem 3.19and Theorem 8.4, ϕ is also injective. Thus, EL R ( G ) ∼ = L R ( G ).50 L R ( G ) ∼ = QL R ( E G ) Q Our next crucial step is to show L R ( G ) ∼ = QL R ( E G ) Q . But, before we do,there are some results we first need to establish analogous to [19, Lemma 5.11,Proposition 5.15, Lemma 5.19, and Theorem 5.22]. By [19, Lemma 4.6 andDefinition 4,7], for each x ∈ E G , there exists a unique path α x in the subgraph F of E G such that r ( α x ) = x and s ( α x ) ∈ W ⊔ Γ (see Remark 2.11). Giventhe enumeration of G , and identifying G as a subset of E G , let P v := t α v t α ∗ v for v ∈ G ,S e n := X x ∈ X ( e n ) t α s ( en ) t ( e n ,x ) t α ∗ x , and S e ∗ n := X x ∈ X ( e n ) t α x t ( e n ,x ) ∗ t α ∗ s ( en ) for e n ∈ G . The following quick claim is useful:
Claim . For e n ∈ G , let x, y ∈ X ( e n ). If x = y , then t α ∗ x t α y = 0 . Proof.
Note that t α ∗ x t α y = 0 if and only if α x extends α y , or vice versa. If x, y ∈ ∆ n , then [19, Lemma 4.6 (2)] implies t α ∗ x t α y = 0. If x, y ∈ G , then [19,Lemma 4.6 (1)] implies t α ∗ x t α y = 0.Finally, WLOG, suppose x ∈ G and y ∈ ∆ n . In this case, [19, Lemma 4.6(1)] implies α y can’t extend α x . And so the only option left is for α y to extend α x . But, for r ′ ( y ) := { v ∈ G : | σ ( v ) | ≥ n, σ ( v ) | n = y } and x ∈ G ∩ X ( e n ) , [19, Lemma 4.6 (3)] implies t α ∗ x t α y = 0. Lemma 9.2.
The set { P v , S e n , S e ∗ n } forms an Exel-Laca G -family in L R ( E G ) = L R ( { t, q } ) .Proof. ( ExL2 ). Using the fact t α v = t α v t α ∗ v t α v (similarly, t α ∗ v = t α ∗ v t α v t α ∗ v ), adirect calculation shows P s ( e n ) S e n = S e n and S e ∗ n = S e ∗ n P s ( e n ) . E G -family, we can show S e n S e ∗ n S e n = X x ∈ X ( e n ) t α s ( en ) t ( e n ,x ) t α ∗ x ! X y ∈ X ( e n ) t α y t ( e n ,y ) ∗ t α ∗ s ( en ) ! X z ∈ X ( e n ) t α s ( en ) t ( e n ,z ) t α ∗ z ! = X x ∈ X ( e n ) t α s ( en ) t ( e n ,x ) t α ∗ x ! X y ∈ X ( e n ) X z ∈ X ( e n ) t α y t ( e n ,y ) ∗ t α ∗ s ( en ) t α s ( en ) t ( e n ,z ) t α ∗ z ! = X x ∈ X ( e n ) t α s ( en ) t ( e n ,x ) t α ∗ x ! X y ∈ X ( e n ) t α y t α ∗ y ! = X x ∈ X ( e n ) X y ∈ X ( e n ) t α s ( en ) t ( e n ,x ) t α ∗ x t α y t α ∗ y = X x ∈ X ( e n ) t α s ( en ) t ( e n ,x ) t α ∗ x t α x t α ∗ x = X x ∈ X ( e n ) t α s ( en ) t ( e n ,x ) t α ∗ x = S e n ; a similar argument shows S e ∗ n S e n S e ∗ n = S e ∗ n .( ExL3 ). First, suppose n = m . We have, t α ∗ s ( en ) t α s ( em ) = t α if α s ( e m ) = α s ( e n ) αt α ′∗ if α s ( e n ) = α s ( e m ) α ′ . If t α ∗ s ( en ) t α s ( em ) = 0, we are done. Otherwise, suppose t α ∗ s ( en ) t α s ( em ) = t α . Then, S e ∗ n S e m = X x ∈ X ( e n ) X y ∈ X ( e m ) t α x t ( e n ,x ) ∗ t α ∗ s ( en ) t α s ( em ) t ( e m ,y ) t α ∗ y = X x ∈ X ( e n ) X y ∈ X ( e m ) t α x t ( e n ,x ) ∗ t α t ( e m ,y ) t α ∗ y = X x ∈ X ( e n ) X y ∈ X ( e m ) t α x t ( e n ,x ) ∗ q s ( e n ) t α t ( e m ,y ) t α ∗ y . Since s ( e n ) ∈ G , [19, Lemma 4.6 (1)] implies | α | = 0 (i.e., t α = q s ( e n ) ). Thus, S e ∗ n S e m = X x ∈ X ( e n ) X y ∈ X ( e m ) t α x t ( e n ,x ) ∗ q s ( e n ) t ( e m ,y ) t α ∗ y . If s ( e n ) = s ( e m ), then q s ( e n ) q s ( e m ) = 0 = ⇒ S e ∗ n S e m = 0. If s ( e n ) = s ( e m ),then S e ∗ n S e m = X x ∈ X ( e n ) X y ∈ X ( e m ) t α x t ( e n ,x ) ∗ t ( e m ,y ) t α ∗ y ;by LP3 , we have t ( e n ,x ) ∗ t ( e m ,y ) = 0 and so S e ∗ n S e m = 0. We can use a similarargument to show this conclusion still holds for the choice t α ∗ s ( en ) t α s ( em ) = t α ′∗ .( ExL4 ). For E G = ( E , E , s E , r E ), note that, for each n , s − E ( s ( e n )) = { ( e n , x ) : x ∈ X ( e n ) } . X v = s ( e n ) S e n S e ∗ n = X v = s ( e n ) X x ∈ X ( e n ) X y ∈ X ( e n ) t α s ( en ) t ( e n ,x ) t α ∗ x t α y t ( e n ,y ) ∗ t α ∗ s ( en ) ! , by Claim 9.1, x = y = ⇒ α ∗ x α y = 0 , and so= X v = s ( e n ) X x ∈ X ( e n ) t α s ( en ) t ( e n ,x ) t ( e n ,x ) ∗ t α ∗ s ( en ) = t α s ( en ) X v = s ( e n ) X x ∈ X ( e n ) t ( e n ,x ) t ( e n ,x ) ∗ ! t α ∗ s ( en ) , by LP2 , for v = s ( e n ),= t α v ( q v ) t α ∗ v = t α v t α ∗ v = P v . ( EL1 & EL2 ). For v, w ∈ G , utilizing [19, Lemma 4.6 (1)], we have t α ∗ v t α w = ( q v if v = w v = w. So, for P v := t α v t α ∗ v and P w := t α w t α ∗ w , P v P w = ( P v if v = w v = w ;meaning, the set { P v } v ∈ G consists of pairwise orthogonal idempotent elements.In fact, elements of the form t α x t α ∗ x are idempotents which pairwise commute(see [31, Equation 2.1]); since a calculation we’ve previously done (see the proofof ExL2 ) shows S e ∗ n S e n = X x ∈ X ( e n ) t α x t α ∗ x , we can conclude { S e ∗ n S e n } e n ∈G is a set of pairwise commuting idempotents.( EL3 ). For n ∈ N and ω ∈ { , } n \ { n } , set r ′ ( ω ) := { v ∈ G : | σ ( v ) | ≥ n, σ ( v ) | n = ω } . Further, let G n denote X ( e n ) ∩ G . Note that for each ω ∈ { , } n \ { n } , wehave r ′ ( ω ) ∩ G n = ∅ since v ∈ G n = ⇒ σ ( ω ) < n . Given ω, ω ′ ∈ { , } n \ { n } such that ω = ω ′ , wehave r ′ ( ω ) ∩ r ′ ( ω ′ ) = ∅ due to the fact v ∈ r ′ ( ω ) = ⇒ σ ( ω ) | n = ω . Further, [19, Lemma 4.4] states r ( e n ) = G n ∪ [ { ω ∈{ , } n : ω n =1 } r ′ ( ω ) ! . v ∈ r ( e n ), v belongs exclusively to G n , or r ′ ( ω ) for aunique ω . Let X ∆ ,n := X ( e n ) ∩ ∆. Then, S e ∗ n S e n P v = X x ∈ X ( e n ) t α x t α ∗ x t α v t α ∗ v = X v ′ ∈ G n t α v ′ t α ∗ v ′ t α v t α ∗ v + X ω ∈ X ∆ ,n t α ω t α ∗ ω t α v t α ∗ v . Now, t α ∗ ω t α v = 0 unless α v extends α ω , or α ω extends α v . However, since v ∈ G , [19, Lemma 4.6 (1)] tells us α ω can’t extend α v . Moreover, [19, Lemma4.6 (3)] tells us there exists a path in F from ω ∈ ∆ to v ∈ G if and only if v ∈ r ′ ( ω ). And so if v / ∈ r ( e n ), then for ω ∈ X ∆ ,n , we have t α ω t α ∗ ω t α v t α ∗ v = 0.By [19, Lemma 4.6 (1)] and [31, Equation 2.1], for v / ∈ r ( e n ) and v ′ ∈ G n , t α v ′ t α ∗ v ′ t α v t α ∗ v = 0. Thus, v / ∈ r ( e n ) = ⇒ S e ∗ n S e n P v = 0 . On the other hand, suppose v ∈ r ( e n ). If v ∈ G n , it means v / ∈ r ′ ( ω ) for any ω ∈ X ∆ ,n . This gives us X ω ∈ X ∆ ,n t α ω t α ∗ ω t α v t α ∗ v = 0 and X v ′ ∈ G n t α v ′ t α ∗ v ′ t α v t α ∗ v = t α v t α ∗ v ;that is, S e ∗ n S e n P v = P v . If v / ∈ G n , then v ∈ r ′ ( ω ′ ) for a unique ω ′ ∈ X ∆ ,n ,which in turn means S e ∗ n S e n P v = t α ω ′ t α ∗ ω ′ t α v t α ∗ v . Because α v must extend α ω ′ , t α ω ′ t α ∗ ω ′ t α v t α ∗ v = t α v t α ∗ v ; again, it must be S e ∗ n S e n P v = P v . A similar argument can be used to show P v S e ∗ n S e n = P v . Thus, P v S e ∗ n S e n = S e ∗ n S e n P v = ( P v if v ∈ r ( e n ) , ExL1 & EL4 ). Finally, let
N, M ⊆ N be finite. Further, suppose N = ∅ , N ∩ M = ∅ , and r ( N, M ) := \ n ∈ N r ( e n ) \ [ m ∈ M r ( e m )is finite. We want to show Y n ∈ N S e ∗ n S e n Y m ∈ M (1 − S e ∗ m S e m ) = X v ∈ r ( N,M ) P v . We will use induction on | N ∪ M | = k . To that end, for k = 1, we have r ( N, M ) = r ( e n ) and Y n ∈ N S e ∗ n S e n Y m ∈ M (1 − S e ∗ m S e m ) = S e ∗ n S e n . Note, by the definitions of ∆ and X ( e n ), | r ( e n ) | < ∞ = ⇒ X ∆ ,n = ∅ = ⇒ r ( e n ) = G n . Thus, S e ∗ n S e n = X x ∈ X ( e n ) t α x t α ∗ x = X v ∈ G n t α v t α ∗ v + X ω ∈ X ∆ ,n t α ω t α ∗ ω = X v ∈ G n t α v t α ∗ v = X v ∈ r ( e n ) P v , | N ∪ M | ≤ k . Let, N ′ , M ′ , besuch that | N ′ ∪ M ′ | = k + 1. There are two cases to consider. Case 1 : N ′ = { n ′ } ∪ N . Note | r ( N, M ) | < ∞ = ⇒ | r ( N, M ′ ) | < ∞ or | r ( e n ′ ) | < ∞ . If | r ( N, M ′ ) | < ∞ , then by our inductive hypothesis, wehave Y n ∈ N S e ∗ n S e n Y m ∈ M ′ (1 − S e ∗ m S e m ) = X v ∈ r ( N,M ′ ) P v . Using
EL3 , we can work out Y n ∈ N ′ S e ∗ n S e n Y m ∈ M ′ (1 − S e ∗ m S e m ) = S e ∗ n ′ S e n ′ Y n ∈ N S e ∗ n S e n Y m ∈ M ′ (1 − S e ∗ m S e m )= S e ∗ n S e n X v ∈ r ( N,M ′ ) P v = X v ∈ r ( N,M ′ ) ∩ r ( e n ′ ) P v = X v ∈ r ( N ′ ,M ′ ) P v . On the other hand, if | r ( e n ′ ) | < ∞ , then Y n ∈ N ′ S e ∗ n S e n Y m ∈ M ′ (1 − S e ∗ m S e m ) = S e ∗ n ′ S e n ′ Y n ∈ N S e ∗ n S e n Y m ∈ M ′ (1 − S e ∗ m S e m )= X v ∈ r ( e n ′ ) P v ! Y n ∈ N S e ∗ n S e n Y m ∈ M ′ (1 − S e ∗ m S e m )= X v ∈ r ( e n ′ ) ∩ (cid:0) T n ∈ N r ( e n ) (cid:1) P v ! Y m ∈ M ′ (1 − S e ∗ m S e m )= X v ∈ r ( e n ′ ) ∩ (cid:0) T n ∈ N r ( e n ) (cid:1) P v − X v ∈ r ( e n ′ ) ∩ (cid:0) T n ∈ N r ( e n ) ∩ T m ∈ M ′ r ( e m ) (cid:1) P v = X v ∈ r ( N ′ ,M ′ ) P v . Case 2 : M ′ = { m ′ } ∪ M . If | r ( N ′ , M ) | < ∞ , then by our inductive hypoth-55sis, Y n ∈ N ′ S e ∗ n S e n Y m ∈ M ′ (1 − S e ∗ m S e m ) = Y n ∈ N ′ S e ∗ n S e n Y m ∈ M (1 − S e ∗ m S e m ) ! (1 − S e ∗ m ′ S e m ′ )= X v ∈ r ( N ′ ,M ) P v ! (1 − S e ∗ m ′ S e m ′ )= X v ∈ r ( N ′ ,M ) P v − X v ∈ r ( N ′ ,M ) ∩ r ( e m ′ ) P v = X v ∈ r ( N ′ ,M ) \ r ( e m ′ ) P v = X v ∈ r ( N ′ ,M ′ ) P v . On the other hand, if | r ( N ′ , M ) | = ∞ , then the fact | r ( N ′ , M ′ ) | < ∞ means | r ( N ′ , { m ′ } ) | < ∞ . Applying our inductive hypothesis, we have Y n ∈ N ′ S e ∗ n S e n (1 − S e ∗ m ′ S e m ′ ) = X v ∈ r ( N ′ , { m ′ } ) P v . And so, by also exploiting
EL2 and
EL3 , Y n ∈ N ′ S e ∗ n S e n Y m ∈ M ′ (1 − S e ∗ m S e m ) = Y n ∈ N ′ S e ∗ n S e n Y m ∈ M (1 − S e ∗ m S e m ) ! (1 − S e ∗ m ′ S e m ′ )= Y n ∈ N ′ S e ∗ n S e n (1 − S e ∗ m ′ S e m ′ ) ! Y m ∈ M (1 − S e ∗ m S e m )= X v ∈ r ( N ′ , { m ′ } ) P v ! Y m ∈ M (1 − S e ∗ m S e m ) , note: v ∈ r ( e m ) = ⇒ P v (1 − S e ∗ m S e m ) = 0 = ⇒ P v Y m ∈ M (1 − S e ∗ m S e m ) = 0 , alternatively, v ∈ \ m ∈ M r ( e m ) c = ⇒ P v Y m ∈ M (1 − S e ∗ m S e m ) = P v , and so,= X v ∈ r ( N ′ , { m ′ } ) P v − X v ∈ r ( N ′ , { m ′ } ) ∩ (cid:0) T m ∈ M r ( e m ) c (cid:1) P v = X v ∈ r ( N ′ ,M ′ ) P v . Thus, our lemma is proved.The following lemma closely follows [19, Lemma 5.11].
Lemma 9.3.
Let G be an ultragraph with no singular vertices. For each ω ∈ ∆ k ⊆ E G , define N, M ⊆ { , , . . . , k } uch that for each n ∈ N , ω n = 1 , and for each m ∈ M , ω m = 0 . Then, Y n ∈ N S e ∗ n S e n Y m ∈ M (1 − S e ∗ m S e m ) = t α ω t α ∗ ω + X v ∈ r ( N,M ) | σ ( v ) | 1. Well, v ∈ r ( N, M ) = ⇒ v ∈ r ( ω ′ ) . By[19, Lemma 3.7], we also have v ∈ r ( σ ( v )). For | ω ′ | = | σ ( v ) | = k − 1, we have σ ( v ) = ω ′ . To see this, suppose σ ( v ) = ω ′ . Since | ω ′ | = | σ ( v ) | = k − σ ( v ) = ω ′ implies there exists j ∈ { , , . . . , k − } such that ω ′ j = σ ( v ) j . WLOG, suppose ω ′ j = 1 and σ ( v ) j = 0; this means r ( ω ′ ) ⊆ r ( e j ) and r ( σ ( v )) ∩ r ( e j ) = ∅ (i.e., r ( ω ′ ) ∩ r ( σ ( v )) = ∅ ). ⇒⇐ Thus, v ∈ r ( N, M ) and | σ ( v ) | = k − ⇒ v ∈ { v ∈ G k : | σ ( v ) | ≥ k − , σ ( v ) | k − = ω ′ } , which in turn means X { v ∈ G k : | σ ( v ) |≥ k − ,σ ( v ) | k − = ω ′ } P v = X v ∈ r ( N,M ) | σ ( v ) | = k − P v . Putting it all together, we have Y n ∈ N S e ∗ n S e n Y m ∈ M (1 − S e ∗ m S e m ) = t α ω t α ∗ ω + X v ∈ r ( N,M ) | σ ( v ) | 1) and N ′′ := N ∪ { k } . Further, let us suppose ω ′′ ∈ ∆ k .By case 1, we have S e ∗ k S e k Y n ∈ N S e ∗ n S e n Y m ∈ M ′ (1 − S e ∗ m S e m ) = Y n ∈ N ′′ S e ∗ n S e n Y m ∈ M ′ (1 − S e ∗ m S e m )= t α ω ′′ t α ∗ ω ′′ + X v ∈ r ( N ′′ ,M ′ ) | σ ( v ) | LP4 , we can deduce q ω ′ = t ω t ω ∗ + t ω ′′ t ω ′′∗ + X { v ∈ G : σ ( v )= ω ′ } P v . We can use the same argument as in case 1) to show { v ∈ G : σ ( v ) = ω ′ } = { v ∈ r ( ω ′ ) : | σ ( v ) | = k − } . And so, for r ( ω ′ ) = r ( N, M ′ ), q ω ′ = t ω t ω ∗ + t ω ′′ t ω ′′∗ + X v ∈ r ( N,M ′ ) | σ ( v ) | = k − P v . Applying [19, Lemma 4.6], we have t α ω ′ t α ∗ ω ′ = t α ω ′ q ω ′ t α ∗ ω ′ = t α ω ′ t ω t ω ∗ + t ω ′′ t ω ′′∗ + X v ∈ r ( N,M ′ ) | σ ( v ) | = k − P v ! t α ∗ ω ′ = t α ω ′ ω t ( α ω ′ ω ) ∗ + t α ω ′ ω ′′ t ( α ω ′ ω ′′ ) ∗ + X v ∈ r ( N,M ′ ) | σ ( v ) | = k − P v which, by the uniqueness of α x , = t α ω t α ∗ ω + t α ω ′′ t α ∗ ω ′′ + X v ∈ r ( N,M ′ ) | σ ( v ) | = k − P v . 59n particular, this means Y n ∈ N S e ∗ n S e n Y m ∈ M (1 − S e ∗ m S e m ) = t α ω t α ∗ ω + X v ∈ r ( N,M ′ ) | σ ( v ) | = k − P v ! + X v ∈ r ( N,M ′ ) | σ ( v ) | 60n the unitization of L R ( E G ) (note, M ′ = { , , . . . , k − } \ N ′ .); and so S e ∗ k S e k = S e ∗ k S e k X N ′ ⊆{ , ,...,k − } Y n ∈ N ′ S e ∗ n S e n Y m ∈ M ′ (1 − S e ∗ m S e m ) !! = S e ∗ k S e k X ∅6 = N ′ ⊆{ , ,...,k − } Y n ∈ N ′ S e ∗ n S e n Y m ∈ M ′ (1 − S e ∗ m S e m ) ! + Y m ∈{ , ,...,k − } (1 − S e ∗ m S e m ) ! . Taking N = { k } , this means Y n ∈ N S e ∗ n S e n Y m ∈ M (1 − S e ∗ m S e m ) = S e ∗ k S e k − X ∅6 = N ′ ⊆{ , ,...,k − } S e ∗ k S e k Y n ∈ N ′ S e ∗ n S e n Y m ∈ M ′ (1 − S e ∗ m S e m )= S e ∗ k S e k − X N ′′ = N ′ ∪{ k } : ∅6 = N ′ ⊆{ , ,...,k − } Y n ∈ N ′′ S e ∗ n S e n Y m ∈ M ′ (1 − S e ∗ m S e m ) , where M ′ = { , , . . . , k } \ N ′′ . Observe that S e ∗ k S e k = X v ∈ G k t α v t α ∗ v + X ω ′′ ∈ X ∆ ,k t α ω ′′ t α ∗ ω ′′ = X v ∈ r ( e k ): | σ ( v ) | Let γ be a path in E G such that s E ( γ ) ∈ W ⊔ Γ and, γ = g ( e n , x ) g ( e n , x ) g . . . ( e n k , x k ) g k , as in [19, Lemma 4.1]. Then, t γ = S e n S e n . . . S e nk t α rE ( γ ) . It’s worth noting the g i ’s can be elements in E G .Proof. We will use induction on k . To that end, suppose k = 0; which is to say, γ = g . For s E ( g ) ∈ W ⊔ Γ , and g a path in F , the uniqueness of α r E ( g ) implies g = α r E ( g ) . Thus, t g = t α rE ( g , completing our base case.Now, for γ = g ( e n , x ) g ( e n , x ) g . . . ( e n k , x i ) g k = γ ′ ( e n k , x k ) g k , we have by our inductive hypothesis, t γ = S e n S e n . . . S e nk − t α rE ( γ ′ ) t ( e nk ,x k ) t g k . For r E ( γ ) = r E ( g k ), the uniqueness of α r E ( γ ) implies α r E ( γ ) = α x k g k , which inturn implies t α rE ( γ ) = t α xk t g k . And so S e nk t α rE ( γ ) = S e nk t α xk t g k = X x ∈ X ( e nk ) t α s ( enk ) t ( e nk ,x ) t α ∗ x t α xk t g k , which, by Claim 9.1,= t α s ( enk ) t ( e nk ,x k ) t g k , and since r E ( γ ′ ) = s ( e n k ),= t α rE ( γ ′ ) t ( e nk ,x k ) t g k . Hence, t γ = S e n S e n . . . S e nk t α rE ( γ ) . 63e are now equipped to prove one of our main results. Theorem 9.5. Let G be an ultragraph with no singular vertices, then L R ( G ) ∼ = QL R ( E G ) Q. Proof. By Theorem 8.8, it suffices to show EL R ( G ) ∼ = QL R ( G ) Q . To that end,we apply Lemma 9.2 to get an Exel-Laca G -family { P v , S e , S e ∗ } in L R ( E G ). Weknow from Lemma 3.5 and Proposition 3.14.1. L R ( E G ) = span R { t α t β ∗ : r E ( α ) = r E ( β ) } , and has a Z -graded structure with L R ( E G ) i = span R { t α t β ∗ : r E ( α ) = r E ( β ) , | α | − | β | = i } . Given Q ∈ M ( L R ( E G )), it is straightforward to see QL R ( G ) Q = span R { t α t β ∗ : r E ( α ) = r E ( β ) } with s E ( α ) , s E ( β ) ∈ W ⊔ Γ } . Further, by the definition of { P v , S e , S e ∗ } , we can see { P v , S e , S e ∗ } ⊆ QL R ( G ) Q .And so, by Lemma 8.3 and Theorem 8.4, we have an injective map φ : EL R ( G ) → ( L R ( E G )) such that im φ ⊆ QL R ( G ) Q . Finally, by [19, Lemma 4.1], every path α in E G can be expressed uniquely as α = g ( e n , x ) g ( e n , x ) g . . . ( e n k , x k ) g k , where each e n i is in G , and each g i is a path in F (possibly an element of E G ).Then, by Lemmas 9.3 and 9.4, we have QL R ( G ) Q ⊆ im φ ; and so φ is surjectiveas well. Thus, EL R ( G ) ∼ = QL R ( E G ) Q. Theorem 9.6. Let G be an ultragraph with no singular vertices. Then, L R ( G ) is Morita equivalent to L R ( E G ) . Proof. This follows directly from Theorem 9.5 and Corollary 6.8. 10 Desingularization and Morita Equivalence The goal of this section is mostly to extend Theorem 9.6 to ultragraphs withsingular vertices. That is, even in the case where G may contain singular ver-tices, L R ( G ) is Morita equivalent to the Leavitt path algebra of a graph. Thetechniques and ideas used here are from [4], but modified to work for ultra-graphs. We begin with the notion of desingularization , which was introducedfor ultragraphs by Tomforde in [30]. To that end, let G be an ultragraph con-taining singular vertices. If v ∈ G is a sink, we add a tail at v by introducingedges { f i } ∞ i =1 , and vertices { v i } ∞ i =1 , such that s ( f i ) = v i − and r ( f i ) = v i .64 v v v i − f f . . . f i . . .Figure 7 v v v v i − r ( g ) = r ( e ) r ( g ) = r ( e ) r ( g i ) = r ( e i ) f f . . . f i . . .. . . g g g i Figure 8If v ∈ G is an infinite emitter, let { e i } ∞ i =1 be an enumeration of s − ( v ).We add a tail starting at v by adding edges { f i , g i } ∞ i =1 , and vertices { v i } ∞ i =1 ,such that s ( g i ) = s ( f i ) = v i − , r ( f i ) = v i , and r ( g i ) = r ( e i ).The ultragraph, F , obtained by adding the appropriate tail to each singularvertex of G is called the desingularization of G . Our task now is to show L R ( G )is Morita equivalent to L R ( F ). In the case of a graph E , it is straightforwardto show L R ( E ) ∼ = M v ∈ E L R ( E ) q v as left L R ( E )-modules. In order for us to make use of the techniques availableto us to prove our result, we need a similar fact to hold for ultragraphs. Onecan quickly check L R ( G ) ≇ M v ∈ G L R ( G ) p v , and L R ( G ) ≇ M A ∈G L R ( G ) p A , as left L R ( G )-modules. And so we need something a little different. Lemma 10.1. Let G be an ultragraph. Then,1) G is countable,2) there exists a collection of sets { A i } i ∈ N ⊆ G such that A i ∩ A j = ∅ when i = j ,and each A ∈ G is contained in the union of finitely many A i ’s.Proof. 1) Let { A ′ i } i ∈ N be an enumeration of the set { v ∈ G } ∪ { r ( e ) : e ∈ G } .The smallest lattice L of P ( G ) generated by { A ′ i } i ∈ N is given by L = (cid:26) \ i ∈ X A ′ i ∪ \ i ∈ X A ′ i ∪ · · · ∪ \ i ∈ X n A ′ i : X , . . . , X n are finite subsets of N (cid:27) [30, Lemma 2.12].For a given set X , let P f ( X ) denote the subset of P ( X ) consisting of finitesubsets of X . It is an established fact that if X is countable, P f ( X ) is countable65s well. This in turn means P f ( P f ( X )) is countable, and so on. Since there isa surjective map P f ( P f ( N )) → L given by { X , X , . . . , X n } 7→ \ i ∈ X A ′ i ∪ \ i ∈ X A ′ i ∪ · · · ∪ \ i ∈ X n A ′ i , we can conclude L is countable; as such, let A , A , . . . be an enumeration of L . Now, consider the set (cid:26) n [ k =1 ( A i (cid:15) A j ) k : i, j, n ∈ N (cid:27) . Using elementary properties of finite unions and intersections, relative comple-ments, and the fact L is a lattice, we can show the set given above is an algebrawithin P ( G ); in fact, it is a subset of G . Since, by definition, G is the smallestalgebra containing { v ∈ G } ∪ { r ( e ) : e ∈ G } , it must be that G = (cid:26) n [ k =1 ( A i (cid:15) A j ) k : i, j, n ∈ N (cid:27) . Since there is a surjective map P f ( N × N ) → (cid:8) S nk =1 ( A i (cid:15) A j ) k : i, j, n ∈ N (cid:9) given by { ( i, j ) , ( i, j ) , . . . , ( i, j ) n } 7→ n [ k =1 ( A i (cid:15) A j ) k , G is countable.2) Now that we have established G is countable, let B , B , . . . be an enu-meration of G . Let A = B A = B \ B ... A k = B k \ (cid:16) k − S i =1 B i (cid:17) ...For i = j , we have A i ∩ A j = ∅ ; moreover, for each k , B k ⊆ (cid:16) k S i =1 A i (cid:17) = (cid:16) k S i =1 B i (cid:17) .What we now want to show is L R ( G ) ∼ = L A ∈G L R ( G ) p A as left L R ( G )-modules.We will do this by first showing { A i } i ∈ N can be used to construct a σ -unit for L R ( G ). 66 emma 10.2. Let { A i } i ∈ N be as in Lemma 10.1. For k ∈ N , let t k := P i ≤ k p A i .Then, { t k } ∞ k =1 is a σ -unit for L R ( G ) . Hence, L R ( G ) is a σ -unital ring.Proof. We will start by showing t k is an idempotent for each k . To that end,by uLP1 and the fact A i ∩ A j = ∅ for i = j , we have t k = (cid:16) X i ≤ k p A i (cid:17)(cid:16) X j ≤ k p A j (cid:17) = X i ≤ k p A i = X i ≤ k p A i = t k . Now, suppose l, k ∈ N with l ≤ k . Then, t k t l = (cid:16) X i ≤ k p A i (cid:17)(cid:16) X j ≤ l p A j (cid:17) = X j ≤ l p A j = X j ≤ l p A j = t l . A similar argument shows t l t k = t l ; and so it is certainly true t k t k +1 = t k +1 t k = t k for all k . Finally, to show { t k } ∞ k =1 is a σ -unit, we need to show L R ( G ) = ∞ [ k =1 t k L R ( G ) t k . So, let x = n P j =1 r j s α j p A j s β ∗ j ∈ L R ( G ), and let V = { B ∈ G : B = s ( α j ) , or B = s ( β j ) , for 1 ≤ j ≤ n } . Take care to note B ∈ V can be a set consisting of multiple vertices. Since V is a finite set, A = S B ∈ V B is an element of G as well. By Lemma 10.1, A iscontained in the union of finitely many A i ’s. This in turn means there existsa k such that A ⊆ S i ≤ k A i . Since the A i ’s are pairwise disjoint, uLP1 gives us t k = P i ≤ k p A i = p S i ≤ k A i ; then, uLP1 and uLP2 imply t k x = (cid:16) p S i ≤ k A i (cid:17)(cid:16) n X j =1 r j s α j p A j s β ∗ j (cid:17) = n X j =1 r j p S i ≤ k A i s α j p A j s β ∗ j = n X j =1 r j s α j p A j s β ∗ j = x. Similarly, xt k = n X j =1 r j s α j p A j s β ∗ j p S i ≤ k A i = n X j =1 r j s α j p A j s β ∗ j = x. This gives us x = t k xt k . Thus, L R ( G ) = ∞ S k =1 t k L R ( G ) t k ; meaning { t k } ∞ k =1 is a σ -unit of L R ( G ).We have the following useful corollary.67 orollary 10.3. Let { A i } i ∈ N be as in Lemma 10.1. Then, L R ( G ) ∼ = L i ∈ N L R ( G ) p A i as left L R ( G ) modules.Proof. Since the A i ’s are pairwise disjoint, we have L R ( G ) p A i ∩ L R ( G ) p A j = { } for i = j . By Lemma 10.2, for any x ∈ L R ( G ), there exists a k such that x = P i ≤ k xp A i . Thus, L R ( G ) ∼ = M i ∈ N L R ( G ) p A i as a left L R ( G )-module.To make our intention explicit, the goal of this section is to prove L R ( G )is Morita equivalent to L R ( F ). Which, by Theorem 9.6, would mean L R ( G )is Morita equivalent to L R ( E F ), thereby establishing our desired result. Wewill prove the Morita equivalence for L R ( G ) and L R ( F ) using the only toolavailable to us: Theorem 5.26; a key component we’ll need to use this theoremis the existence of an injective R -algebra homomorphism φ : L R ( G ) → L R ( F )which commutes nicely with inclusion maps within the setting of direct limits—that is what we will do in Proposition 10.4.1. In order to establish the existenceof such a map φ , we will need to first show that for any 0 = x ∈ L R ( G ), thereexist a, b ∈ L R ( G ) such that:1) axb = rp v , with r ∈ R \ { } , or2) axb = n P i =1 r i s iα , for some cycle α ∈ G ∗ .Note, case 2) is stating axb is a polynomial in s α over R for some cycle α ; wewill prove this assertion in Lemma 10.4. Finally, in order to prove Lemma 10.4,we will need to first show, given any 0 = x ∈ L R ( G ), we can find v ∈ G suchthat xp v = 0—this bit is straightforward in the case of graphs, not so much inthe case of ultragraphs. Proposition 10.3.1. Let x ∈ L R ( G ) such that x = 0 . Then, there exists v ∈ G such that xp v = 0 .Proof. Suppose x = 0. We will break up our proof into three cases. Beforewe start, recall x can be expressed as x = n P i =1 r i s α i p A i s β ∗ i (see Theorem 3.19).For our first case, we will consider the possibility where | s ( β i ) | < ∞ for each i . Again, we will reiterate the fact s ( β i ) need not be a vertex when | β i | = 0( | β i | = 0 = ⇒ β i = A ∈ G , with s β ∗ i = p A , in which case s ( β i ) = A ). Case 1: Let x = n P i =1 r i s α i p A i s β ∗ i with s ( β i ) finite for each i . Set V = { v ∈ G : v ∈ s ( β i ) for 1 ≤ i ≤ n } . s ( β i ) is finite, V must be finite as well. Suppose xp v = 0 for every v ∈ G ; because xp V = P v ∈ V xp v , it must be xp V = 0. But, xp V = n X i =1 r i s α i p A i s β ∗ i p V = n X i =1 r i s α i p A i s β ∗ i = x, which means x = 0. ⇒⇐ Thus, there exists v ∈ G such that xp v = 0.Now, we must consider the possibility that not every s ( β i ) is finite. Saidmore specifically, we must consider the possibility that, for some i , β i = A ∈ G ,a countably infinite subset of G . We will prove it in the next case; to makethings easier we will assume | β i | = 0 for all 1 ≤ i ≤ n . This means, by exploiting uLP1 , we can express x as x = n P i =1 r i s α i p A i . Case 2: Let x = n P i =1 r i s α i p A i with A i possibly infinite. Pick a vertex v ∈ r ( α ) ∩ A . If xp v = 0, we have, by applying uLP1 and uLP2 , xp v = n X i =1 r i s α i p A i ∩ v = X { i : v ∈ r ( α i ) ∩ A i } r i s α i p v = 0 . The sum on the right-hand side has at least one term since v ∈ r ( α ) ∩ A .Rewrite xp v as P k P | α i | = k r i s α i p v . By the Z -grading on L R ( G ) (see Theorem3.19), X k X | α i | = k r i s α i p v = 0 ⇐⇒ X | α i | = k r i s α i p v = 0 for each k. And so, for what follows, we may as well assume all the α i ’s have the samelength in the sum X { i : v ∈ r ( α i ) ∩ A i } r i s α i p v . Then, by uLP3 , we have0 = s α ∗ xp v = X { i : v ∈ r ( α i ) } r i s α ∗ s α i p v = r p v ;but we know r p v = 0 (recall that rp A = 0 for A ∈ G and r ∈ R \ { } , seeparagraph following Definition 3.18). ⇒⇐ Thus, xp v = 0.For our final, and most general, case, we will consider the possibility thatnot all s ( β i )’s are finite, and not all β i ’s have length 0. By grouping termsaccordingly, we can express x as x = n X i =1 r i s α i p A i + m X j =1 r ′ j s α ′ j p A ′ j s β ′∗ j A i ’s are infinite, and the s ( β ′ j )’s are finite. Moreover, by uLP1 and uLP2 , we have n X i =1 r i s α i p A i = n X i =1 r i s α i p r ( α i ) p A i = n X i =1 r i s α i p r ( α i ) ∩ A i . Now, if r ( α i ) ∩ A i is finite for each i , then by replacing each A i by r ( α i ) ∩ A i in our expression of x , and by our assumption each s ( β ′ j ) is finite, we reduce tocase 1. Thus, for our final case, we may as well assume r ( α i ) ∩ A i is infinite foreach i . Case 3: Let x = n P i =1 r i s α i p A i + m P j =1 r ′ j s α ′ j p A ′ j s β ′∗ j with r ( α i ) ∩ A i infinite foreach i and s ( β ′ j ) finite for each j . Let V = { v ∈ G : v ∈ s ( β ′ j ) for 1 ≤ j ≤ m } .V is a finite set, and since A is countably infinite, this means A \ V = ∅ .Further, since s ( β ′ j ) ∩ ( A \ V ) = ∅ for each j , we have xp A \ V = n X i =1 r i s α i p A i ∩ ( A \ V ) = X { i : r ( α i ) ∩ A i ∩ ( A \ V ) = ∅} r i s α i p A i ∩ ( A \ V ) . Since, r ( α ) ∩ A ∩ ( A \ V ) = ∅ ( V is finite, r ( α ) ∩ A is not), the sum on theright has at least one term. Now, suppose X { i : r ( α i ) ∩ A i ∩ ( A \ V ) = ∅} r i s α i p A i ∩ ( A \ V ) = 0 . By the Z -grading on L R ( G ), we may assume the α i ’s have the same length foreach i in { i : r ( α i ) ∩ A i ∩ ( A \ V ) = ∅} . This means0 = s α ∗ xp A \ V = r p r ( α ) ∩ A ∩ ( A \ V ) ⇒⇐ . So, xp A \ V = X { i : r ( α i ) ∩ A i ∩ ( A \ V ) = ∅} r i s α i p A i ∩ ( A \ V ) = 0 , reducing us to case 2; for which we know there exists a v such that xp v = 0.Since every x ∈ L R ( G ) fits into one of the three cases, we have our desiredresult.With our previous result in hand, we are now ready to set the next step-ping stone. We invite the reader to compare the following lemma with [24,Proposition 3.1]. Lemma 10.4. Let x ∈ L R ( G ) such that x = 0 . Then, there exist a, b ∈ L R ( G ) such that axb = rp v , for some r ∈ R \ { } , or axb = n P i =0 r i s iα for some cycle α ∈ G ∗ . roof. Let x ∈ L R ( G ) such that x = 0. Our first goal is to show that there exists µ ∈ G ∗ such that xs µ can be expressed entirely in real edges and is nonzero; putmore succinctly, xs µ can be expressed as xs µ = n X i =1 r i s α i p A i = 0 . To that end, let v ∈ G such that xp v = 0; by Proposition 10.3.1, such a v mustexist. By grouping terms based on the presence of ghost edges, we can write xp v as xp v = n X i =1 r i s α i p A i s β ∗ i + m X j =1 r ′ j s ζ j p B j with | β i | ≥ i ; further, we may also assume the degree of n P i =1 r i s α i p A i s β ∗ i in ghost edges is minimal. That is, given any other expression of xp v , xp v = X k r ′′ k s α ′′ k p A k s β ′′∗ k , there exists some k such that | β ′′ k | ≥ max {| β i |} . Note that xp v = xp v = n X i =1 r i s α i p A i s β ∗ i p v + m X j =1 r ′ j s ζ j p B j p v . Since, by Lemma 3.17, r i s α i p A i s β ∗ i p v = (cid:26) r i s α i p A i s β ∗ i if v = s ( β i ) , , and r ′ j s ζ j p B j p v = (cid:26) r ′ j s ζ j p v if v ∈ r ( ζ j ) ∩ B j , , we may express xp v as xp v = n X i =1 r i s α i p A i s β ∗ i + m X j =1 r ′ j s ζ j p v where s ( β i ) = v and v ∈ r ( s ζ j ) for each i, j ; because multiplying an expres-sion by p v can’t increase the degree in ghost edges, the expression xp v = n P i =1 r i s α i p A i s β ∗ i + m P j =1 r ′ j s ζ j p v is in minimal ghost edge degree as well. Letting e i denote the first edge of β i , and β ′ i the path such that β i = e i β ′ i (it’s possible β i = e i , in which case we can take β ′ i = p r ( e i ) ), set Y i = r i s α i p A i s β ′∗ i and Y = m P j =1 r ′ j s ζ j p v , and so xp v = n X i =1 Y i s e i ∗ + Y. xp v s e i = 0 for each i , we have Y i = − Y s e i . In which case, xp v = n X i =1 − Y s e i s e i ∗ + Y = Y (cid:18) n X i =1 − s e i s e i ∗ + p v (cid:19) . Since xp v = 0, n P i =1 − s e i s e i ∗ + p v = 0. By uLP4 , there exists f ∈ G such that s ( f ) = v and f = e i for any i . It then follows xp v s f = xs f = n X i =1 − Y s e i s e i ∗ s f + Y s f = Y s f . Suppose Y s f = m X j =1 r ′ j s ζ j p v s f = m X j =1 r ′ j s ζ j s f = m X j =1 r ′ j s ζ j f = 0 . By the Z -grading on L R ( G ) (see case 2 in proof of Proposition 10.3.1), we mayassume | ζ j f | is the same for each j . We then have r ′ j p r ( f ) = r ′ j s ( ζ j f ) ∗ s ζ j f = s ( ζ j f ) ∗ (cid:18) m X j =1 r ′ j s ζ j f (cid:19) = 0for each fixed j , which is only possible if r ′ j = 0. This would then mean r ′ j = 0for each j , implying Y = 0, which in turn would mean xp v = 0 . ⇒⇐ Thus, Y s f = 0, and setting µ = s f , we have our desired result in the case where xp v s e i = 0 for all i . On the other hand, there is the possibility xp v s e k = 0 forsome 1 ≤ k ≤ n . Which is to say, xp v s e k = xs e k = n X i =1 Y i s e i ∗ s e k + Y s e k = Y k + Y s e k = 0 . Notice that xs e k has a strictly less degree in ghost edges than xp v . We canapply the same process to xs e k as we did to x ; getting either an edge f suchthat xs e k s f = 0 and can be expressed using only real edges, or an edge e k ′′ suchthat xs e k s e k ′′ = 0 and has a degree in ghost edges which is strictly less thanthat of xs e k . Repeating this process a finite number of times, we will arrive ata µ ∈ G ∗ such that xs µ can be expressed using only real edges.With that, we will show there exist a, b ∈ L R ( G ) such that axb = rp v forsome v ∈ G and r = 0, or axb = n P i =0 r i s iα for some cycle α ∈ G ∗ . To that end,let x ∈ L R ( G ). If x can be expressed in only real edges we are fine. If not, as wehave shown above, x = 0, there exists µ ∈ G ∗ such that xs µ can be expressedusing only real edges. Replacing x by xs µ if necessary, we may assume x canbe expressed in solely in real edges. Which is to say, x can be expressed as x = n X i =1 r i s α i p A i . v ∈ G such that xp v = 0. Replacing x by xp v if necessary, we may also assume x = n X i =1 r i s α i p v for some v ∈ G . We will prove our assertion by induction on n . For n = 1, x = r s α p v . In which case, taking a = s α ∗ , we have, by uLP3 , ax = r s α ∗ s α p v = r p r ( α ) p v = r p v ;setting b = p v , we have concluded our base case.Now, assuming our hypothesis holds for all numbers less than n , let x = n X i =1 r i s α i p v with the summands arranged such that | α j | ≤ | α k | for all j, k with 1 ≤ j ≤ k ≤ n . Muliplying x on the left by s α ∗ , we have s α ∗ x = n X i =1 r i s α ∗ s α i p v = r p v + n X i =2 r i s α ∗ s α i p v , which is nonzero since. To see why, note that if | α j | = | α k | , for all j, k , uLP3 implies s α ∗ x = r p v = 0. Otherwise, by uLP3 , there is some 2 ≤ k ≤ n suchthat s α ∗ x = r p v + n X i = k r i s α ∗ s α i p v with | α ∗ α i | ≥ k ≤ i ≤ n ; meaning, given k ≤ i , s α ∗ s α i = 0, or s α ∗ s α i ∈ L R ( G ) m for some m > 0. And so, by the Z -grading on L R ( G ), we againhave s α ∗ x = 0 (see case 2 in proof of Proposition 10.3.1). Now, if s α ∗ s α k = 0for some 2 ≤ k ≤ n , we reduce to the case of less than n summands, and by ourinductive hypothesis, we can find a, b such that axb is in one of the two desiredforms. The only other possibility is s α ∗ s α i = 0 for all 2 ≤ i ≤ n . In which case,there exists a path β i for each 2 ≤ i ≤ n , with | β i | ≥ 1, such that α i = α β i .We then have s α ∗ x = n X i =1 r i s α ∗ s α i p v = r p v + n X i =2 r i s β i p v ;multiplying s α ∗ by p v on the left, we have p v s α ∗ x = r p v + n X i =2 r i p v s β i p v = r p v + n X i =2 r i p v s β i p v = 0 . p v s β k p v = 0 for some 2 ≤ k ≤ n , we again reduce to less than n summandsand we are done. Otherwise, we are left with the case s α ∗ x = r p v + n X i =2 r i s β i p v where s ( β i ) = v for each i ; which is to say, β i is a cycle for each i . Up to thispoint, having started with an arbitrary x ∈ L R ( G ), we have have whittled ourway down to showing our assertion holds for x , or s α ∗ x = r p v + n X i =2 r i s β i p v with each β i a cycle. Now, for s α ∗ x = r p v + n P i =2 r i s β i p v , suppose there is a ν ∈ G ∗ such that s ν ∗ s β k = 0 for some 2 ≤ k ≤ n but not all such k . This means s ( ν ) = v (otherwise s ν ∗ s β i = 0 for all i ). In turn, this implies s ν ∗ r p v s ν = r p r ( ν ) = 0 . Note, s ν ∗ s α ∗ xs ν = r p r ( ν ) + n X i =2 r i s ν ∗ s β i s ν . And, by our assumption, there is at least one k such that s ν ∗ s β k s ν = 0, forwhich it must be ν ∗ β k ν ∈ G ∗ . Since | ν ∗ β k ν | = | β k | ≥ 1, by the Z -grading of L R ( G ), we have s ν ∗ s α ∗ xs ν = r p r ( ν ) + n X i =2 r i s ν ∗ s β i s ν = 0 . Because we are assuming s ν ∗ s β k = 0 for some, but not all, 2 ≤ k ≤ n (and sosome, but not all, the summands in the above expression are 0), our inductivehypothesis tells us there exist a ′ , b ′ such that a ′ s ν ∗ s α ∗ xs ν b ′ satisfies one of thespecified forms. Setting a = a ′ s ν ∗ s α ∗ and b = s ν b ′ , we prove our assertion for x in the case s ν ∗ s β k = 0 for some, but not all, 2 ≤ k ≤ n . On the other hand,if ν ∈ G ∗ is such that s ν ∗ s β k = 0 for some k , then it must be s ν ∗ s β i = 0 forall i . In particular, this means s β ∗ j s β k = 0 for all j, k (since s β ∗ j s β j = 0), fromwhich we can deduce | β j | < | β k | for j < k , since | β j | = | β k | and β j = β k implies s β ∗ j s β k = 0. Moreover, since all the β i ’s are cycles, this means for j < k thereexists a cycle τ , with | τ | ≥ 1, such that β k = β j τ .There is an important fact we should note about cycles: for β ∈ G ∗ a cycle,we have β = γ ...γ m where each γ i = e i ...e ik is a cycle with s ( e i ) / ∈ r ( e ij ) for j < k ; we will callsuch cycles simple cycles . To see this, let β = e ...e n be a cycle and let i be74he smallest number such that s ( e ) ∈ r ( e i ). If i = n , we are done. Otherwise,letting γ = e ..e i , we have β = γ β ′ where β ′ is also a cycle with | β ′ | < | β | .Taking β ′ and applying the same process a finite number of times, we willeventually have β = γ ...γ m where each γ i has the desired property.So, by the observation made in the previous paragraph, and the fact β k = β j τ , for some cycle τ , when j < k , we can express the β i ’s, for 2 ≤ i ≤ n , asfollows. We first express β as β = γ ...γ m where each of the γ ’s are simple cycles. To clear up any confusion, the su-perscripts in the expression above are purely for indexing purposes and not tosuggest exponentiation. Now, since β = β τ for some cycle τ , and τ in turncan be expressed as τ = γ ...γ m , we have β = γ ...γ m γ ...γ m where the γ ’s are again simple cycles. Continuing in this manner, we mayexpress β n as β n = γ ...γ m γ ...γ m · · · γ n ...γ nm n in simple cycles. Thus, s α ∗ x = r p v + n P i =2 r i s β i p v = r p v + r s γ ...γ m p v + r s γ ...γ m γ ...γ m p v + · · · + r n s γ ...γ m γ ...γ m ··· γ n ...γ nmn p v .Now, if the γ ’s aren’t all the same cycle, it must be γ = γ jk for some j, k . Sinceall the γ ’s are simple cycles starting at v , it must then be s γ j ∗ k s γ = 0 and so s γ ∗ jk s α ∗ xs γ jk = r p v ;setting a = s γ j ∗ k s α ∗ and b = s γ jk , we have axb = r p v . If, on the other hand, the γ ’s are the same simple cycle, setting α = γ , a = s α ∗ , and b = p v , we have anexpression of axb as a polynomial in s α .Armed with Lemma 10.4, we are now ready to establish the last key piecewe need in order to prove the main result of this section. From here on out wewill denote the source and range maps of G by s G and r G , and of F by s F and r F , when necessary. Otherwise, we will simply use r and s to avoid clutterednotation and hope it’s clear which is which from context. Proposition 10.4.1. Let G be an ultragraph and F its desingularization. Thereexists an injective R-algebra homomorphism φ : L R ( G ) → L R ( F ) . (See [4,Proposition 5.5]) Proof. We will prove our assertion by constructing a Leavitt G -family, { P A , S e , S ∗ e } ,in L R ( F ). Then, by the universal mapping property of L R ( G ), we will have an R -algebra homomorphism φ : L R ( G ) → L R ( F ) . e ∈ G is such that s ( e ) is an infinite emitter, we will write “ e i ” for e to indicate it is the i -th edge in the enumeration of s − ( s ( e )). To that end, for L R ( F ) = L R ( { q, t } ) (i.e., L R ( F ) is generated by a universal Leavitt F -family { q A , t e , t ∗ e } ), define { P A , S e , S ∗ e } ⊆ L R ( F ) by: P A := q A , for A ∈ G , S e := t e , S e ∗ := t e ∗ , for e ∈ G such that s ( e ) is not an infinite emitter, S e i := t α , S e ∗ i := t α ∗ , for s ( e i ) an infinite emitter, where α = f f ...f i − g i is a portion of the tail added to s ( e i ) (see Figure 8).We will show { P A , S e , S ∗ e } is indeed a Leavitt G -family in L R ( F ).( uLP1. ) The fact { P A , S e , S ∗ e } satisfies uLP1 follows immediately from thefact { q A , t e , t ∗ e } satisfies uLP1 .( uLP2. ) Suppose e ∈ G such that | s − ( s ( e )) | < ∞ . Then, the fact P s ( e ) S e = S e P r ( e ) = S e and P r ( e ) S e ∗ = S e ∗ = S e ∗ follows directly from the fact { q A , t e , t ∗ e } satisfies uLP2 . On the other hand,suppose e i ∈ G with | s − ( s ( e i )) | = ∞ . For S e i := t α , and S e ∗ i := t α ∗ , where α = f f ...f i − g i is a portion of the tail added to s ( e i ), we have, by againexploiting the fact { q A , t e , t ∗ e } satisfies uLP2 , P s ( e i ) S e i = q s ( e i ) t α = q s ( f ) t α = t α = S e i and S e i P r ( e i ) = t α q r ( e i ) = t α q r ( g i ) = t α = S e i . We can similarly show, P r ( e i ) S e ∗ i = S e ∗ i P s ( e i ) = S e ∗ i .( uLP3. ) First, suppose e, e ′ ∈ G such that | s − ( s ( e )) | , | s − ( s ( e ′ )) | < ∞ .Then, for S e := t e and S e ′ := t e ′ the fact S e ∗ S e = δ e,e ′ P r ( e ) follows directly fromthe fact { q A , t e , t ∗ e } satisfies uLP3 . Second, suppose e i , e ∈ G are such that | s − ( s ( e )) | < ∞ and | s − ( s ( e i )) = ∞ . Since it is then the case e i = e , it sufficesto show S e ∗ S e i = S e ∗ i S e = 0. To that end, for S e ∗ i := t α ∗ = t g ∗ i t f ∗ i − ...t f ∗ , wehave S e ∗ i S e = t g ∗ i t f ∗ i − ...t f ∗ t e . Since f = e , and { q A , t e , t ∗ e } satisfies uLP3 , t f ∗ t e = 0; and so S e ∗ i S e = 0.A similar argument shows S e ∗ S e i = 0. Lastly, suppose e i , e j ∈ G such that | s − ( s ( e i )) | = | s − ( s ( e j )) | = ∞ ; and so S e i := t α , S e j := t α ′ , with α = f f . . . f i − g i and α ′ = f ′ f ′ . . . f ′ j − g ′ j . Now, if e i = e j , it must be, by con-struction, α = α ′ . In which case, we have S e ∗ i S e i = t α ∗ t α = q r ( g i ) = q r ( e i ) = P r ( e i ) . It remains to show S e ∗ i S e j = S e ∗ j S e i = 0 when e i = e j . To that end, suppose e i = e j . Consider first the case when s ( e i ) = s ( e j ). Then f = f ′ , and so wehave S e ∗ i S e j = t α ∗ t α ′ = t g ∗ i t f ∗ i − ... (cid:0) t f ∗ t f ′ (cid:1) ...t f ′ j − t g ′ j = t g ∗ i t f ∗ i − ...t f ∗ t f ′ ...t f ′ j − t g ′ j = 0 . 76n the other hand, should s ( e i ) = s ( e j ), e i = e j implies i = j . WLOG, let j < i . In which case we have f k = f ′ k for k ∈ { , . . . , j − } , but g ′ j = f j , andso S e ∗ i S e j = t α ∗ t α ′ = 0. These same arguments also show S e ∗ j S e i = 0. Thus, S e ∗ i S e j = S e ∗ j S e i = 0 when e i = e j . Putting it all together, we have { P A , S e , S ∗ e } satisfies uLP3 .( uLP4. ) Suppose v ∈ G such that | s − ( v ) | < ∞ . Then, for each e ∈ s − ( v ),we have S e = t e and S e ∗ = t e ∗ ; also note that in this case s − G ( v ) = s − F ( v )( s − G ( v ) = s − F ( v ) if and only if v is an infinite emitter in G ). Since { q A , t e , t ∗ e } satisfies uLP4 , we have P v := q v = X e ∈ s − F ( v ) t e t e ∗ = X e ∈ s − G ( v ) t e t e ∗ = X e ∈ s − G ( v ) S e S e ∗ . And so { P A , S e , S ∗ e } satisfies uLP4 as well.Now that we have established { P A , S e , S ∗ e } is a Leavitt G -family in L R ( F ),for L R ( G ) = L R ( { p, s } ), the universal mapping property of L R ( G ) give an R -algebra homomorphism φ : L R ( G ) → L R ( F )such that φ ( p A ) = P A , φ ( s e ) = S e , and φ ( s e ∗ ) = S e ∗ for all A ∈ G and e ∈ G .Our goal now is to show φ is injective. Up to this point, we have exclusivelyrelied on Theorem 3.20 to show a map out of a Leavitt path algebra is injective.However, for s ( e i ) an infinite emitter in G , the fact S e i = t α with | α | = i means φ is not a Z -graded homomorphism. This means we will have to show injectivityby a different manner. To that end, suppose 0 = x ∈ ker φ . By Lemma 10.4,there exist a, b ∈ L R ( G ) such that1) 0 = axb = rp v for some v ∈ G and r ∈ R \ { } , or2) 0 = axb = n P i =0 r i s iα for some cycle α ∈ G ∗ .And, certainly, x ∈ ker φ = ⇒ axb ∈ ker φ . Now, if axb = rp v , we have0 = φ ( axb ) = φ ( rp v ) = rP v = rq v . But, as we have previously seen, rq v = 0 for r ∈ R \ { } . ⇒⇐ The remainingalternative is 0 = φ ( axb ) = φ (cid:0) n X i =0 r i s iα (cid:1) = n X i =0 r i φ ( s α ) i = n X i =0 r i S iα , for some cycle α ∈ G ∗ . At this point, we call the reader’s attention to the fact,due to how { P A , S e , S e ∗ } is defined, S α = t α ′ for some cycle α ′ ∈ F ∗ ;and, since L R ( F ) = L R ( { q, t } ), rt α ′ = 0 ⇐⇒ r = 0 (to see this, simply note rt α ′∗ t α ′ = rq r ( α ′ ) ). So, we have0 = n X i =0 r i S iα = n X i =0 r i t iα ′ . axb = 0, all the r i ’s can’t be zero in the expression n P i =0 r i t iα ′ ; moreover,a summand r i t iα ′ with r i = 0 uniquely belongs in L R ( F ) i ·| α ′ | (with respect tothe Z -grading on L R ( F )). But then, as we have seen before, the Z -grading on L R ( F ) implies n X i =0 r i t iα ′ = 0 . ⇒⇐ Thus, there can’t exist 0 = x ∈ ker φ , so φ is injective.We are now ready to prove the main result of this section. As the proof islong, we will outline the main idea before proceeding. As mentioned before, wewish to leverage Theorem 5.26. Toward that goal, we’ll take { A i } i ∈ N ⊆ G ⊆ F to be as in Lemma 10.1. Then, for L R ( G ) = L R ( { p, s } ), L R ( F ) = L R ( { q, t } ),and k ∈ N we set t k := X i ≤ k p A i and t ′ k := X i ≤ k q A i . We will first show the injective map from Proposition 10.4.1 restricts to anisomorphism from t k L R ( G ) t k onto t ′ k L R ( F ) t ′ k for each k , a fact we will use toshow lim −→ k ∈ N t ′ k L R ( F ) t ′ k ∼ = lim −→ k ∈ N t k L R ( G ) t k . Having established the isomorphism of the direct limits above, we will show L R ( G ) ∼ = lim −→ k ∈ N t k L R ( G ) t k . Finally, for L R ( F ) t ′ k (a finitely generated, projective, left L R ( F )-module) wewill show the existence of a compatible set { L R ( F ) t ′ k , ϕ, ψ, N } such that lim −→ k ∈ N L R ( F ) t ′ k is a generator of L R ( F ) − M OD andlim −→ k ∈ N (cid:16) End L R ( F ) ( L R ( F ) t ′ k ) (cid:17) op ∼ = lim −→ k ∈ N t ′ k L R ( F ) t ′ k . And so, forlim −→ k ∈ N (cid:16) End L R ( F ) ( L R ( F ) t ′ k ) (cid:17) op ∼ = lim −→ k ∈ N t ′ k L R ( F ) t ′ k ∼ = lim −→ k ∈ N t k L R ( G ) t k ∼ = L R ( G ) , Theorem 5.26 establishes the Morita equivalence of L R ( F ) and L R ( G ). Now, onto showing what we have outlined. The proof is essentially that of [4, Theorem5.6], but modified to work in the case of ultragraph Leavitt path algebras. Theorem 10.5. Let G be an ultragraph and F its desingularization. Then, L R ( G ) and L R ( F ) are Morita equivalent. roof. Let { A i } i ∈ N ⊆ G ⊆ F be as in Lemma 10.1, and set t k := X i ≤ k p A i and t ′ k := X i ≤ k q A i . The first crucial step in our proof is establishing the fact the map φ fromProposition 10.4.1 restricts to an isomorphism from t k L R ( G ) t k onto t ′ k L R ( F ) t ′ k for each k . Since φ ( t k ) = t ′ k , it followsim φ | t k L R ( G ) t k ⊆ t ′ k L R ( F ) t ′ k . That φ is injective has already been established. What we need to show now isthat it is surjective. To that end, we have t ′ k L R ( F ) t ′ k = span R { t α q A t β ∗ : α, β ∈ F ∗ , r F ( α ) ∩ A ∩ r F ( β ) = ∅ , and s F ( α ) , s F ( β ) ⊆ [ i ≤ k A i ∈ G } , which follows primarily from Theorem 3.19. And so, to establish the surjectivityof φ | t k L R ( G ) t k , we need only show t α q A t β ∗ ∈ im φ | t k L R ( G ) t k for α, β ∈ F ∗ with s F ( α ) , s F ( β ) ⊆ [ i ≤ k A i . As one might guess, understanding what α ∈ F ∗ , with s F ( α ) ∈ G , looks like isimportant to our endeavor. We will show such a path α takes one of two forms:1) α = α . . . α n , where, for each i , α i ∈ G ∗ , or α i = f ...f j − g j in some tailadded to an infinite emitter in G , or2) α = α . . . α n f . . . f m , where each α i satisfies one of the conditions listedabove, and f . . . f m is an initial segment in a tail added to a singular vertex(either a sink or an infinite emitter) in G .To see this, notice that the construction of F makes it so that, for α ∈ F ∗ , r F ( α ) ∈ G , or r F ( α ) = r F ( f ) ∈ F \ G , where f is an edge along a tail added to some singular vertex. For α such that s F ( α ) ∈ G , consider first the case where r F ( α ) ∈ G . Now, should α ∈ G ∗ , weare done. Otherwise, α must contain an edge f in a tail added to a singularvertex, since s F ( α ) ∈ G , this in turn means it contains an edge whose sourceis a singular vertex in G . And so we can express α as α = α α ′ where α is a path in G (possibly an element of G ), and α ′ ∈ F ∗ such that s F ( α ′ ) is a singular vertex in G . But then the fact r F ( α ′ ) = r F ( α ) ∈ G forces α ′ to have an initial segment of the form f ...f j − g j ; i.e., α ′ can be expressed as α ′ = α α ′′ where α = f ...f j − g j in a tail added to s F ( α ′ ). Notice α ′′ ∈ F ∗ is such that s F ( α ′′ ) , r F ( α ′′ ) ∈ G , just like α . Applying the same argument to α ′′ as we didwith α , and so on, we have α = α . . . α n i , α i ∈ G ∗ , or α i = f ...f j − g j in some tail added to aninfinite emitter in G . Alternatively, suppose r F ( α ) = r F ( f ) = v ∈ F \ G forsome edge f in a tail added to singular vertex. Since s F ( α ) ∈ G , we can express α as α = α ′ f ...f m where f m = f and α ′ is a path (possibly an element of G ) such that s F ( α ′ ) ∈ G and s F ( f ) ∈ r F ( α ′ ) ∈ G . But then, by what we have previously seen, α = α ′ f . . . f m = α . . . α n f . . . f m where each α i satisfies one of the given two conditions. In light of what we havejust seen, consider again t ′ k L R ( F ) t ′ k = span R { t α q A t β ∗ : α, β ∈ F ∗ , r F ( α ) ∩ A ∩ r F ( β ) = ∅ , and s F ( α ) , s F ( β ) ⊆ [ i ≤ k A i ∈ G } . For α, β ∈ F ∗ such that t α q A t β ∗ ∈ t ′ k L R ( F ) t ′ k , it must be α = α ..α n , or α = α ..α n f ...f m , and β = β ...β k , or β = β ...β k f ′ ...f ′ l as described above.There’s more we can say, notice if α = α ..α n f ...f m and β = β ...β k (similarly α = α ..α n and β = β ...β k f ′ ...f ′ l ), t α q A t β ∗ = 0 since r F ( α ) ∩ r F ( β ) = ∅ .Therefore, we only need to consider t α q A t β ∗ where α = α ..α n and β = β ...β k ,or α = α ..α n f ...f m and β = β ...β k f ′ ...f ′ l .In the case where α and β are such that α = α ..α n and β = β ...β k , we cansee t α q A t β ∗ ∈ im φ | t k L R ( G ) t k based on how φ is defined (see Proposition 10.4.1). What’s left is to show thesame holds when α = α ..α n f ...f m and β = β ...β k f ′ ...f ′ l . To that end, itsuffices to show t f ...f m q A t ( f ′ ...f ′ l ) ∗ ∈ im φ | t k L R ( G ) t k . Based on the construction of F , and uLP2 , t f m q A t ( f ′ l ) ∗ = t f m q r F ( f m ) t ( f ′ l ) ∗ = 0 ⇐⇒ f m = f ′ l . This in turn means t f ...f m q A t ( f ′ ...f ′ l ) ∗ = 0 ⇐⇒ f ...f m = f ′ ...f ′ l , since f m = f ′ l implies m = l , and the fact t f ...f m q A t ( f ′ ...f ′ l ) ∗ = 0 means f i = f ′ i for each 1 ≤ i ≤ m . For t f ...f m q A t ( f ...f m ) ∗ ∈ t ′ k L R ( F ) t ′ k , there are two casesto consider: s F ( f ) is a sink, or s F ( f ) is an infinite emitter in G . In the case s F ( f ) is a sink, a quick calculation shows t f ...f m q A t ( f ...f m ) ∗ = q s F ( f ) = φ ( p s F ( f ) ) ∈ im φ | t k L R ( G ) t k . On the other hand, suppose s F ( f ) is an infinite emitter and let { e i } i ∈ N andan enumeration of s − F ( s F ( f )). By uLP4 , we have t f m t f ∗ m = q s F ( f m ) − t g m t g ∗ m ,80nd so t f ...f m q A t ( f ...f m ) ∗ = t f ..f m − (cid:0) q s F ( f m ) − t g m t g ∗ m (cid:1) t ( f ...f m − ) ∗ = t f ...f m − (cid:0) t f m − t f ∗ m − (cid:1) t ( f ...f m − ) ∗ − t f ..f m − g m t ( f ...f m − g m ) ∗ , applying uLP4 to t f m − t f ∗ m − , = t f ...f m − (cid:0) q s F ( f m − ) − t g m − t g ∗ m − (cid:1) t ( f ...f m − ) ∗ − t f ..f m − g m t ( f ...f m − g m ) ∗ = t f ...f m − t ( f ...f m − ) ∗ − (cid:0) t f ..f m − g m − t ( f ...f m − g m − ) ∗ + t f ..f m − g m t ( f ...f m − g m ) ∗ (cid:1) , applying uLP4 to t f m − t f ∗ m − and so on, we have= t f t f ∗ − m X i =1 t f ...f i − g i t ( f ...f i − g i ) ∗ ! = (cid:0) q s F ( f ) − t g t g ∗ (cid:1) − m X i =1 t f ...f i − g i t ( f ...f i − g i ) ∗ ! = φ ( p s F ( f ) ) − φ ( t e t e ∗ ) − m X i =1 φ ( t e i t e ∗ i ) ! ∈ im φ | t k L R ( G ) t k . Thus, t ′ k L R ( F ) t ′ k ⊆ im φ | t k L R ( G ) t k , and so, as desired, φ | t k L R ( G ) t k is surjective.Now, for k ≤ l , let ϕ k,l : t k L R ( G ) t k → t l L R ( G ) t l and ϕ ′ k,l : t ′ k L R ( F ) t ′ k → t ′ l L R ( F ) t ′ l be the inclusion maps; one can then easily check (cid:10) t k L R ( G ) t k , ϕ k,l (cid:11) and (cid:10) t ′ k L R ( G ) t ′ k , ϕ ′ k,l (cid:11) are direct systems over N (note, we are taking the directlimit in the category of rings). Moreover, we have the following commutingdiagram: t k L R ( G ) t k t ′ k L R ( F ) t ′ k t k +1 L R ( G ) t k +1 t ′ k +1 L R ( F ) t ′ k +1 φ | t k L R ( G ) t k φ | t k +1 L R ( G ) t k +1 ϕ k,k +1 ϕ ′ k,k +1 Since φ | t k L R ( G ) t k is an isomorphism for each k , we havelim −→ k ∈ N t k L R ( G ) t k ∼ = lim −→ k ∈ N t ′ k L R ( F ) t ′ k as rings. For verification of the stated isomorphism above see 24.4 in [32] where itis proved within the context of modules, but the argument remains unchanged81or any category which admits direct limits. Further, let i k : t k L R ( G ) t k → L R ( G ) be the inclusion map for each k . For l ≥ k , we have i k = i l ◦ ϕ k,l . By theuniversal mapping property of lim −→ k ∈ N t k L R ( G ) t k , there exists a ring homomorphismΦ : lim −→ k ∈ N t k L R ( G ) t k → L R ( G ) . Since { t k } k ∈ N is a set of local units for L R ( G ), Φ is a surjective. Further, supposeΦ( x ) = 0. Since x = η k ( y ), where η k : t k L R ( G ) t k → lim −→ k ∈ N t k L R ( G ) t k is as in the definition of a direct limit, we have Φ ◦ η k ( y ) = 0. But, by theproperties of a direct limit, i k = Φ ◦ η k , meaningΦ ◦ η k ( y ) = i k ( y ) = 0 . Well, i k is the inclusion map, so it must be y = 0, and so x = η k ( y ) = 0. Thus,Φ is injective. All in all, this meanslim −→ k ∈ N t k L R ( G ) t k ∼ = L R ( G ) . We have established, up to this pointlim −→ k ∈ N t ′ k L R ( F ) t ′ k ∼ = lim −→ k ∈ N t k L R ( G ) t k ∼ = L R ( G ) . To complete our proof, it remains to show L R ( F ) t ′ k ) is a finitely generated,projective, left L R ( F )-module for each k , and that there exists a compatible set { L R ( F ) t ′ k , ϕ, ψ, N } such that lim −→ k ∈ N L R ( F ) t ′ k is a generator of L R ( F ) − M OD andlim −→ k ∈ N (cid:16) End L R ( F ) ( L R ( F ) t ′ k ) (cid:17) op ∼ = lim −→ k ∈ N t ′ k L R ( F ) t ′ k . To that end, first note that, for each k , L R ( F ) t ′ k is generated by t ′ k as an L R ( F )-module; thus, it is a finitely generated module. Further, by Claim 6.6, we have L R ( F ) t ′ k is projective. We now want to show lim −→ k ∈ N L R ( F ) t ′ k is a generator of L R ( F ) − M OD . As showing this fact is a bit convoluted, we will first sketchout the steps. We will first show L R ( F ) is a generator of L R ( F ) − M OD ;a fact which we will in turn leverage to show L i ∈ N L R ( F ) q A i is a generator of L R ( F ) − M OD . And, finally, we will establish M i ∈ N L R ( F ) q A i ∼ = lim −→ k ∈ N L R ( F ) t ′ k , −→ k ∈ N L R ( F ) t ′ k is a generator for L R ( F ) − M OD .By Claim 5.23, to show L R ( F ) is a generator of L R ( F ) − M OD , it sufficesto show, for every non-zero L R ( F )-module homomorpshim f : M → N, thereexists an L R ( F )-module homomorpshim h : L R ( F ) → M such that f ◦ h = 0.To that end, let 0 = f : M → N be an L R ( F )-module homomorpshim. Thismeans there exists m ∈ M such that f ( m ) = 0, and since M = L R ( F ) M , thereexists m ′ such that m = xm ′ for some x ∈ L R ( F ). Defining h : L R ( F ) → M by x xm ′ , we have have f ◦ h = 0. Meaning, L R ( F ) is a generator of L R ( F ) − M OD .Now, let { A i } i ∈ N ⊆ G ⊆ F be as in Lemma 10.1. For any A ∈ F , onecan see A = A ′ ∪ { v j } kj =1 for some A ′ ∈ G and { v j } kj =1 ⊆ F \ G . Since A ′ iscontained in finitely many of the A i ’s, if after enumerating F \ G we set B i := A i and B i − := { v i } , for v i ∈ F \ G , we can check { B i } i ∈ N satisfies the hypothesis stated in Lemma 10.1. Thus, byCorollary 10.3, L R ( F ) ∼ = M i ∈ N L R ( F ) q B i ∼ = (cid:18) M i ∈ N L R ( F ) q A i (cid:19) M (cid:18) M v ∈ F \ G L R ( F ) q v (cid:19) . Our next task is to use the fact established above to show, for some set S ,there is a surjective module homomorphism ρ : M s ∈ S (cid:18) M i ∈ N L R ( F ) q A i (cid:19) → L R ( F ) , thereby establishing L i ∈ N L R ( F ) q A i is a generator of L R ( F ) − M OD as well. Withthis in mind, suppose v is a singular vertex in G , and v j ∈ F \ G is a vertexalong a portion of a tail, α = f ...f j , added at v (see figure below). v v v v j − v j f f . . . f j Figure 9We can define a module homomorphism, ρ ∗ j : L R ( F ) q v j → L R ( F ) q v , by ρ ∗ j ( y ) = yα ∗ ; similarly, we can define ρ j : L R ( F ) q v → L R ( F ) q v j by ρ j ( x ) = xα . Since α ∗ α = v j , we can conclude ρ j ◦ ρ ∗ j = Id L R ( F ) q vj , which inturn means ρ j is surjective. Suppose now v ∈ A i . For L R ( F ) q A i ∼ = L R ( F ) q A i \{ v } M L R ( F ) q v , 83e can take the projection onto L R ( F ) q v , and by composing with ρ j , we havea surjective homomorphsim from L R ( F ) q A i onto L R ( F ) q v j . In turn, by com-posing with the projection of L i ∈ N L R ( F ) q A i onto L R ( F ) q A i , we can conclude L R ( F ) q v j is the homomorphic image of L i ∈ N L R ( F ) q A i for each v j ∈ F \ G .Thus, defining homomorphisms corrdinate wise, we get a surjective homomor-phism M v j ∈ F \ G (cid:18) M i ∈ N L R ( F ) q A i (cid:19) → M v j ∈ F \ G L R ( F ) q v j , giving us a surjective homomorphism (cid:18) M i ∈ N L R ( F ) q A i (cid:19) ⊕ (cid:18) M v j ∈ F \ G (cid:18) M i ∈ N L R ( F ) q A i (cid:19)(cid:19) → (cid:18) M i ∈ N L R ( F ) q A i (cid:19) ⊕ (cid:18) M v j ∈ F \ G L R ( F ) q v j (cid:19) . Thus, for L R ( F ) ∼ = (cid:18) L i ∈ N L R ( F ) q A i (cid:19) L (cid:18) L v ∈ F \ G L R ( F ) q v (cid:19) , we have a set S and a surjective homomorphism ρ : M s ∈ S (cid:18) M i ∈ N L R ( F ) q A i (cid:19) → L R ( F ) , and since L R ( F ) is a generator of L R ( F ) − M OD , this means L i ∈ N L R ( F ) q A i isa generator of L R ( F ) − M OD as well.For k ≤ l , let ϕ ′ k,l : L R ( F ) t ′ k → L R ( F ) t ′ l be the inclusion map and considerthe direct system of L R ( F )-modules, D L R ( F ) t ′ k , ϕ ′ k,l E , over N . First, notice themap given by x ( xq A , ..., xq A k )defines an isomorphism from L R ( F ) t ′ k onto L i ≤ k L R ( F ) q A i ; then, by composingwith the inclusion map from L i ≤ k L R ( F ) q A i into L i ∈ N L R ( F ) q A i , we get an injectivemap, i k : L R ( F ) t ′ k → M i ∈ N L R ( F ) q A i , such that i k = i l ◦ ϕ ′ k,l for each k ≤ l . And so, by the universal mapping propertyof direct limits, we have an L R ( F )-module homomorphism θ : lim −→ k ∈ N L R ( F ) t ′ k → M i ∈ N L R ( F ) q A i such that i k = θ ◦ i k for each k . Since for each y ∈ L i ∈ N L R ( F ) q A i , there exists a k and x ∈ L R ( F ) t ′ k such that i k ( x ) = y , θ is surjective. Moreover, since, beingthe inclusion map, ϕ ′ k,l is injective for each k ≤ l , θ is injective as well (see [32,24.3, 1) and 2)]). Thus, lim −→ k ∈ N L R ( F ) t ′ k ∼ = M i ∈ N L R ( F ) q A i , −→ k ∈ N L R ( F ) t ′ k is a generator of L R ( F ) − M OD as well. Finally, tocomplete our proof, we need to showlim −→ k ∈ N (cid:16) End L R ( F ) ( L R ( F ) t ′ k ) (cid:17) op ∼ = lim −→ k ∈ N t ′ k L R ( F ) t ′ k . To that end, let φ ∈ End L R ( F ) ( L R ( F ) t ′ k ). Since φ is an L R ( F )-module homo-morphism, note φ ( xt ′ k ) = xφ ( t ′ k ). Moreover, t ′ k φ ( t ′ k ) = φ (( t ′ k ) ) = φ ( t ′ k ), andsince φ ( t ′ k ) is already an element of L R ( F ) t ′ k , this means φ ( t ′ k ) ∈ t ′ k L R ( F ) t ′ k .Thus, we can conclude φ is given by right multiplication by an element of t ′ k L R ( F ) t ′ k . Define, then, a mapΦ k : t ′ k L R ( F ) t ′ k → End L R ( F ) ( L R ( F ) t ′ k )where Φ k ( x ) ∈ End L R ( F ) ( L R ( F ) t ′ k ) is right multiplication by x ; since everyelement of End L R ( F ) ( L R ( F ) t ′ k ) arises this way, Φ k is surjective. Also, L R ( F )is a ring with local units (meaning xy = 0 for all x if and only if y = 0), andso Φ k is injective as well. Lastly, we can easily check Φ k ( xy ) = Φ k ( y ) ◦ Φ k ( x ),which means Φ k is an anti-isomorphism; put differently, t ′ k L R ( F ) t ′ k ∼ = (cid:16) End L R ( F ) ( L R ( F ) t ′ k ) (cid:17) op . By Proposition 5.24.1, D(cid:16) End L R ( F ) ( L R ( F ) t ′ k ) (cid:17) op , ϕ ′ k,l E is a direct system of rings over N . For k ≤ l , one can check the following diagramcommutes: 85 ′ k L R ( F ) t ′ k (cid:16) End L R ( F ) ( L R ( F ) t ′ k ) (cid:17) op t ′ l L R ( F ) t ′ l (cid:16) End L R ( F ) ( L R ( F ) t ′ l ) (cid:17) op Φ k Φ l ϕ ′ k,l ϕ ′ k,l Since each Φ k is an isomorphism, we havelim −→ k ∈ N (cid:16) End L R ( F ) ( L R ( F ) t ′ k ) (cid:17) op ∼ = lim −→ k ∈ N t ′ k L R ( F ) t ′ k ∼ = L R ( G ) . Thus, by Theorem 5.26, L R ( G ) and L R ( F ) are Morita equivalent.Theorems 10.5 and 9.6 allow us to concisely state the main result of thisthesis: Theorem 10.6. Let G be any ultragraph and let R be any commutative unitalring. Then, there exists a graph E such that L R ( G ) is Morita equivalent to L R ( E ) . 11 Simplicity conditions for L R ( G ) In [3], Abrams and Pino give conditions on a row-finite graph E which willguarantee L K ( E ) is a simple algebra, where K is a field—a graph is row-finite if { v ∈ E : | s − E ( v ) | = ∞} = ∅ . We will extend their result to ultragraph Leavittpath algebras using Morita equivalence. It’s worth noting the result we willprove has already been shown in [13]; however, the authors achieve their resultrelying on entirely different techniques. Moreover, we will show, unlike the casewith graphs, we need not impose the row-finite condition. Tomforde proves ananalogous result for ultragraph C ∗ -algebras in [29].To begin, recall that a cycle in an ultragraph G is a path α = e . . . e n such that s ( α ) ∈ r ( α ); an edge e is an exit for α if there exists an i such that s ( e ) = s ( e i ) with e = e i . An ultragraph G satisfies Condition (L) if everycycle α = e . . . e n in G has an exit, or there is an i such that r ( e i ) contains asink. Similarly, for a graph E , a cycle is a path α = e . . . e n in E such that86 E ( α ) = s E ( α ); an edge e is an exit for α if there exists an i such that s ( e ) = s ( e i )with e = e i . A graph E satisfies Condition (L) if every cycle in E has an exit. Condition (L) is a necessary condition, for graphs and ultragraphs, in order fortheir respective Leavitt path algebras over K to be simple. Beside Condition(L) , there is one more necessary condition in order to guarantee simplicity. Definition 11.1. Let G be an ultragraph. H ⊆ G is hereditary if:1) for each e ∈ G , s ( e ) ∈ H = ⇒ r ( e ) ∈ H .2) for all A, B ∈ H , A ∪ B ∈ H .3) given A ∈ H and B ∈ G , B ⊆ A = ⇒ B ∈ H .Also, H ⊆ G is saturated if for any non-singular vertex v ∈ G , { r ( e ) } e ∈G : s ( e )= v ⊆ H = ⇒ { v } ∈ H. In the case of a graph E , H ⊆ E is hereditary if, for each e ∈ H , s E ( e ) ∈ H = ⇒ r E ( e ) ∈ H ; it is saturated if, for every non-singular vertex v ∈ E , { r ( e ) } e ∈ E : s E ( e )= v ⊆ H = ⇒ v ∈ H. Abrams and Pino show in [3], assuming E is row-finite, L K ( E ) is simple if and only if E is such that:1) E satisfies Condition (L) , and2) the only saturated hereditary subsets of E are E and ∅ .We will show that for any ultragraph G , L K ( G ) is simple if and only if G issuch that:1) G satisfies Condition (L) , and2) the only saturated hereditary subsets of G are G and ∅ . We will start byfirst proving it for an ultragraph G with no singular vertices. Proposition 11.1.1. Let G be an ultragraph with no singular vertices and E G its associated graph. Then, G and ∅ are the only saturated hereditary subsetsof G if and only if E G and ∅ are the only saturated hereditary subsets of E G .Proof. Since G doesn’t contain any singular vertices, [18, Theorem 6.12] impliesthere is a one-to-one correspondence between the gauge-invariant ideals of C ∗ ( G )and the saturated hereditary subsets of G . By [19, Proposition 3.14], E G doesn’tcontain singular vertices if and only if G doesn’t contain any singular vertices;which, by [8, Theorem 3.6], means there is a one-to-one correspondence betweenthe saturated hereditary subsets of E G and the gauge-invariant ideals of C ∗ ( E G ).Finally, in section 6 of [19], it’s shown there is a bijection between the gauge-invariant ideals of C ∗ ( G ) and the gauge-invariant ideals of C ∗ ( E G ). Thus, for G with no singular vertices, there is a bijection between the saturated hereditarysubsets of G and the saturated hereditary subsets of E G . Proposition 11.1.2. Let G be an ultragraph and E G its associated graph. G satisfies Condition (L) if and only if E G satisfies Condition (L).Proof. ( ⇒ ) Suppose G satisfies Condition (L) . Let α be a cycle in E G . Thegraph F from Remark 2.11 doesn’t contain any cycles. Thus, α must containan edge of the form ( e n , x ); which in turn means α must pass through a vertex v ∈ G . We will take v to be the base point of α . Recall that α can be expressedas α = g ( e n , x ) g ( e n , x ) g . . . ( e n k , x k ) g k i , e n i ∈ G , x i ∈ X ( e n i ), and g i ∈ F ∗ ; because we are assuming v to be the base point, we can safely assume s E ( g ) = v = r E ( g k ). Then, [19,Lemma 4.9] implies γ = e n e n . . . e n k is a cycle in G with v = s ( e n ) ∈ r ( e n k ).Since we are assuming G satisfies Condition (L) , one of two things must hold. Case 1: γ has an exit e . In this case, there is an edge, ( e, x ), which is anexit for α . Case 2: γ contains an edge e n i , for some i , such that there exists a sink w ∈ r ( e n i ). There are two possibilities here. First, if w ∈ X ( e n i ), then ( e n i , w )is an edge in E G . Further, by [19, Proposition 3.14], w is a sink in E G as well,which means ( e n i , w ) = ( e n i , x i ) since x i = s E ( g i ). And so ( e n i , w ) is an exitfor α . If, on the other hand, w / ∈ X ( e n i ), then we have, by [19, Lemma 4.4], ω ∈ X ( e n i ) ∩ ∆ such that w ∈ r ′ ( ω ). Should ω = x i , we would have ( e n i , ω )as an exit for α . So, suppose ω = x i . By [19, Lemma 4.6 (3)], there exists apath g in F such that s E ( g ) = ω and r E ( g ) = w . Now, note s E ( g i ) = ω and r E ( g i ) = s ( e n i +1 ) = w . This means g = g i ; however, since s E ( g ) = s E ( g i ), thisimplies α must have an exit.Taking the two cases together, we have that E G must satisfy Condition (L) as well.( ⇐ ) Suppose E G satisfies Condition (L) . Let γ = e n e n . . . e n k be a cyclein G . Suppose there is an i such that | r ( e n i ) | > 1. If r ( e n i ) contains a sink, weare done. If not, it must be that there exists an edge e such that s ( e ) ∈ r ( e n i )and s ( e ) = s ( e n i +1 ); in which case, e is an exist for γ . If, on the other hand, | r ( e n i ) | = 1 for each i , it must be r ( e n i ) = X ( e n i ) = { s ( e n i +1 ) } for 1 ≤ i ≤ k − r ( e n k ) = { s ( e n ) } . And so we get a cycle α = ( e n , s ( e n ))( e n , s ( e n )) . . . ( e n k , s ( e n ))in E G . Since E G satisfies Condition (L) , α has an exit; further, by [19, Lemma4.6 (1)], the exit must be of the form ( e, x ). Moreover, for s ( e ) = s ( e n i ), it mustbe that e = e n i . This is because e n i = e , and ( e, x ) = ( e n i , s ( e n i +1 )), imply x = s ( e n i +1 ), but then this contradicts the fact X ( e n i ) = { s ( e n i +1 ) } . And so,for ( e, x ) an exit for α , we have that e is an exit for γ . Thus, γ has an exist, orthere exists an i such that r ( e n i ) contains a sink; meaning G satisfies Condition(L) . Theorem 11.2. Let G be an ultragraph with no singular vertices, and K a field.Then, L K ( G ) is simple if and only if G satisfies Condition (L) and G and ∅ arethe only saturated hereditary subsets of G .Proof. Since L K ( G ) and L K ( E G ) are Morita equivalent, Proposition 5.16.1 thenimplies L K ( G ) and L K ( E G ) have isomorphic lattice of ideals. Since G doesn’thave any singular vertices, E G doesn’t have any singular vertices either [19,Proposition 3.14]. Which, by the work of Abrams and Pino (see [3]), means L K ( E G ) is simple if and only if E G satisfies Condition (L) and the only saturatedhereditary subsets of E G are E G and ∅ . But, by propositions 11.1.1 and 11.1.2, E G satisfies Condition (L) , with E G and ∅ being the only saturated hereditarysubsets of E G , if and only if G satisfies Condition (L) and G and ∅ are the88nly saturated hereditary subsets of G . Thus, L K ( G ) is simple if and only if G satisfies Condition (L) and G and ∅ are the only saturated hereditary subsetsof G .We now want to extend the previous theorem to the case where G maycontain singular vertices. In order to do that, we need the following proposition. Proposition 11.2.1. Let G be an ultragraph, and F its desingularization.Then,1) G and ∅ are the only saturated hereditary subsets of G if and only if F and ∅ are the only saturated hereditary subsets of F .2) G satisfies Condition (L) if and only if F satisfies Condition (L).Proof. 1) Suppose G and ∅ are the only hereditary subsets of G . By [29,Theorem 3.10], C ∗ ( G ) is a simple C ∗ -algebra. Further, by [30, Proposition6.6], C ∗ ( G ) and C ∗ ( F ) are Morita equivalent as C ∗ -algebras. This means, by[27, Theorem 3.22], there is a lattice isomorphism between the closed ideals of C ∗ ( G ) and the closed ideals of C ∗ ( F ); which in turn means C ∗ ( F ) is also asimple C ∗ -algebra. Applying [29, Theorem 3.10] again, we have F and ∅ arethe only saturated hereditary subsets of F . The same argument establishes theopposite direction. Thus, G and ∅ are the only saturated hereditary subsets of G if and only if F and ∅ are the only saturated hereditary subsets of F .2) ( ⇒ ) Suppose G satisfies Condition (L) , and let α be a cycle in F . If α doesn’t pass through an infinite emitter in G , then α is also a cycle in G and weare done. Otherwise, α must pass through an infinite emitter, v , in G . By theconstruction of F , there are edges f and g such that s F ( f ) = s F ( g ) = v with f = g ; thus, α must have an exit. And so, all in all, if G satisfies Condition(L) , F must satisfy Condition (L) as well.( ⇐ ) Suppose F satisfies Condition (L) , and let α be a cycle in G . If α doesn’tpass through an infinite emitter, then α is a cycle in F and we are done. If, onthe other hand, α does pass through an infinite emitter, then it certainly has anexit. Thus, F satisfying Condition (L) implies G satisfies Condition (L) . Theorem 11.3. Let G be an ultragraph and K a field. Then, L K ( G ) is simpleif and only if G satisfies Condition (L) and G and ∅ are the only saturatedhereditary subsets of G .Proof. Let G be an ultragraph, and F its desingularization. By Theorem 10.5and Proposition 5.16.1, L K ( G ) is simple if and only if L K ( F ) is simple. But, byTheorem 11.2, L K ( F ) is simple if and only if F satisfies Condition (L) and F and ∅ are the only saturated hereditary subsets of F . Finally, applyingProposition 11.2.1, we can conclude L K ( G ) is simple if and only if G satisfies Condition (L) and G and ∅ are the only saturated hereditary subsets of G .Now, one might ask what sort of simplicity conditions exist for L R ( G ), where R is any unital commutative ring. In general, studying the ideal structure of L R ( G ) purely in terms of the properties of G is difficult, if not impossible. Themain reason being, if R is not a field, it can have a rich ideal structure of its89wn, which in turn influences the ideal structure of L R ( G ); obviously, the idealstructure of R has nothing to do with the properties of G . There are, however,types of ideals we can study in terms of the properties of G . Definition 11.4. Let G be an ultragraph. I ⊳ L R ( G ) is called a basic ideal iffor any r ∈ R \ { } and any A ∈ G , rp A ∈ I = ⇒ p A ∈ I . Further, L R ( G ) is basically simple if { } and L R ( G ) are its only basic ideals.In the case of a graph E , an ideal I in L R ( E ) is basic if rq v ∈ I , for any r ∈ R \ { } , implies q v ∈ I . Similarly, we say L R ( E ) is basically simple if { } and L R ( E ) are the only basic ideals of L R ( E ). Let H ⊆ G be saturated andhereditary. We set B H := n v ∈ G : | s − ( v ) | = ∞ with 0 < | s − ( v ) ∩ { e : r ( e ) / ∈ H }| < ∞ o . The elements of B H are called the breaking vertices of H . An admissible pair in G is a pair ( H, S ) where H is a saturated hereditary subset of G and S ⊆ B H .Interestingly, there is a bijection between the set of admissible pairs of G andthe graded basic ideals of L R ( G ) [16, Theorem 4.4 (2)]. In the case of a graph E , this reduces to a bijection between the set of saturated hereditary subsets of E and the graded basic ideals of L R ( E ) (see [31, Theorem 7.9]).Tomforde shows in [31], given a row-finite graph E , L R ( E ) is basically simpleif and only if E satisfies Condition (L) and ∅ and E are the only saturatedhereditary subsets of E . One can easily be led to believe a similar result shouldhold for L R ( G ). One might further be tempted to use Morita equivalence, aspreviously done, to establish such a result. 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