A Geometric Reverse To The Plus Construction And Some Examples Of Pseudo-Collars On High-Dimensional Manifolds
aa r X i v : . [ m a t h . G T ] A ug A GEOMETRIC REVERSE TO THE PLUS CONSTRUCTION ANDSOME EXAMPLES OF PSEUDO-COLLARS ONHIGH-DIMENSIONAL MANIFOLDS
JEFFREY J. ROLLAND
Abstract.
In this paper, we develop a geometric procedure for producing a “re-verse” to Quillen’s plus construction, a construction called a or semi-h-cobordism . We then use this reverse to the plus construction to pro-duce uncountably many distinct ends of manifolds called pseudo-collars , which arestackings of 1-sided h-cobordisms. Each of our pseudo-collars has the same bound-ary and pro-homology systems at infinity and similar group-theoretic properties fortheir pro-fundamental group systems at infinity. In particular, the kernel group ofeach group extension for each 1-sided h-cobordism in the pseudo-collars is the samegroup. Nevertheless, the pro-fundamental group systems at infinity are all distinct.A good deal of combinatorial group theory is needed to verify this fact, includingan application of Thompson’s group V.The notion of pseudo-collars originated in Hilbert cube manfold theory, where itwas part of a necessary and sufficient condition for placing a Z -set as the boundaryof an open Hilbert cube manifold. We are interested in pseudo-collars on finite-dimensional manifolds for the same reason, attempting to put a Z -set as the bound-ary of an open high-dimensional manifold. Contents
1. Introduction and Main Results 22. A Handlebody-Theoretic Reverse to the Plus Construction 53. Some Preliminaries to Creating Pseudo-Collarable High-DimensionalManifolds 144. Some Algebraic Lemmas, Part 1 185. Some Algebraic Lemmas, Part 2 206. Some Algebraic Lemmas, Part 3 247. Manifold Topology 27References 28
Date : August 14, 2015.1991
Mathematics Subject Classification.
Primary 57R65, 57R19; Secondary 57S30 57M07.
Key words and phrases. plus construction, 1-sided h-cobordism, 1-sided s-cobordism, pseudo-collar, Thompson’s group V. Introduction and Main Results
In this paper, we develop a geometric procedure for producing a “reverse” toQuillen’s plus construction, a construction called a or semi-h-cobordism . We then use this reverse to the plus construction to produce uncountablymany distinct ends of manifolds called pseudo-collars , which are stackings of 1-sidedh-cobordisms. Each of our pseudo-collars has the same boundary and pro-homologysystems at infinity and similar group-theoretic properties for their pro-fundamentalgroup systems at infinity. In particular, the kernel group of each group extension foreach 1-sided h-cobordism in the pseudo-collars is the same group. Nevertheless, thepro-fundamental group systems at infinity are all distinct. A good deal of combinato-rial group theory is needed to verify this fact, including an application of Thompson’sgroup V.The notion of pseudo-collars originated in Hilbert cube manifold theory, where itwas part of a necessary and sufficient condition for placing a Z -set as the boundaryof an open Hilbert cube manifold. We are interested in pseudo-collars on finite-dimensional manifolds for the same reason, attempting to put a Z -set as the boundaryof an open high-dimensional manifold.We work in the category of smooth manifolds, but all our results apply equally wellto the categories of PL and topological manifolds. The manifold version of Quillen’splus construction provides a way of taking a closed smooth manifold M of dimension n ≥ G = π ( M ) contains a perfect normal subgroup P which is the normal closure of a finite number of elements and producing a compactcobordism ( W, M, M + ) to a manifold M + whose fundamental group is isomorphic to Q = G/P and for which M + ֒ → W is a simple homotopy equivalence. By duality,the map f : M → M + given by including M into W and then retracting onto M + induces an isomorphism f ∗ : H ∗ ( M ; Z Q ) → H ∗ ( M + ; Z Q ) of homology with twistedcoefficients. By a clever application of the s-Cobordism Theorem, such a cobordismis uniquely determined by M and P (see [8] P. 197).In “Manifolds with Non-stable Fundamental Group at Infinity I” [11], Guilbaultoutlines a structure to put on the ends of an open smooth manifold N with finitelymany ends called a pseudo-collar , which generalizes the notion of a collar on the end ofa manifold introduced in Siebenmann’s dissertation [25]. A pseudo-collar is definedas follows. Recall that a manifold U n with compact boundary is an open collar if U n ≈ ∂U n × [0 , ∞ ); it is a homotopy collar if the inclusion ∂U n ֒ → U n is a homotopyequivalence. If U n is a homotopy collar which contains arbitrarily small homotopycollar neighborhoods of infinity, then we call U n a pseudo-collar . We say that an open n -manifold N n is collarable if it contains an open collar neighborhood of infinity, andthat N n is pseudo-collarable if it contains a pseudo-collar neighborhood of infinity.Each pseudo-collar admits a natural decomposition as a sequence of compact cobor-disms ( W, M, M − ), where W is a 1-sided h-cobordism (see Definition 1 below). If a GEOMETRIC REVERSE TO THE PLUS CONSTRUCTION 3
W, M − , M ) is a a plus cobordism. (This somewhat justifiesthe use of the symbol “ M − ” for the right-hand boundary of a 1-sided h-cobordism, aplay on the traditional use of M + for the right-hand boundary of a plus cobordism.)The general problem of a reverse to Quillen’s plus construction in the high-dimensional manifold category is as follows. Problem 1 (Reverse Plus Problem) . Suppose G and Q are finitely-presented groupsand Φ : G ։ Q is an onto homomorphism with ker(Φ) perfect. Let M n ( n ≥ ) be aclosed smooth manifold with π ( M ) ∼ = Q .Does there exist a compact cobordism ( W n +1 , M, M − ) with ✲ ker( ι ) ✲ π ( M − ) ι ✲ π ( W ) ✲ equivalent to ✲ ker(Φ) ✲ G Φ ✲ Q ✲ and M ֒ → W a (simple) homotopy equivalence. Notes: • The fact that G and Q are finitely presented forces ker(Φ) to be the normalclosure of a finite number of elements. (See, for instance, [11] or [25].) • Closed manifolds M n ( n ≥
5) in the various categories with π ( M ) isomorphicto a given finitely presented group Q always exist. In the PL category, onecan simply take a presentation 2-complex for Q , K , embed K in S n +1 , take aregular neighborhood N of K in S n +1 , and let M = ∂N . Similar proceduresexist in the other categories.The following terminology was first introduced in [17]. Definition 1.
Let N n be a compact smooth manifold. A ( W, N, M ) is a cobordism with either N ֒ → W or M ֒ → W is a homotopy equivalence(if it is a simple homotopy equivalence, we call ( W, N, M ) a ).[A 1-sided h-cobordism ( W, N, M ) is so-named presumably because it is “one side ofan h-cobordism”]. One wants to know under what circumstances 1-sided h-cobordisms exists, and, ifthey exist, how many there are. Also, one is interested in controlling the torsion andwhen it can be eliminated.There are some cases in which 1-sided h-cobordisms are known not to exist. Forinstance, if P is finitely presented and perfect but not superperfect, Q = h e i , and M = S n , then a solution to the Reverse Plus Problem would produce an M − that isa homology sphere. But it is a standard fact that a manifold homology sphere must JEFFREY J. ROLLAND have a a superperfect fundamental group! (See, for instance, [20].) (The definition ofsuperperfect will be given in Definition 2.)The key point is that the solvability of the Reverse Plus Problem depends notjust upon the group data, but also upon the manifold M with which one begins.For instance, one could start with a group P which is finitely presented and perfectbut not superperfect, let N − be a manifold obtained from the boundary of a regularneighborhood of the embedding of a presentation 2-complex for P in S n +1 , and let( W, N − , N ) be the result of applying Quillen’s plus construction to to N − with respectto all of P . Then again Q = h e i and Φ : P ։ Q but N clearly admits a 1-sideds-cobordism, namely ( W, N, N − ) (however, of course, we cannot have N a sphere or N − a homology sphere).Here is a statement of our main results. Theorem 1.1 (Existence of 1-sided s-cobordisms) . Given → S → G → Q → where S is a finitely presented superperfect group, G is a semi-direct product of Q by S , and any n -manifold N with n ≥ and π ( M ) ∼ = Q , there exists a solution ( W, N, N − ) to the Reverse Plus Problem for which N ֒ → W is a simple homotopyequivalence. One of the primary motivations for Theorem 1.1 is that it provides a “machine”for constructing interesting pseudo-collars. As an application, we use it to prove:
Theorem 1.2 (Uncountably Many Pseudo-Collars on Closed Manifolds with theSame Boundary and Similar Pro- π ) . Let M n be a closed smooth manifold ( n ≥ ) with π ( M ) ∼ = Z and let S be the finitely presented group V ∗ V , which is thefree product of 2 copies of Thompson’s group V . Then there exists an uncountablecollection of pseudo-collars { N n +1 ω | ω ∈ Ω } , no two of which are homeomorphic atinfinity, and each of which begins with ∂N n +1 ω = M n and is obtained by blowing upcountably many times by the same group S . In particular, each has fundamental groupat infinity that may be represented by an inverse sequence Z ✛✛ α G ✛✛ α G ✛✛ α G ✛✛ α . . . with ker( α i ) = S for all i . An underlying goal of papers [11], [13], and [14] is to understand when non-compactmanifolds with compact (possibly empty) boundary admit Z -compactifications. In[5], it is shown that a Hilbert cube manifold admits a Z -compactification if and only ifit is pseudo-collarable and the Whitehead torsion of the end can be controlled. In [10],Guilbault asks whether the universal cover of a closed, aspherical manifold ( n ≥ n ≥
6) toadmit a Z -compactification. Still further, he shows that any two Z -boundaries of anANR must be shape equivalent. Finally, he and Ancel show in [1] that if two closed, GEOMETRIC REVERSE TO THE PLUS CONSTRUCTION 5 contractible manifolds M n and N n ( n ≥
6) admit homeomorphic boundaries, then M is homeomorphic to N . This is most interesting when the contractible manifoldsare universal covers of closed aspherical manifolds. In that case, these questions maybe viewed as an approach to the famous Borel Conjecture, which asks whether twoaspherical manifolds with isomorphic fundamental group are necessarily homeomor-phic.The author would like to thank would also Jeb Willenbring, Marston Conder, andChris Hrusk for their helpful conversations.The work presented in this paper is part of the author’s Ph.D. dissertation writtenunder Craig Guilbault at the University of Wisconsin - Milwaukee.2. A Handlebody-Theoretic Reverse to the Plus Construction
In this section, we will describe our partial solution to the Reverse Plus Problem.Our solution only applies to superperfect (defined in Definition 2 below), finitelypresented kernel groups where the total group G of the group extension 1 → K → G → Q → M and S are fixed, we are able to analyze the various solutions to theReverse Plus Problem by studying the algebraic problem of computing semi-directproducts of Q by S ; this is supposed to be the goal of algebraic topology in general. Definition 2.
A group G is said to be superperfect if its first two homology groupsare 0, that is, if H ( G ) = H ( G ) = 0 . (Recall a group is perfect if its first homologygroup is 0.) Example 1.
A perfect group is superperfect if it admits a finite, balanced presentation,that is, a finite presentation with the same number of generators as relators. (Theconverse is false.)
Lemma 2.1.
Let S be a superperfect group. Let K be a cell complex which hasfundamental group isomorphic to S . Then all elements of H ( K ) can be killed byattaching 3-cells. Proof
By Proposition 7.1.5 in [9], there is a K ( S,
1) which is formed from K by attaching cells of dimension 3 and higher. Let L be such a K ( S, L is formed from K by attaching only 3-cells, and H ( L ) ∼ = H ( L ), as L is formedfrom L by attaching cells of dimension 4 and higher, which cannot affect H . But H ( L ) ∼ = H ( S ) by definition and H ( S ) ∼ = 0 by hypothesis. Thus, all elements of H ( K ) can be killed by attaching 3-cells. (cid:3) JEFFREY J. ROLLAND
Definition 3.
A group extension ✲ K ι ✲ G σ ✲ Q ✲ is a semi-direct product if there is a left-inverse τ (which is a homomorphism)to σ . Note that in this case, • there are “slide relators” qk = kφ ( q )( k ) q , where φ is the outer action of Q on K , which “represent the price of sliding k across q ”. • every word k q k q · . . . · k n q n admits a normal form k ′ q ′ where all elementsfrom K come first on the left and all elements of Q come last on the right. • there is a presentation for G in terms of the presentations for K and Q andthe slide relators; to wit, Claim 1. h α , . . . , α k , β , . . . , β k | r , . . . , r l , s , . . . , s l ,β α ( α φ ( β )( α ) β ) − , . . . , β k α k ( α k φ ( β k )( α k ) β k ) − i (1) is a presentation for G . (The β j α i ( α i φ ( β j )( α i ) β j ) − are “slide relators”). Proof
There is clearly a homomorphism from the group presented above to G.From this, it follows that G = KQ and that K ∩ Q = { } . From this, it follows thatthe kernel is trivial (in the finite case, just check orders). (cid:3) Lemma 2.2 (Equivariant Attaching of Handles) . Let M n be a smooth manifold, n ≥ , with M one boundary component of W with π ( M ) ∼ = G . Let P E G and Q = G/P . Let M be the cover of M with fundamental group P and give H ∗ ( M ; Z ) the structure of a Z Q -module. Let k + 1 ≤ n and let S be a finite collection ofelements of H k ( M ; Z ) which all admit embedded spherical representatives which havetrivial tubular neighborhoods. If k = 1 , assume all elements of S represent elementsof P .Then one can equivariantly attach ( k + 1) -handles across S , that is, if S = { s j,q | q ∈ Q } is the collection of lifts of elements of S to M , one can attach ( k + 1) -handles across tubular neighborhoods of the s j,q so that each lift s j,q projects downvia the covering map p to an element s j of S and so that the covering map extendsto send each ( k + 1) -handle H j,q attached across a tubular neighborhood of s j,q in M bijectively onto a handle attached across the projection via the covering map of thetubular neighborhood of the element s j in M . Proof H k ( M ; Z ) has the structure of a Z Q -module. The action of Q on S per-mutes the elements of S . For each embedded sphere s j in S , lift it via its inverseimages under the covering map to a pairwise disjoint collection of embedded spheres s j,q . (This is possible since a point of intersection or self-intersection would haveto project down to a point of intersection or self-intersection (respectively) by theevenly-covered neighborhood property of covering spaces.) The s j,q all have trivial GEOMETRIC REVERSE TO THE PLUS CONSTRUCTION 7 tubular neighborhoods. Attach an ( k + 1)-handle across the tubular neighborhood ofthe elements s j of the S . For each j ∈ { , . . . , | S |} and q ∈ Q attach an ( k + 1)-handleacross the spherical representative s j,q ; extend the covering projection so it projectsdown in a bijective fashion from the handle attached along s j,q onto the handle weattached along s j . (cid:3) Lemma 2.3.
Let
A, B, and C be R -modules, with B a free R -module (on the basis S ), and let Θ : A L B → C be an R -module homomorphism. Suppose Θ | A is onto.Then ker(Θ) ∼ = ker(Θ | A ) L B . Proof
Define φ : ker(Θ | A ) L B → ker(Θ) as follows. For each s ∈ S , where S is abasis for B , choose α ( s ) ∈ A with Θ( α ( s ) ,
0) = Θ(0 , s ), as Θ | A is onto. Extend α toa homomorphism from B to A , and note that α has the same property for all b ∈ B .Then set φ ( x, b ) = ( x − α ( b ) , b ).(Well-defined) Let x ∈ ker(Θ | A ) and b ∈ B . Then Θ( φ ( x, b )) = Θ( x − α ( b ) , b ) =Θ( x, − α ( b ) , , b ) = 0+ − Θ( α ( b ) , , b ) = 0+ − Θ(0 , b )+Θ(0 , b ) = 0.So, φ is well-defined.Define ψ : ker(Θ) → ker(Θ | A ) L B by ψ ( z ) = ( π ( z ) + α ( π ( z )) , π ( z )), where π : A L B → A and π : A L B → B are the canonical projections.(Well-defined) Let z ∈ ker(Θ). It is clear that π ( z ) ∈ B , so it remains to prove that π ( z ) + α ( π ( z )) ∈ ker(Θ | A ). [Note Θ( z ) = Θ | A ( π ( z )) + Θ | B ( π ( z )) ⇒ Θ | A ( π ( z )) = − Θ | B ( π ( z )). Note also, by definition of α , Θ( α ( π ( z ))) = Θ(0 , π ( z ))]. We computeΘ | A ( π ( z ) + α ( π ( z ))) = Θ | A ( π ( z )) + Θ( α ( π ( z ))) = − Θ | B ( π ( z )) + Θ(0 , π ( z )) = − Θ(0 , π ( z )) + Θ(0 , π ( z )) = 0. So, ψ is well-defined.(Homomorphism) Clear.(Inverses) Let ( x, b ) ∈ ker(Θ | A ) L B . The ψ ( φ ( x, b )) = ψ ( x − α ( b ) , b ) = ( π ( x − α ( b ) , b ) + α ( π ( x − α ( b ) , b )) , π ( x − α ( b ) , b )) = ( x − α ( b ) + α ( b ) , b ) = ( x, b ).Let z ∈ ker(Θ). Then φ ( ψ ( z )) = φ ( π ( z ) + α ( π ( z )) , π ( z )) = ( π ( z ) + α ( π ( z )) − α ( π ( z )) , π ( z )) = ( π ( z ) , π ( z )) = z .So, φ and ψ are inverses of each other, and the lemma is proven. (cid:3) Definition 4. A k -handle is said to be trivially attached if and only if it is possibleto attach a canceling ( k + 1) -handle. Here is our solution to the Reverse Plus Problem in the high-dimensional manifoldcategory.
Theorem 1.1 (An Existence Theorem for Semi-s-Cobordisms) . Given → S → G → Q → where S is a finitely presented superperfect group, G is a semi-direct productof Q by S , and any closed n -manifold N with n ≥ and π ( N ) ∼ = Q , there exists JEFFREY J. ROLLAND a solution ( W, N, N − ) to the Reverse Plus Problem for which N ֒ → W is a simplehomotopy equivalence. Proof
Start by taking N and crossing it with I . Let Q ∼ = h α , . . . , α k | r , . . . , r l i be a presentation for Q . Let S ∼ = h β , . . . , β k | s , . . . , s l i be a presentation for S .Take a small n -disk D inside of N × { } . Attach a trivial 1-handle h i for each β i in this disk D . Note that because they are trivially attached, there are canceling2-handles k i , which may also be attached inside the disk together with the 1-handles D ∪ { h i } . We identify these 2-handles now, but do not attach them yet. They willbe used later.Attach a 2-handle h j across each of the relators s j of the presentation for S in thedisk together with the 1-handles D ∪ { h i } , choosing the framing so that it is triviallyattached in the manifold that results from attaching h i and k i (although we have notyet attached the handles k i ). Note that because they are trivially attached, there arecanceling 3-handles k j , which may also be attached in the portion of the manifoldconsisting of the disk D together with the 1-handles { h i } and the 2-handles { k i } . Weidentify these 3-handles now, but do not attach them yet. They will be used later.Attach a 2-handle f i,j for each relator β j α i β − j φ ( α i ) − , choosing the framing so thatit is trivially attached in the result of attaching the h i , k i , h j and k j . This is possiblesince each of the relators becomes trivial when the k i ’s and k i ’s are attached. Notethat because the f i,j are trivially attached, there are canceling 3-handles g i,j . Weidentify these 3-handles now, but do not attach them yet. They will be used later.Call the resulting cobordism with only the h i ’s, the h j ’s, and the f i,j ’s attached( W ′ , N, M ′ ) and call the right-hand boundary M ′ .Note that we now have π ( N ) ∼ = Q , π ( W ′ ) ∼ = G , and ι : π ( M ′ ) → π ( W ) anisomorphism because, by inverting the handlebody decomposition, we are startingwith M ′ and adding ( n − n − π as n ≥ W ′ of W ′ corresponding to S . Then the right-hand boundary ofthis cover, M ′ , also has fundamental group isomorphic to S by covering space theory.Also, the left-hand boundary of this cover, e N , has trivial fundamental group.Consider the handlebody chain complex C ∗ ( W ′ , e N ; Z ). This is naturally a Z Q -module complex. It looks like0 ✲ C ( W ′ , e N ; Z ) ✲ C ( W ′ , e N ; Z ) ∂ ✲ C ( W ′ , e N ; Z ) ✲ C ( W ′ , e N ; Z ) ✲
0= = ∼ = ∼ = = =0 ✲ ✲ l M i =1 Z Q ⊕ k ∗ k M j =1 Z Q ∂ ✲ k M i =1 Z Q ✲ ✲ GEOMETRIC REVERSE TO THE PLUS CONSTRUCTION 9 where C ( W ′ , e N ; Z ) decomposes as A = L l i =1 Z Q , which has a Z Q -basis obtainedby arbitrarily choosing one lift of the 2-handles for each of the h j , and B = L k · k j =1 Z Q ,which has a Z Q -basis obtained by arbitrarily choosing one lift of the 2-handles for eachof the f i,j . Set C = C ( W ′ , e N ; Z ) ∼ = L k i =1 Z Q (as Z Q -modules). Choose a preferredbasepoint ∗ and a preferred lift of the the disk D to a disk D in M . Decompose ∂ as ∂ , = ∂ | A and ∂ , = ∂ | B Since S is perfect, we must have l ≥ k , as we must have as many or more relatorsas we have generators in the presentation for S to have no 1-dimensional homology.We examine the contribution of ∂ , to H ( W ′ , e N ; Z ). It will be useful to firstlook downstairs at the Z -chain complex for ( W ′ , N ). Let A ′ be the submodule of C ( W ′ , N ; Z ) determined by the h j ’s and let C ′ be C ( W ′ , N ; Z ), which is generatedby the h i ’s. Then A ′ is a finitely generated free abelian group, so, the kernel K ′ of ∂ ′ , : A ′ → C ′ is a subgroup of a finitely generated free abelian group, and thus K ′ isa finitely generated free abelian group, say on the basis { k , . . . , k a } Claim 2. ker( ∂ , ) is a free Z Q -module on a generating set of cardinality | a | . Proof
The disk D has | Q | lifts of itself to M , `a la Lemma 2.2. Now, Q acts asdeck transformations on M , transitively permuting the lifts of D as the cover M isa regular cover. A preferred basepoint ∗ and a preferred lift of the the disk D toa disk D in M have already been chosen for the identification of C ∗ ( W ′ , e N ; Z ) withthe Z Q -module C ∗ ( W ′ , N ; Z Q ). Let the handles attached inside the preferred lift D be our preferred lifts h i and let the lifts of the h j s that attach to D ∪ ( ∪ h i ) be ourpreferred 2-handles h j .Note that none of the qh i spill outside the disk qD and none of the qh j spill outsidethe disk qD ∪ ( ∪ qh i ). This implies ∂ , ( h j ) ∈ { z i h i | z i ∈ Z } ≤ { z i q i h i | z i q i ∈ Z Q } and so ∂ , ( qh j ) ∈ { z i qh i | z i ∈ Z q t ≤ Z Q } . This mean if q = q are in Q and c and c are lifts of chains in A to D , then ∂ , ( q c + q c ) = 0 ∈ Z Q if and only if ∂ , ( c ) = ∂ , ( c ) = 0 ∈ Z ; ( ‡ ).With this in mind, let k i be a lift of the chain k i in a generating set for K ′ in thedisk D to D . Then ∂ , ( k i ) = 0. Moreover, Q transitively permutes each k i with theother lifts of k i to the other lifts of D . Now, suppose ∂ , ( c ) = 0, with c an element of C ( W ′ , N ; Z Q ). By ( ‡ ), we must have c = P mt =1 n t q t k t with n t ∈ Z and q t ∈ Q . Thisproves the k t ’s generate ker( ∂ , ).Finally, suppose some linear combination P ai =1 ( P n t q t ) k i is zero. Then, as q t k t and q t k t cannot cancel if q t = q t , it follows that all n t are zero. This proves the k i ’s are a free Z Q -basis for ker( ∂ , ). This proves the claim. (cid:3) Now, we have ∂ : A L B → C . Recall S is a finitely presented, superperfectgroup, and W’ contains a 1-handle for each generator and a 2-handle for each relatorin a chosen finite presentation for S . It then follows that ker( ∂ ) / im( ∂ | A ) ∼ = 0,as if Λ contains the collection of lifts of 1-handles for each generator of S and thecollection of lifts of 2-handles for each relator of S , then Λ = 0 as a Z Q -modules andΛ = ker( ∂ ) / im( ∂ | A ). But C ( W ′ , e N ; Z ) = 0, so ker( ∂ ) = C . This implies ∂ | A isonto. By Lemma 2.3, we have that ker( ∂ ) ∼ = ker( ∂ | A ) L B . By the previous claim,ker( ∂ | A ) is a free and finitely generated Z Q -module. Clearly, B is a free and finitelygenerated Z Q -module. Thus, ker( ∂ ) ∼ = H ( W ′ , e N ; Z ) is a free and finitely generated Z Q -module.By Lemma 2.1, we may choose spherical representatives for all elements of H ( W ′ ; Z ).By the Long Exact Sequence in homology for ( W ′ , e N ), we have · · · ✲ H ( W ′ ; Z ) ✲ H ( W ′ , e N ; Z ) ✲ H ( e N ; Z ) ✲ · · · = = ∼ = · · · ✲ H ( W ′ ; Z ) ✲✲ H ( W ′ , e N ; Z ) ✲ ✲ · · · so any element of H ( f W ′ , N ; Z ) also admits a spherical representative.So, we may choose spherical representatives for any element of H ( W ′ , N ; Z Q ). Let { s k } be a collection of embedded, pair-wise disjoint 2-spheres which form a free, finite Z Q -basis for H ( W ′ , N ; Z Q ).Note that the { s k } can be arranged to live in right-hand boundary M ′ of W ′ . Todo this, view W ′ upside-down, so that it has ( n − − and ( n − − handles attached.For each s k , make it transverse to the (2-dimensional) co-core of each ( n − − handle,then blow it off the handle by using the product structure of the handle less the co-core; do the same thing with the ( n − − handles. Finally, use the product structureof N × I to push s k into the right-hand boundary.If we add the k i , h j and g i,j to W ′ , and similarly make sure the k i s, k j s, and g i,j s donot intersect the { s k } s, and call the resulting cobordism W ′′ , we can think of the { s k } as living in the right-hand boundary of ( W ′′ , N, M ′′ ). Note that W ′′ is diffeomorphicto N × I .We wish to attach 3-handles along the collection { s k } and, later, 4-handles com-plimentary to those 3-handles. A priori , this may be impossible; for instance, thereis a framing issue. To make this possible, we borrow a trick from [15] to alter the2-spheres to a usable collection without changing the elements of H ( W ′ , N ; Z Q ) theyrepresent. GEOMETRIC REVERSE TO THE PLUS CONSTRUCTION 11
Claim 3.
For each s k , we may choose a second embedded 2-sphere t k with the propertythat • t k represents the same element of π ( M ′′ ) as s k (as elements of π ( W ′ ) , theywill be different) • each t k misses the attaching regions of all the { h i } , { k i } , { h j } , { k j } , { f i,j } and { g i,j }• the collection of { t k } are pair-wise disjoint and disjoint from the entire collec-tion { s k } Proof
Note that each canceling (2,3)-handle pairs h j and k j and f i,j and g i,j forman ( n + 1)-disks attached along an n -disk which is a regular neighborhood of a 2-diskfilling the attaching sphere of the 2-handle. Also, each canceling (1,2)-handle h i and k i forms an ( n + 1)-disk in N × { } attached along an n -disk which is a regularneighborhood of a 1-disk filling the attaching sphere of the 1-handle. We may push agiven s k off the (2,3)-handle pairs and then off the (1,2)-handle pairs, making sure notto pass back into the (2,3)-handle pairs. Let t k be the end result of the pushes. Makethe collection { t k } pair-wise disjoint and disjoint from the { s k } ’s by tranversality,making sure not to pass back into the (1,2)- or (2,3)-handle pairs. (cid:3) Replace each s k with s k − t k ), an embedded connected sum of s k with a copy of t k with its orientation reversed.Since the t k ’s miss all the handles attached to the original collar N × I , theycan be pushed into the right-hand copy of N . Thus, s k and s k − t k ) representthe same element of H ( W ′ , N ; Z Q ). Hence, the collection { s k − t k ) } is still a freebasis for H ( W ′ , N ; Z Q ). Furthermore, each s k − t k ) bounds an embedded 3-diskin the boundary of W ′′ . This means each s k − t k ) has a product neighborhoodstructure, and we may use it as the attaching region for a 3-handle h l . Choosethe framing of h l so that it is a trivially attached 3-handle with respect to W ′′ ,and choose a canceling 4-handle k l . We identify these 4-handles now, but do notattach them yet. They will be used later. Call the resulting cobordism with the h i , h j , f i,j , and h l attached ( W ′′′ , N, M ). Let W ( iv ) be M × I with the k i , k j , and k k ’s attached. Then W ′′′ S M W ( iv ) has all canceling handles and so is diffeomorphicto N × I . Clearly, W ′′′ S M W ( iv ) strong deformation retracts onto the right-handboundary N . Despite all the effort put into creating ( W ′′′ , M, N ), ( W ( iv , M, N ), or,more precisely, ( W ( iv , N, M ) (modulo torsion) will be seen to satisfy the conclusionof the theorem.We are note yet finished with ( W ′′′ , N, M ) yet. In order to prove ( W ( iv , M, N )satisfies the desired properties, we must study W ′′′ more carefully. Note that sinceker( ∂ ) is a free, finitely generated Z Q -module and { h k } is a set whose attachingspheres are a free Z Q -basis for ker( ∂ ), ∂ : C ( W ′′′ , N ; Z Q ) → C ( W ′′′ , N ; Z Q ) is ontoand has no kernel. This means H ( W ′′′ , N ; Z Q ) ∼ = 0. Clearly, H ∗ ( W ′′′ , N ; Z Q ) ∼ = 0for ∗ ≥ C ∗ ( W ′′′ , N ; Z Q ) ∼ = 0 for ∗ ≥ Thus, H ∗ ( W ′′′ , e N ; Z ) ∼ = 0, i.e., H ∗ ( W ′′′ , N ; Z Q ) ∼ = 0. (*)However, this is not the only homology complex we wish to prove acyclic; we alsowish to show that H ∗ ( W ′′′ , M ; Z Q ) ∼ = 0.By non-compact Poincare duality, we cando this by showing that the relative cohomology with compact supports is 0, i.e., H ∗ c ( W ′′′ , e N ; Z ) ∼ = 0.By the cohomology with compact supports, we mean to take the chain complexthat has linear functions f : C i ( W ′′′ , e N ; Z ) → Z from the relative handlebody complexof the intermediate cover of W ′′′ with respect to K to Z relative to e N , that is, thatsends all of the handles of the universal cover of N to 0 and that is nonzero ononly finitely many of the qh j ’s. The fact that δ is not well-defined, that is, that g has compact supports depends on the fact that C ∗ ( W ′′′ , e N ; Z ) is locally finite, whichin turn depends on the fact that W ′′′ is a covering space of a compact manifold,with finitely many handles attached.The co-boundary map δ ∗ will send a co-chain f in C ic ( W ′′′ , e N ; Z ) to the co-chain g in C i +1 c ( W ′′′ , e N ; Z ) which sends g ( ∂ ( n j q i h j ) to δ ( f )( n j q i h j ) for q i ∈ Q and n j h j ∈ C i ( W ′′′ , N ; Z ).Clearly, δ : C c ( W ′′′ , e N ; Z ) → C c ( W ′′′ , e N ; Z ) and δ : C c ( W ′′′ , e N ; Z ) → C c ( W ′′′ , e N ; Z ) are the zero maps. This means we must show ker( δ ) = 0, i.e., δ is1-1, and im ( δ ) = C , i.e., δ is onto. Finally, we must show exactness at C c , that is,we must show im ( δ ) = ker( δ ).Consider the acyclic complex0 ✲ C ( W ′′′ , e N ; Z ) ✲ C ( W ′′′ , e N ; Z ) ∂ ✲ C ( W ′′′ , e N ; Z ) ✲ ι : C ( W ′′′ , e N ; Z ) → C ( W ′′′ , e N ; Z ) with the property that ∂ ( C ) L ι ( C ) = C (ker( δ ) = 0) Let f ∈ C c ( W ′′′ , e N ; Z ) be non-zero, that is, let f : C ( W ′′′ , e N ; Z ) → c ∈ C ( W ′′′ , e N ; Z ) with c = 0 and f ( c ) = 0. As ∂ is onto, choose c ∈ C ( W ′′′ , e N ; Z ) with c = 0 and ∂ ( c ) = c . The δ ( f )( c ) = f ( ∂ ( c )) = f ( c ) = 0, and δ ( f ) is not the zero co-chain.( im ( δ ) = C ) Let g ∈ C c ( W ′′′ , e N ; Z ) be a basis element with g ( qh i ) = 1 and allother g ( q ′ h i ′ ) = 0. We must show there is an f ∈ C C ( W ′′′ , e N ; Z ) with δ ( f ) = g .Consider ∂ ( qh i ). This is a basis element for C ( W ′′′ , e N ; Z ).Choose f k,l ∈ C c ( W ′′′ , e N ; Z ) to have f k,l ( ∂ ( qh i )) = 1 and 0 otherwise. Then δ ( f )( q i h j ) = f ( ∂ ( q i h j )) = 1 = g ( qh i ).This proves δ ( f ) = g , and δ is onto.( im ( δ ) = ker( δ )) GEOMETRIC REVERSE TO THE PLUS CONSTRUCTION 13
Clearly, if f ∈ im ( δ ), then δ ( f ) = 0, as δ is a chain map.Suppose δ ( f ) = 0 but f = 0. Consider ι ( qh i ) = c ,i ∈ C ( W ′′′ , e N ; Z ). This is abasis element for C ( W ′′′ , e N ; Z ).Set g ( qh i ) = f ( c ,i ).Then δ ( g )( c ,i ) = g ( ∂ ( c ,j )) = g ( qh i ) = f ( c ,i ), and we are done.So, H ∗ C ( W ′′′ , e N ; Z ) ∼ = 0, so H ∗ ( W ′′′ , M ; Z ) ∼ = 0 by Theorem 3.35 in [16], and H ∗ ( W ′′′ , M ; Z ) ∼ = 0Note that we again have π ( N ) ∼ = Q , π ( W ′′′ ) ∼ = G , and ι : π ( M ) ∼ = π ( W ′′′ )an isomorphism, as attaching 3-handles does not affect π , and, dually, attaching( n − π for n ≥ W ( iv ) from right to left. This is (almost) the cobordism we desire. (Wewill need to deal with torsion issues below.) Note that the left-hand boundary of W ( iv ) read right to left is N and the right-hand boundary of W ( iv ) read right to leftis M . Moreover, W ( iv ) read right to left is N × I with [( n + 1) − n + 1) − n + 1) − n ≥
6, addingthese handles does not affect π ( W ( v ) ). Thus, we have ι : π ( N ) → π ( W ( v ) ) is anisomorphism; as was previously noted, π ( M ) ∼ = G .Let H : W ′′′ S M W ( iv ) → W ′′′ S M W ( iv ) a strong deformation retraction onto theright-hand boundary N . We will produce a retraction r : W ′′′ S M W ( iv ) → W ( iv ) .Then r ◦ H will restrict to a strong deformation retraction of W ( iv ) onto its right-hand boundary N . This, in turn, will yield a strong deformation retraction of W ( iv ) read right to left onto its left-hand boundary N .Note that by (*), H ∗ C ( W ′′′ , e N ; Z ) ∼ = 0. By Theorem 3.35 in [16], we have that H ∗ ( W ′′′ , M ; Z ) ∼ = 0, and H ∗ ( W ′′′ , M ; Z Q ) ∼ = 0, respectively, by the natural Z Q struc-ture on C ∗ ( W ′′′ ; Z ).To get the retraction r , we will use the following Proposition from [12]. Proposition 2.4.
Let ( X, A ) be a CW pair for which A ֒ → X induces a π iso-morphism. Suppose also that L E π ( A ) and A ֒ → X induces Z [ π ( A ) /L ] -homologyisomorphisms in all dimensions. Next suppose α , . . . , α k is a collection of loops in A that normally generates L . Let X ′ be the complex obtained by attaching a 2-cellalong each α l and let A ′ be the resulting subcomplex. Then A ′ ֒ → X ′ is a homotopyequivalence. (Note: In the above situation, we call A ֒ → X a mod L homotopyequivalence .) Since H ∗ ( W ′′′ , M ; Z Q ) = 0, we have that by Proposition 2.4, W ′′′ union the 2-handles f j strong deformation retracts onto M union the 2-handles f j . One maynow extend via the identity to get a strong deformation retraction r : W ′′′ S M S W ( iv ) → W ( iv ) . Now r ◦ H is the desired strong deformationretraction, of both W ( iv ) onto its right-hand boundary N and W ( iv ) read backwardsonto its left-hand boundary N .Now, suppose, for the cobordism ( W ( iv ) , N, M ), we have τ ( W ( iv ) , N ) = A = 0. Asthe epimorphism η : G → Q admits a left inverse ζ : Q → G , by the functoriality ofWhitehead torsion, we have that W h ( η ) : W h ( G ) → W h ( q ) is onto and admits a leftinverse W h ( ζ ) : W h ( q ) → W h ( G ). Let B have A + B = 0 in W h ( Q ) and set B ′ = W h ( ζ )( B ). By The Realization Theorem from [24], there is a cobordism ( R, M, N − )with τ ( R, M ) = B ′ . If W = ( W ( iv ) ∪ M R ), by Theorem 20.2 in [6], τ ( W, N ) = τ ( W ( iv ) , N ) + τ ( W, W ( iv ) ). By Theorem 20.3 in [6], τ ( W, W ( iv ) ) = W h ( η )( τ ( R, M )).So, τ ( W ( iv ) , N ) + W h ( η )( τ ( R, M ) = A + W h ( η )( B ′ ) = A + B = 0, and ( W, N, N − )is a 1-sided s-cobordism. (cid:3) Some Preliminaries to Creating Pseudo-CollarableHigh-Dimensional Manifolds
Our goal in this section is to display the usefulness of 1-sided s-cobordisms by usingthem to create large numbers of topologically distinct pseudo-collars (to be definedbelow), all with similar group-theoretic properties.We start with some basic definitions and facts concerning pseudo-collars.
Definition 5.
Let W n +1 be a 1-ended manifold with compact boundary M n . Wesay W is inward tame if W admits a co-final sequence of “clean” neighborhoods ofinfinity ( N i ) such that each N i is finitely dominated. [A neighborhood of infinity is a subspace the closure of whose complement is compact. A neighborhood of infinity N is clean if (1) N is a closed subset of W (2) N ∩ ∂W = ∅ (3) N is a codimension-0submanifold with bicollared boundary.] Definition 6.
A manifold N n with compact boundary is a homotopy collar if ∂N n ֒ → N n is a homotopy equivalence. Definition 7.
A manifold is a pseudo-collar if it is a homotopy collar whichcontains arbitrarily small homotopy collar neighborhoods of infinity. A manifold is pseudo-collarable if it contains a pseudo-collar neighborhood of infinity.
Pseudo-collars naturally break up as 1-sided h-cobordisms. That is, if N ⊆ N arehomotopy collar neighborhoods of infinity of an end of a pseudo-collarable manifold,the cl ( N \ N ) is a cobordism ( W, M, M − ), where M ֒ → W is a homotopy equivalence.Taking an decreasing chain of homotopy collar neighborhoods of infinity yields adecomposition of a pseudo-collar as a “stack” of 1-sided h-cobordisms. GEOMETRIC REVERSE TO THE PLUS CONSTRUCTION 15
Conversely, if one starts with a closed manifold M and uses the techniques of chap-ter 3 to produce a 1-sided h-cobordisms ( W , M, M − ), then one takes M − and againuses the techniques of Chapter 3 to produce a 1-sided h-cobordisms ( W , M − , M −− ),and so on ad infinitum, and then one glues W ∪ W ∪ . . . together to produce an( n + 1) − dimensional manifold N n +1 , then N is a pseudo-collar.So, 1-sided h-cobordisms are the “correct” tool to use when constructing pseudo-collars. Definition 8.
The fundamental group system at ∞ , π ( ǫ ( X ) , r ) , of an end ǫ ( X ) of a non-compact topological space X , is defined by taking a cofinal sequence ofneighborhoods of ∞ of the end of X , N ⊇ N ⊇ N ⊇ . . . , , a proper ray r : [0 , ∞ ) → X , and looking at its related inverse sequence of fundamental groups π ( N , p ) ← π ( N , p ) ← π ( N , p ) ← . . . (where the bonding maps are induced by inclusion andthe basepoint change isomorphism, induced by the ray r ). Such a fundamental group system at infinity has a well-defined associated pro-fundamental group system at infinity, given by its equivalence class inside the categoryof inverse sequences of groups under the below equivalence relation.
Definition 9.
Two inverse sequences of groups ( G i , α i ) and ( H i , β i ) are said to be pro-isomorphic if there exists subsequences of each, which may be fit into a com-muting ladder diagram as follows: G i ✛ α i G i ✛ α i G i ✛ α i G i ✛ α j . . .H j ✛ β j ✛ g i ✛ f j H j ✛ β j ✛ g i ✛ f j H j ✛ β j ✛ g i ✛ f j H j ✛ β j . . . ✛ f j A more detailed introduction to fundamental group systems at infinity can be foundin [9] or [10].
Definition 10.
An inverse sequence of groups is stable if is it pro-isomorphic to aconstant sequence G ← G ← G ← G . . . with the identity for bonding maps.
The following is a theorem of Brown from [4].
Theorem 3.1.
The boundary of a manifold M is collared, i.e., there is a neighborhood N of ∂M in M such that N ≈ ∂M × I . The following is from Siebenmann’s Thesis, [25].
Theorem 3.2.
An open manifold W n +1 ( n ≥ ) admits a compactification as an n + 1 -dimensional manifold with an n -dimensional boundary manifold M n if(1) W is inward tame(2) π ( ǫ ( W )) is stable for each end of W , ǫ ( W ) (3) σ ∞ ( ǫ ( W )) ∈ e K [ Z π ( ǫ ( W ))] vanishes for each end of W , ǫ ( W ) Definition 11.
An inverse sequence of groups is semistable or Mittag-Leffler ifis it pro-isomorphic to a sequence G և G և G և G . . . with epic bonding maps. Definition 12.
An inverse sequence of finitely presented groups is perfectlysemistable if and only if is it pro-isomorphic to a sequence G և G և G և G . . . with epic bonding maps and perfect kernels. The following two lemmas show that optimally chosen perfectly semistable inversesequences behave well under passage to subsequences.
Lemma 3.1.
Let ✲ K ι ✲ G σ ✲ Q ✲ be a short exact sequence of groups with K, Q perfect. Then G is perfect. Proof
Follows from Lemma 1 in [11]. Let g ∈ G . Then σ ( g ) ∈ Q , so σ ( g ) =Π ki =1 [ x i , y i ] , x i , y i ∈ Q , as Q is perfect. But, now, σ is onto, ∃ u i ∈ G with σ ( u i ) = x i and v i ∈ G with σ ( v i ) = y i . Set g ′ = Π ki =1 [ u i , v i ]. Then σ ( g · ( g ′ ) − ) = σ ( g ) · σ ( g ′ ) − = Π ki =1 [ x i , y i ] · (Π ki =1 [ x i , y i ]) − = 1 ∈ Q. Thus, g · ( g ′ ) − ∈ ι ( K ), and ∃ r j , s j ∈ K with g · ( g ′ ) − = ι (Π lj =1 [ r j , s j ], as K isperfect. But, finally, g = [ g · ( g ′ ) − ] · g ′ = Π lj =1 [ ι ( r j ) , ι ( s j )] · Π ki =1 [ u i , v i ], which proves g ∈ [ G, G ]. (cid:3) Lemma 3.2. If α : A → B and β : B → C are both onto and have perfect kernels,the ( β ◦ α ) : A → C is onto and has perfect kernel. Proof
It suffices to show the composition has perfect kernel. Set K = ker( α ) , Q =ker( β ) , G = ker( β ◦ α ) Claim 4. K = ker( α | G ) : G → B Proof ( ⊆ ) Let g ∈ G have α ( g ) = e ∈ B Then g ∈ A and α ( g ) = e ∈ B , so G ∈ K ( ⊇ ) Let k ∈ K . Then α ( k ) = e ∈ B , so β ( α ( k )) = β ( e ) = e ∈ Q . Thus( β ◦ α )( k ) = e ∈ C , and k ∈ G . Since α ( k ) = e ∈ B , this shows k ∈ ker( α | G ). (cid:3) This completes the proof. (cid:3)
The following is a result from [14].
Theorem 3.3 (Guilbault-Tinsley) . A non-compact manifold W n +1 with compact(possibly empty) boundary ∂W = M is pseudo-collarable if and only if(1) W is inward tame(2) π ( ǫ ( W )) is perfectly semistable for each end of W , ǫ ( W ) (3) σ ∞ ( ǫ ( W )) ∈ e K [ Z π ( ǫ ( W ))] vanishes for each end of W , ǫ ( W ) GEOMETRIC REVERSE TO THE PLUS CONSTRUCTION 17
So, the pro-fundamental group system at infinity of a pseudo-collar is perfectlysemistable. As is outlined in Chapter 4 of [10], the pro-fundamental group system atinfinity is independent of base ray for ends with semistable pro-fundamental group atinfinity, and hence for 1-ended pseudo-collars.
Theorem 1.2 (Uncountably Many Pseudo-Collars on Closed Manifolds with the SameBoundary and Similar Pro- π ) . Let M n be a closed smooth manifold ( n ≥ ) with π ( M ) ∼ = Z and let S be the finitely presented group V ∗ V , which is the free productof 2 copies of Thompson’s group V . Then there exists an uncountable collection ofpseudo-collars { N n +1 ω | ω ∈ Ω } , no two of which are homeomorphic at infinity, andeach of which begins with ∂N n +1 ω = M n and is obtained by blowing up countably manytimes by the same group S . In particular, each has fundamental group at infinity thatmay be represented by an inverse sequence Z ✛✛ α G ✛✛ α G ✛✛ α G ✛✛ α . . . with ker( α i ) = S for all i . We give a brief overview of our strategy. For convenience, we will start with themanifold S × S n − , which has fundamental group Z . We let S be the free product of2 copies of Thompson’s group V , which is a finitely presented, superperfect group forwhich Out ( S ) has torsion elements of all orders (see [19]). Then we will blow Z upby S to semi-direct products G p , G p , G p , ..., in infinitely many different ways usingdifferent outer automorphisms φ p i of prime order. We will then use the theorem oflast chapter to blow up S × S n − to a manifolds M p , M p , M p , ..., by cobordisms W p , W p , W p , ... We will then use different automorphisms, each with order a primenumber strictly greater than the prime order used in the last step, from the infinitegroup Out ( S ) to blow up each of G p , G p , G p , ..., to a different semi-direct productsby S , and will then use the theorem of last chapter to extend each of W p , W p , W p ,..., in infinitely many different ways.Continuing inductively, we will obtain increasing sequences ω of prime numbersdescribing each sequence of 1-sided s-cobordisms. We will then glue together all thesemi-s-cobordisms at each stage for each unique increasing sequence of prime numbers ω , creating for each an ( n + 1)-manifold N n +1 ω , and show that there are uncountablymany such pseudo-collared ( n + 1)-manifolds N ω , one for each increasing sequence ofprime numbers ω , all with the same boundary S × S n − , and all the result of blowingup Z to a semi-direct product by copies of the same superperfect group S at eachstage. The fact that no two of these pseudo-collars are homeomorphic at infinity willfollow from the fact that no two of the inverse sequences of groups are pro-isomorphic.Much of the algebra in this chapter is aimed at proving that delicate result. Remark 1.
There is an alternate strategy of blowing up each the fundamental group G i at each stage by the free product G i ∗ S i ; using a countable collection of freely indecomposible kernel groups { S i } would then allow us to create an uncountable col-lection of pseudo-collars; an algebraic argument like that found in [26] or [7] wouldcomplete the proof. However, they would not have the nice kernel properties that ourconstruction has. It seems likely that other groups than Thompson’s group V would work for thepurpose of creating uncountably many pseudo-collars, all with similar group-theoreticproperties, from sequences of 1-sided s-cobordisms. But, for our purposes, V possessesthe ideal set of properties.4. Some Algebraic Lemmas, Part 1
In this section, we go over the main algebraic lemmas necessary to do our strategyof blowing up the fundamental group at each stage by a semi-direct product with thesame superperfect group S .Thompson’s group V is finitely presented, superperfect, simple, and contains tor-sion elements of all orders. Note that simple implies V is centerless, Hopfian, andfreely indecomposable.An introduction to some of the basic properties of Thompson’s group V can befound in [19], There, it is shown that V is finitely presented and simple. It is alsonoted in [19] that V contains torsion elements of all orders, as V contains a copy ofevery symmetric group on n letters, and hence of every finite group. In [3], it is notedthat V is superperfect. We give proofs of some of the simpler properties. Lemma 4.1.
Every non-Abelian simple group is perfect
Proof
Let G be a simple, non-Abelian group, and consider the commutator sub-group K of G . This is not the trivial group, as G is non-Abelian, and so by simplicity,must be all of G . This shows every element of G can be written as a product of com-mutator of elements of G , and so G is perfect. (cid:3) Definition 13.
A group G is Hopfian if every onto map from G to itself is anisomorphism. Equivalently, a group is Hopfian if it is not isomorphic to any of itsproper quotients. Lemma 4.2.
Every simple group is Hopfian.
Proof
Clearly, the trivial group is Hopfian. So, let G be a non-trivial simple group.Then the only normal subgroups of G are G itself and h e i , so the only quotients of G are h e i and G , respectively. So, the only proper quotient of G is h e i , which cannotbe isomorphic to G as G is nontrivial. (cid:3) Let S = P ∗ P be the free product of 2 copies of V with itself. This is clearly finitelypresented, perfect (by Meyer-Vietoris), and superperfect (again, by Meyer-Vietoris).Note that S is a free product of non-trivial groups, so S is centerless. In [18], it isnoted that free products of Hofpian, finitely presented, freely indecomposable groups GEOMETRIC REVERSE TO THE PLUS CONSTRUCTION 19 are Hopfian, so S = V ∗ V is Hopfian. S (and not V itself) will be the superperfectgroup we use in our constructions.We need a few lemmas. Lemma 4.3.
Let
A, B, C, and D be non-trivial groups. Let φ : A × B → C ∗ D be asurjective homomorphism. Then one of φ ( A × { } ) and φ ( { } × B ) is trivial and theother is all of C ∗ D Proof
Let x ∈ φ ( A × { } ) ∩ φ ( { } × B ). Then x ∈ φ ( A × { } ), so x commutes witheverything in φ ( { } × B ). But x ∈ φ ( { } × B ), so x commutes with everything in φ ( A × { } ). As φ is onto, this implies φ ( A × { } ) ∩ φ ( { } × B ) ≤ Z ( C ∗ D ).But, by a standard normal forms argument, the center of a free product is trivial!So, φ ( A × { } ) ∩ φ ( { } × B ) ≤ Z ( C ∗ D ) = 1. However, this implies that φ ( A ×{ } ) × φ ( { } × B ) = C ∗ D . By a result in [2], a non-trivial direct product cannot bea non-trivial free product. (If you’d like to see a proof using the Kurosh SubgroupTheorem, that can be found in many group theory texts, such as Theorem 6.3.10 of[23]. An alternate, much simpler proof due to P.M. Neumann can be found in [21] inthe observation after Lemma IV.1.7). Thus, φ ( A ×{ } ) = C ∗ D or φ ( { }× B ) = C ∗ D and the other is the trivial group. The result follows. (cid:3) Corollary 4.4.
Let A , . . . , A n be non-trivial groups and let C ∗ D be a free productof non-trivial groups. Let φ : A × . . . × A n → C ∗ D be a surjective homomorphism.Then one of the φ ( { } × . . . A i × . . . × { } ) is all of C ∗ D and the rest are all trivial. Proof
Proof is by induction.( n = 2) This is Lemma 4.3.(Inductive Step) Suppose the result is true for n −
1. Set B = A × . . . × A n − . ByLemma 4.3, either φ ( B × { } ) is all of C ∗ D and φ ( { } × A n ) is trivial or φ ( B × { } )is trivial and φ ( { } × A n ) is all of C ∗ D .If φ ( B × { } ) is trivial and φ ( { } × A n ) is all of C ∗ D , we are done.If φ ( B × { } ) is all of C ∗ D and φ ( { } × A n ) is trivial, then, by the inductivehypothesis, we are also done. (cid:3) Corollary 4.5.
Let S , S , . . . , S n all be copies of the same non-trivial free product,and let ψ : S × S × . . . × S n → S × S × . . . × S n be a isomorphism. Then ψ decomposes as a “matrix of maps” ψ i,j , where each ψ i,j = π S j ◦ ψ | S i (where π S j isprojection onto S j ), and there is a permutation σ on n indices with the property thateach ψ σ ( j ) ,j : S σ ( j ) → S j is an isomorphism, and all other ψ i,j ’s are the zero map. Proof
By Lemma 4.4 applied to π S j ◦ ψ , we clearly have a situation where each π S j ◦ ψ | S i is either trivial or onto. If we use a schematic diagram with an arrow from S i to S j to indicate non-triviality of a map ψ i,j , we obtain a diagram like the following. S × S × S × S × S × S × S × . . . S n . . .S ❄ × S ✲ × S ✲ × S ❄ × S ❄ × S ✛ × S ✲ × ✛ . . . ✛ S n ❄ where a priori some of the S i ’s in the domain may map onto multiple S j ’s in thetarget, and there are no arrows emanating from some of the S i ’s in the domain.By the injectivity of ψ , there must be at least one arrow emanating from each S i , while by surjectivity of ψ , there must be at least one arrow ending at each S j .Corollary 4.4 prevents more than one arrow from ending in a given S j . By thePidgeonhole Principle, the arrows determine a one-to-one correspondence betweenthe factors in the domain and those in the range. A second application of injectivitynow shows each arrow represents an isomorphism. (cid:3) Note that the ψ i,j ’s form a matrix where each row and each column contain exactlyone isomorphism, and the rest of the maps are trivial maps - what would be a permu-tation matrix (see page 100 in [22], for instance) if the isomorphisms were replacedby “1”’s and the trivial maps were replaced by “0”’s. Corollary 4.6.
Let S , S , . . . , S n all be copies of the same non-trivial Hopfian freeproduct, and let ψ : S × S × . . . × S n → S × S × . . . × S m be a epimorphism with m < n . Then ψ decomposes as a “matrix of maps” ψ i,j = π S j ◦ ψ | S i , and there isa 1-1 function σ from the set { , . . . , m } to the set { , . . . , n } with the property that ψ σ ( j ) ,j : S σ ( j ) → S j is an isomorphism, and all other ψ i,j ’s are the zero map. Proof
Begin with a schematic arrow diagram as we had in the previous lemma. Bysurjectivity and Lemma 4.4, each of the m factors in the range is at the end of exactly1 arrow. From there, we may conclude that each arrow represents an epimorphism,and, hence, by Hopfian, an isomorphism.To complete the proof, we must argue that at most one arrow can emanate from an S i factor. Suppose to the contrary, that two arrows emanate from a given S i factor.Then we have an epimorphism of S i onto a non-trivial direct product in which eachcoordinate function is a bijection. This is clearly impossible. (cid:3) Some Algebraic Lemmas, Part 2
Let Ω be the uncountable set consisting of all increasing sequences of prime numbers( p , p , p , . . . ) with p i < p i +1 . For ω ∈ Ω and n ∈ N , define ( ω, n ) to be the finitesequence consisting of the first n entries of ω . GEOMETRIC REVERSE TO THE PLUS CONSTRUCTION 21
Let p i denote the i th prime number, and for the group S = P ∗ P , where each P i is Thompson’s group V , choose u i ∈ P to have order ( u i ) = p i .Recall, if K is a group, Aut ( K ) is the automorphism group of K . Define µ : K → Aut ( K ) to be µ ( k )( k ′ ) = kk ′ k − . Then the image of µ in Aut ( K ) is called the innerautomorphism group of K , Inn ( K ). The inner automorphism group of a group K isalways normal in Aut ( K ). The quotient group Aut ( K ) /Inn ( K ) is called the outerautomorphism group Out ( K ). The kernel of µ is called the center of K , Z ( K ); it isthe set of all k ∈ K such that for all k ′ ∈ K, kk ′ k − = k ′ . One has the exact sequence1 ✲ Z ( K ) ✲ K µ ✲ Aut ( K ) α ✲ Out ( K ) ✲ P → Out ( P ∗ P ) by Φ( u ) = φ u , where φ u ∈ Out ( P ∗ P ) is theouter automorphism defined by the automorphism φ u ( p ) = ( p if p ∈ P upu − if p ∈ P ( φ u is called a partial conjugation .) Claim 5.
Φ : P → Out ( P ∗ P ) is an embedding Proof
Suppose Φ( u ) is an inner automorphism for some u not e in P . Since Φ( u )acts on P by conjugation by u , to be an inner automorphism, Φ( u ) must also act on P by conjugation by u . Now, Φ( u ) acts on P trivially for all p ∈ P , which implies u is in the center of P . But P is centerless! Thus, no Φ( u ) is an inner automorphismfor any u ∈ P . (cid:3) So, for each u i with prime order the i th prime p i , φ u i has prime order p i , as doesevery conjugate of φ u i in Out ( P ∗ P ), as Φ is an embedding. Lemma 5.1.
For any finite collection of groups A , A , . . . , A n , Π ni =1 Out ( A i ) embedsin Out (Π ni =1 A i ) . Proof
The natural map from Π ni =1 Aut ( A i ) to Aut (Π ni =1 A i ) which sends a Cartesianproduct of automorphism individually in each factor to that product considered asan automorphism of the product is clearly an embedding. Now, Inn ( A × . . . × A n )is the image under this natural map of Inn ( A i ) × . . . × Inn ( A n ), because if b i ∈ A i ,then ( b , . . . , b n ) − ( a , . . . , a n )( b , . . . , b n ) = ( b − a b , . . . , b − n a n b n ). So, the inducedmap on quotient groups, from Π ni =1 Out ( A i ) to Out (Π ni =1 A i ), is also a monomorphism. (cid:3) Now, because the quotient map Ψ : Π ni =1 Out ( A i ) → Out (Π ni =1 A i ) is an embedding, order ( φ , . . . , φ n ) in Out (Π ni =1 A i ) is just lcm ( order ( φ ) , . . . , order ( φ n )), which is justits order in Π ni =1 Out ( A i ). Moreover each conjugate of ( φ , . . . , φ n ) in Out (Π ni =1 A i )has the same order lcm ( φ , . . . , φ n ). Finally, note that if each φ i has prime order andeach prime occurs only once, then order ( φ , . . . , φ n ) = order ( φ ) × . . . × order ( φ n ). Lemma 5.2.
Let K be a group and suppose Θ : K ⋊ φ Z → K ⋊ ψ Z is an isomorphismthat restricts to an isomorphism Θ : K → K . Then φ and ψ are conjugate as elementsof Out ( K ) Proof
We use the presentations h gen ( K ) , a | rel ( K ) , ak i a − = φ ( k i ) i and h gen ( K ) , b | rel ( K ) , bkb − = ψ ( k ) i of the domain and range respectively, Since Θinduces an isomorphism on the infinite cyclic quotients by K , there exists c ∈ K withΘ( a ) = cb ± . We assume Θ( a ) = cb , with the case Θ( a ) = cb − being similar.For each k ∈ K , we haveΘ( φ ( k )) = Θ( aka − )= Θ( a )Θ( k )Θ( a ) − = cb Θ( k ) b − c − = cψ (Θ( k )) c − If we let ι c : K → K denote conjugation by c , we have Θ φ = ι c ψ Θ in
Aut ( K ).Quotienting out by Inn ( K ) and abusing notation slightly, we have Θ φ = ψ Θ orΘ φ Θ − = ψ in Out ( K ). (cid:3) Lemma 5.3.
For any finite, strictly increasing sequence of primes ( s , s , . . . , s n ) , de-fine φ ( s ,...,s n ) : S × . . . × S n → S × . . . S n by φ ( s ,...s n ) ( x , . . . , x n ) = ( φ u ( x ) , . . . , φ u n ( x n )) ,where φ u i is the partial conjugation outer automorphism associated above to the ele-ment u i with prime order s i . Let ( s , . . . , s n ) and ( t i , . . . , t n ) be increasing sequencesof prime numbers of length n . Let G ( s ,...,s n ) = ( S × . . . × S n ) ⋊ φ ( s ,...,sn ) Z and G ( t i ,...,t n ) = ( S × . . . × S n ) ⋊ φ ( t ,...,tn ) Z be two semidirect products with such outeractions. Then G ( s ,...,s n ) is isomorphic to G ( t i ,...,t n ) if and only if for the underlyingsets { s , . . . , s n } = { t , . . . , t n } . Proof ( ⇒ ) Let θ : G ( s ,...,s n ) → G ( t i ,...,t n ) be an isomorphism. There are n factorsof S in the kernel group of each of G ( ω,n ) and G ( η,n ) . Then θ must preserve thecommutator subgroup, as the commutator subgroup is a characteristic subgroup, andso induces an isomorphism of the perfect kernel group K = S × S × . . . × S n , say θ .By Corollary 4.5, it must permute the factors of K , say via σ .Now, the isomorphism θ must take the (infinite cyclic) abelianization G ( s ,...,s n ) /K ( s ,...,s n ) of the one to the (infinite cyclic) abelianization G ( t i ,...,t n ) /K ( t i ,...,t n ) of the other, and hence takes a generator of G ( s ,...,s n ) /K ( s ,...,s n ) (say aK ( s ,...,s n ) ) to a generator of G ( t i ,...,t n ) /K ( t i ,...,t n ) (say b e K ( t i ,...,t n ) , where bK ( t i ,...,t n ) is a given generator of G ( t i ,...,t n ) /K ( t i ,...,t n ) and e = ± θ takes K ( s ,...,s n ) = [ G ( s ,...,s n ) , G ( s ,...,s n ) ] to [ G ( t i ,...,t n ) , G ( t i ,...,t n ) ] = K ( t i ,...,t n ) ,it follows that θ takes a to a multiple of b e , say c − b e where c lies in K ( t i ,...,t n ) and e = ± φ ( s ,...,s n ) is conjugate in Out ( K ) to φ ( t ,...,t n ) , θ ( φ ( s ,...,s n ) ) θ − = φ ( t ,...,t n ) .But Ψ is an embedding by Lemma 5! This shows that order ( φ ( s ,...,s n ) ) = Π ni =1 s i and order ( φ ( t ,...,t n ) ) = Π ni =1 t i are equal, so, as each s i and t i is prime and occurs GEOMETRIC REVERSE TO THE PLUS CONSTRUCTION 23 only once in each increasing sequence, by the Fundamental Theorem of Arithmetic, { s , . . . , s n } = { t , . . . , t n } ( ⇐ ) Clear. (cid:3) Lemma 5.4.
Let ( ω, n ) = ( s , . . . , s n ) and ( η, m ) = ( t , . . . , t m ) be increasing se-quences of prime numbers with n > m .Let G ( ω,n ) = ( S × . . . × S n ) ⋊ φ ( ω,n ) Z and G ( η,m ) = ( S × . . . × S m ) ⋊ φ ( η,m ) Z be twosemidirect products. Then there is an epimorphism g : G ( ω,n ) → G ( η,m ) if and only if { t , . . . , t m } ⊆ { s , . . . , s n } . Proof
The proof in this case is similar to the case n = m , except that the epi-morphism g must crush out n − m factors of K ( ω,n ) = S × . . . × S n by Corollary 4.6and the Pidgeonhole Principle and then is an isomorphism on the remaining factors.( ⇒ ) Suppose there is an epimorphism g : G ( ω,n ) → G ( η,m ) . Then g must send thecommutator subgroup of G ( ω,n ) onto the commutator subgroup of G ( η,m ) . By Corollary4.6, g must send m factors of K ( ω,n ) = S × . . . × S n in the domain isomorphicallyonto the m factors of K ( η,m ) = S × . . . × S m in the range and sends the remaining n − m factors of K ( ω,n ) to the identity. Let { i , . . . , i m } be the indices in { , . . . , n } offactors in K ( ω,n ) which are sent onto a factor in K ( η,m ) and let { j , . . . , j n − m } be theindices in { , . . . , n } of factors in K ( ω,n ) which are sent to the identity in K ( η,m ) . Then g induces an isomorphism between S i × . . . × S i m and K ( η,m ) . Set L m = S i × . . . × S i m Also, by an argument similar to Lemmas 5.2 and 5.3, g sends sends the infinite cyclicgroup G ( ω,n ) /K ( ω,n ) isomorphically onto the infinite cyclic quotient G ( η,m ) /K ( η,m ) .Note that L m ⋊ φ ( si ,...,sim ) Z is a quotient group of G ( ω,n ) by a quotient map whichsends S j × . . . × S j n − m to the identity. Consider the induced map g ′ : L m ⋊ φ ( si ,...,sim ) Z → G ( η,m ) . By the facts that g ′ maps L m isomorphically onto K ( η,m ) and preservesthe infinite cyclic quotients, we have that the kernel of g must equal exactly S j × . . . × S j n − m ; thus, by the First Isomorphism Theorem, we have that g ′ is an isomorphism.Finally, g ′ is an isomorphism of L m ⋊ φ ( si ,...,sim ) Z with G ( ω,n ) which restricts toan isomorphism of L m with S t × . . . × S t m , so, by Lemma 5.2, we have φ ( s i ,...,s im ) is conjugate to φ ( t ,...,t m ) , so, in Out (Π ni =1 A ), order ( φ ( s i ,...,s im ) ) = order ( φ ( t ,...,t m ) ),and thus, as each s i and t i is prime and appears at most once, by an argumentsimilar to Lemma 5.3 using the Fundamental Theorem of Arithmetic, { t , . . . , t m } ⊆{ s , . . . , s n } .( ⇐ ) Suppose { t , . . . , t m } ⊆ { s , . . . , s n } . Choose a ∈ G ( ω,n ) with aK ( ω,n ) gener-ating the infinite cyclic quotient G ( ω,n ) /K ( ω,n ) and choose b ∈ G ( η,m ) with bK ( η,m ) generating the infinite cyclic quotient G ( η,m ) /K ( η,m ) . Set g ( a ) = b .Send each element of S i (where S i uses an element of order t i in its semidirectproduct definition in the domain) to a corresponding generator of S i (where S i uses an element of order t i in its semidirect product definition in the range) under g . Sendthe elements of all other S j ’s to the identity.Then g : G ( ω,n ) → G ( η,m ) is an epimorphism. Clearly, g is onto by construction. Itremains to show g respects the multiplication in each group.Clearly, g respects the multiplication in each S i and in Z Finally, if α i ∈ S i and a ∈ Z , g ( aα i ) = g ( a ) g ( α i ) g ( φ s i ( α i ) a ) = φ t i ( g ( α i )) g ( a )using the slide relators for each group and the fact that s i = t i , which implies φ s i = φ t i . So, g respects the multiplication in each group. This completes the proof. (cid:3) Some Algebraic Lemmas, Part 3
Recall Ω is an uncountable set consisting of increasing sequences of prime numbers( p , p , p , . . . ) with p i < p i +1 . For ω ∈ Ω and n ∈ N , recall we have defined ( ω, n ) tobe the finite sequence consisting of the first n entries of ω .Recall also that p i denotes the i th prime number, and for the group S = P ∗ P ,where each P i is Thompson’s group V , we have chosen u i ∈ P to have order ( u i ) = p i .Recall finally we have define a map Φ : P → Out ( P ∗ P ) (where each P i is acopy of Thompson’s group V ) by Φ( u ) = φ u , where φ u ∈ Out ( P ∗ P ) is the outerautomorphism defined by the automorphism φ u ( p ) = ( p if p ∈ P upu − if p ∈ P (Recall φ u is called a partial conjugation .)Set G ( ω,n ) = ( S × S × . . . × S ) ⋊ φ ( ω,n ) Z . Lemma 6.1. G ( ω,n ) ∼ = S ⋊ φ wsn G ( ω,n − , where φ w sn is partial conjugation by u s n . Proof
First, note that there is a short exact sequence1 ✲ S ι ✲ G ( ω,n ) α n ✲ G ( ω,n − ✲ ι takes S identically onto the n th factor, and α crushes out factor, as describedin Lemma 5.4.Next, note that there is a left inverse j : G ( ω,n − → G ( ω,n ) to α given by (1) sendingthe generator a of the Z from its image γ n − ( a ) in the semi-direct product GEOMETRIC REVERSE TO THE PLUS CONSTRUCTION 25 ✲ ( S × . . . × S ) ι n − ✲ G ( ω,n − β n − ✲ Z ✲ γ n − is a left inverse to β n − , to its image γ n ( a ) in1 ✲ ( S × . . . × S ) ι n ✲ G ( ω,n ) β n ✲ Z ✲ γ n is a left inverse to β n and (2) sending each of the images ι n − ( t i ) of the elements t i of the S i associated with φ w si in G ( ω,n − in1 ✲ ( S × . . . × S ) ι n − ✲ G ( ω,n − β n − ✲ Z ✲ ι n ( t i ) of the elements t i of the S i associated with φ s i in G ( ω,n ) in1 ✲ ( S × . . . × S ) ι n ✲ G ( ω,n ) β n ✲ Z ✲ i ∈ { , . . . , n − } The existence of a left inverse proves the group extension is a semi-direct product.The needed outer action for the final copy of S in G ( ω,n ) may now be read off thedefining data for G ( ω,n ) in the definition G ( ω,n ) = ( S × S × . . . × S ) ⋊ φ ( ω,n ) Z , showingthat it is indeed partial conjugation by u s n .(Alternately, one may note there is a presentation for ( S × S × . . . × S ) ⋊ φ ( ω,n ) Z that contains a presentation for S ⋊ φ wsn G ( ω,n − Generators: z , the generator of Z , together with the generators of the first copy of S , the generators of the second copy of S , ..., and the generators of the n th copy of S .Relators defining P i ’s: the relators for the copy of P in the first copy of S , therelators for the copy of P in the first copy of S , the relators for the copy of P in thesecond copy of S, the relators for the copy of P in the second copy of S , , ..., and therelators for the copy of P in the n th copy of S , the relators for the copy of P in the n th copy of S .Slide Relators: The slide relators between z and the generators of P in the first copyof S due to the semi-direct product, the slide relators between z and the generatorsof P in the second copy of S due to the semi-direct product, ..., the slide relatorsbetween z and the generators of P in the n th copy of S due to the semi-direct product,and the slide relators between z and the generators of P in the n th copy of S due tothe semi-direct product.) (cid:3) Now, this way of looking at G ( ω,n ) as a semi-direct product of S with G ( ω,n − yieldsan inverse sequence ( G ( ω,n ) , α n ), which looks like G ( ω, ✛ α G ( ω, ✛ α G ( ω, ✛ α . . . with bonding maps α i : G ( ω,i +1) → G ( ω,i ) that each crush out the most recentlyadded copy of S .A subsequence will look like G ( ω,n ) ✛ α n G ( ω,n ) ✛ α n G ( ω,n ) ✛ α n . . . with bonding maps α n i : G ω,n j ) → G ω,n i ) that each crush out the most recentlyadded n j − n i copies of S . Lemma 6.2.
If, for inverse sequences ( G ( ω,n ) , α n ) , where α n : G ( ω,n ) → G ( ω,n − isthe bonding map crushing out the most recently-added copy of S , ω does not equal η ,then the two inverse sequences are not pro-isomorphic. Proof
Let ( G ( ω,n ) , α n ) and ( G ( η,m ) , β m ) be two inverse sequences of group exten-sions, assume there exists a commuting ladder diagram between subsequences of thetwo, as shown below. By discarding some terms if necessary, arrange that ω and η do not agree beyond the term n . G ( ω,n ) ✛ α G ( ω,n ) ✛ α G ( ω,n ) ✛ α . . .. . .G ( η,m ) ✛ β ✛ g n ✛ f m G ( η,m ) ✛ β ✛ g n ✛ f m . . . By the commutativity of the diagram, all f ’s and g ’s must be epimorphisms, as allthe α ’s and β ’s are.Now, it is possible that g ( ω,n ) is an epimorphism; by Lemma 5.4, ( η, m ) might be asubset of ( ω, n ) when considered as sets. But, f ( η,m ) cannot also be an epimorphism,since ( ω, n ) cannot be a subset of ( η, m ) when considered as sets. Since the twosequences can only agree up to n , if ( η, m ) is a subset of ( ω, n ) when consideredas sets, then there must be an prime p i in ( ω, n ) in between some of the primesof ( η, m ). This prime p i now cannot be in ( η, m ) and is in ( ω, n ), so we cannothave ( ω, n ) a subset of ( η, m ) when considered as sets, so f ( η,m ) cannot be anepimorphism. (cid:3) GEOMETRIC REVERSE TO THE PLUS CONSTRUCTION 27 Manifold Topology
We now begin an exposition of our example.
Theorem 1.2 (Uncountably Many Pseudo-Collars on Closed Manifolds with the SameBoundary and Similar Pro- π ) . Let M n be a closed smooth manifold ( n ≥ ) with π ( M ) ∼ = Z and let S be the finitely presented group V ∗ V , which is the free productof 2 copies of Thompson’s group V . Then there exists an uncountable collection ofpseudo-collars { N n +1 ω | ω ∈ Ω } , no two of which are homeomorphic at infinity, andeach of which begins with ∂N n +1 ω = M n and is obtained by blowing up countably manytimes by the same group S . In particular, each has fundamental group at infinity thatmay be represented by an inverse sequence Z ✛✛ α G ✛✛ α G ✛✛ α G ✛✛ α . . . with ker( α i ) = S for all i . Proof
For each element ω ∈ Ω, the set of all increasing sequences of prime num-bers, we will construct a pseudo-collar N n +1 ω whose fundamental group at infinity isrepresented by the inverse sequence ( G ( ω,n ) , α ( ω,n ) ). By Lemma 6.2, no two of thesepseudo-collars can be homeomorphic at infinity, and the Theorem will follow.To form one of the pseudo-collars, start with M = S × S n − with fundamentalgroup Z and then blow it up, using Theorem 1.1, to a cobordism ( W ( s ) , M, M ( s ) )corresponding to the group G ( s ) ( s a prime)..We then blow this right-hand boundaries up, again using Theorem 1.1 and Lemma6.1, to cobordisms ( W ( s ,s ) , M ( s ) , M ( s ,s ) ) corresponding to the group G ( s ,s ) above.We continue in the fashion ad infinitum .The structure of the collection of all pseudo-collars will be the set Ω describedabove.We have shown that the pro-fundamental group systems at infinity of each pseudo-collar are non-pro-isomorphic in Lemma 6.2, so that all the ends are non-diffeomorphic(indeed, non-homeomorphic).This proves we have uncountably many pseudo-collars, each with boundary M ,which have distinct ends. (cid:3) Remark 2.
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GEOMETRIC REVERSE TO THE PLUS CONSTRUCTION 29
Department of Mathematics, Computer Science, and Statistics, Marquette Uni-versity, Milwaukee, Wisconsin 53201
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