Homotopy lifting property for actions of finite abelian groups on Hausdorff topological spaces
HHOMOTOPY LIFTING PROPERTY FOR ACTIONS OFFINITE ABELIAN GROUPS ON HAUSDORFFTOPOLOGICAL SPACES
EDUARDO BLANCO-G ´OMEZ
Abstract.
In this paper we prove the homotopy lifting propertyfor actions of finite abelian groups on Hausdorff topological spaces. Introduction
Lifting a continuous function between topological spaces is a questionthat depends on the topological properties of the two spaces. There areknown cases where the lifting can be done relatively easy. One of thesesituations is when the continuous function is defined over a coveringspace (cid:101) X of a topological space X (see Proposition 1.30 page 69 of [6] andTheorem 17.6 page 157 of [7]). In that case one can prove the importanthomotopy lifting property, but even there, lifting a continuous functionbetween any topological space and a covering space depends stronglyon the homotopy type of the spaces (see Proposition 1.33 page 70 of [6]and theorem 21.2 page 174 of [7]). Another case one has the homotopylifting property is when working with fibrations. Fibrations in the senseof Hurewicz (see definition 1.2 page 393 of [4] and page 66 of [11]) havethe homotopy lifting property by definition, beyond this, fibrationsin the sense of Hurewicz are equivalent to fibrations for which onehas the homotopy lifting property (see Curtis and Hurewicz’s theorempage 396 of [4]). Again, as it happens with covering maps, lifting acontinuous function between any topological space and a fiber spacedepends strongly on the homotopy type of the spaces (see theorem 5page 76 of [11]). In fact, fibrations can be treated in most situations ascovering maps (something natural as covering spaces are fiber spaceswith discrete fiber -see theorem 3 page 67 of [11]-). Out of that cases,there are few general situations but specific ones where a kind of liftingcan be done. One of this specific cases is that of the paper of Papaky-riakopoulos [10]; he uses the lifting of some maps whose image lies inwhat he calls prismatic neighborhood and defined in the universal co-vering of such neighborhood; with these liftings Papakyriakopoulos isable to prove Dehn’s lemma and the sphere theorem. The lifting pro-blem from the general point of view can be studied using obstructiontheory, Postnikov towers and Moore-Postnikov towers (see pages 410and 415 of [6]). a r X i v : . [ m a t h . A T ] J un EDUARDO BLANCO-G ´OMEZ
In paper [1] we proved the homotopy lifting property for the sym-metric products SP m ( X ) and F m ( X ) of a topological space, developingthe theory of topological puzzles for the m -cartesian product X m . Infact, we are going to use some of the ideas showed in that paper toprove the main result of this article.2. Isotropy groups and decomposition of the quotientspace
Let X be a topological space and G be a group. We define the actionof g ∈ G on x ∈ X as gx ∈ X (remember that the action of G on X isasked to be associative and the identity element in G have to fix every x ∈ X ). The maps,(1) ϕ g : X → Xx (cid:55)→ ϕ g ( x ) = gx, are bijective. This action generates a continuous and onto map, from X over the quotient space X/G , defined by,(2) ϕ : X → X/Gx (cid:55)→ ϕ ( x ) = [ x ] ϕ , with X/G endowed with the quotient topology generates by ϕ . Theequivalence class [ x ] ϕ is defined as,[ x ] ϕ = { x (cid:48) ∈ X : ∃ g ∈ G such that x (cid:48) = gx } . Moreover, if X is a G -space (see definition 5.10 page 40 [7]) one hasthat the maps ϕ g are homeomorphisms and ϕ is also open (see theorem5.12 page 40 [7]). Definition 2.1.
Let X be a topological space and G acting on X . For x ∈ X we define the isotropy group of x like, G x = { g ∈ G : gx = x } . First of all, let’s proof that the map ϕ defined in (2) is continuous,open and onto. Lemma 2.2.
Let ϕ be the map defined by, ϕ : X → X/Gx (cid:55)→ ϕ ( x ) = [ x ] ϕ , with [ x ] ϕ defined in (2). Then ϕ is continuous, open and onto.Proof. We want to prove that ϕ g , defined in (1) is a homeomorphism.1 ϕ g is bijective. Suppose ϕ g ( x ) = ϕ g ( x (cid:48) ), so gx = gx (cid:48) , then g − gx = g − gx (cid:48) and x = x (cid:48) . Take x (cid:48) ∈ X . As G is a group thenit exists an identity element g and for every g ∈ G it exists g − ∈ G such that gg − = g . Defining x = g − x (cid:48) we have, ϕ g ( g − x (cid:48) ) = gg − x (cid:48) = x (cid:48) , LP FOR ACTIONS OF FINITE ABELIAN GROUPS 3 thus ϕ g ( x ) = x (cid:48) for x = g − x (cid:48) and ϕ g is onto.2 ϕ g is continuous. Take V ⊂ X an open subset. It is enoughto prove that ∀ g ∈ G, gV ⊂ X is open if V ⊂ X is open, being gV = { x ∈ X : ∃ x (cid:48) ∈ V with x = gx (cid:48) } . That is easy to see justusing the next property: x ∈ V ⇔ gx ∈ gV .3 ϕ g is an open map. Take V ⊂ X an open subset. This statementproves similarly to 2 .Now let’s prove the identity,(3) ϕ − ( ϕ ( V )) = (cid:91) g ∈ G ϕ g ( V ) , for every V ⊂ X an open subset. As the sets of the two members of (3)are into the same topological space endowed with the same topology,we just have to prove the equality between the two sets. ⊂ Take x ∈ ϕ − ( ϕ ( V )). Then it exists g ∈ G such that gx ∈ V .But then ϕ g ( x ) ∈ V . As ϕ g is a homeomorphism by the first part ofthe proof we have x ∈ ϕ − g ( V ) = ϕ g − ( V ) so, x ∈ (cid:91) g ∈ G ϕ g ( V ) ⇒ ϕ − ( ϕ ( V )) ⊂ (cid:91) g ∈ G ϕ g ( V ) . ⊃ Take x ∈ ∪ g ∈ G ϕ g ( V ). Then it exists g ∈ G such that x ∈ ϕ g ( V ).As ϕ g is a homeomorphism by the first part of the proof we have ϕ − g ( x ) ∈ V . Like ϕ ◦ ϕ g = ϕ for every g ∈ G , then ϕ ( x ) ∈ ϕ ( V ). As ϕ ( V ) is a saturated set then x ∈ ϕ − ( ϕ ( V )), and we conclude that, (cid:91) g ∈ G ϕ g ( V ) ⊂ ϕ − ( ϕ ( V )) . Thus, we have (3). As we have that ϕ is continuous and onto by defini-tion 2, putting together (3) and the fact that ϕ g is a homeomorphismfor every g ∈ G we conclude that ϕ is also an open map. (cid:3) In this paper we will use extensively the theory of passings-through for topological spaces, introduced in section 4 of [1].Our aim now is to decompose our space X into subspaces such thatwe can take a different action (of a quotient group of G by every isotropygroup) over every subspace and with the objective of building a newspace for which the existence of the homotopy lifting property is equiva-lent to the existence of that property for X . Begin with X a topologicalspace and G a finite abelian group acting on X . Let { G j } j ∈ J be theset of all subgroups of G . Define the subspace,(4) X j = { x ∈ X : G x = G j } , EDUARDO BLANCO-G ´OMEZ endowed with the relative topology. The subspaces X j are pairwisedisjoint by definition. Let (cid:101) G j = G/G j be the quotient set defined as,(5) ρ j : G → G/G j g (cid:55)→ ρ j ( g ) = [ g ] ρ j = { g ∈ G : g − g ∈ G j } . Like G is abelian then G j is normal for every j ∈ J and then (cid:101) G j is agroup, see [2] or [3] or [8]. Furthermore, one can define an action of (cid:101) G j on X j that is well defined. Let’s see this. Lemma 2.3.
Let X be a topological space and G a finite abelian groupacting on X . Let G j be a subgroup of G and (cid:101) G j = G/G j be the quotientgroup generated by G j . Then the action, (6) θ j : X j → X j / (cid:101) G j x (cid:55)→ θ j ( x ) = [ x ] θ j = { x ∈ X j : ∃ (cid:101) g ∈ (cid:101) G j for which x = (cid:101) gx } , of the group (cid:101) G j on X j is well defined. One understands that if (cid:101) g = [ g ] ρ j then (cid:101) gx = [ gx ] θ j . Moreover, if X is Hausdorff then θ j is a coveringmap for every j ∈ J .Proof. Take g , g ∈ (cid:101) g = [ g ] ρ j . Then g − g ∈ G j . As x ∈ X j then g − g x = x . Operating on the left by g we get g x = g x so if g , g ∈ [ g ] ρ j then g x, g x ∈ [ gx ] θ j .Let’s prove that θ j is a covering map. We begin to show that theaction θ j is free. Take g , g ∈ G j . Then, for every x ∈ X j , g x = g x ,so g − g x = x thus g − g ∈ G j . This last assertion implies that [ g ] ρ j =[ g ] ρ j so the action θ j is free. All in all, the map θ j is a covering mapbecause it is defined as a free action of a finite group over a Hausdorffspace (see theorems 17.1 and 17.2 page 154 of [7]). (cid:3) Define now the new topological spaces,(7) (cid:98) X = (cid:32)(cid:91) j ∈ J X j , T J (cid:33) , (cid:101)(cid:98) X = (cid:32)(cid:91) j ∈ J ( X j / (cid:101) G j ) , (cid:101) T J (cid:33) with T J and (cid:101) T J the disjoint union topologies. With the definition ofthese new spaces and with the covering maps (6) we can define anotherfunction gluing them with the disjoint union topology (see [12] or [5]or [4]),(8) θ : (cid:98) X → (cid:101)(cid:98) Xx (cid:55)→ θ ( x ) = [ x ] θ = [ x ] θ j . The map θ is continuous and onto. Let’s see that it is also open. Lemma 2.4.
Let { X j } j ∈ J and { Y j } j ∈ J be two families of pairwise dis-joint topological spaces. Let X = ∪ j ∈ J X j and Y = ∪ j ∈ J Y j be topological LP FOR ACTIONS OF FINITE ABELIAN GROUPS 5 spaces both endowed with the disjoint union topology. Suppose we haveopen f j : X j → Y j for every j ∈ J . From them, define the map, f : X → Yx (cid:55)→ f ( x ) = f j ( x ) , Then, f is open ⇔ f j is open ∀ j ∈ J. Proof.
The right implication is obvious. Let’s see the left one. Byan easy argument (see [12] or [5] or [4]) one has that the inclusions i j : Y j → Y are open (and, in fact, closed too). Take U ⊂ X an opensubset. Then, f ( U ) = f ( (cid:91) j ∈ J ( U ∩ X j )) = (cid:91) j ∈ J f ( U ∩ X j ) = (cid:91) j ∈ J i j ( f j ( U )) , that is a union of open sets in Y . (cid:3) Definition 2.5.
Let X be a topological space and n ∈ N . A continuousmap γ : [0 , n → X will be denoted as an n-region .From this point we will lead our efforts to prove that the lifting ofan n-region in X/G to another one in X is equivalent to the lifting ann-region in (cid:101)(cid:98) X to another one in (cid:98) X . Lemma 2.6.
Let X = ( X, T X ) be a topological space, A a set of in-dexes and X α ⊂ X pairwise disjoint subspaces endowed with the relativetopology, α ∈ A . Let (cid:98) X = ( ∪ α ∈ A X α , T A ) be a topological space endowedwith the disjoint union topology. Then,(a) T X ⊂ T A .(b) Define, π : (cid:98) X → Xx (cid:55)→ π ( x ) = x. Then π is bijective, continuous and for every V ∈ T X , π ( V ) = V ∈ T X .(c) Define, i : X → (cid:98) Xx (cid:55)→ i ( x ) = x. Then i is bijective, open and for every V ∈ T X , i − ( V ) = V ∈ T X .(d) Let Y be another topological space. Take (cid:98) f : Y → (cid:98) X a functiondenoting (cid:98) f ( y ) = x . Then (cid:98) f induces a function f : Y → X , with f ( y ) = x , such that:If (cid:98) f is bijective then f is bijective.If (cid:98) f is continuous then f is continuous.If (cid:98) f is open then f is open.(e) Every homeomorphism (cid:98) f : (cid:98) X → (cid:98) X induces a homeomorphism f : X → X such that (cid:98) f | X = i ◦ f . EDUARDO BLANCO-G ´OMEZ
Proof.
To see (a) we just need the following equality for every V ∈ T X , V = (cid:91) α ∈ A ( V ∩ X α ) ∈ T A . Now (b) and (c) are direct consequences of (a). To prove (d) it isenough to define f = π ◦ (cid:98) f and use (b). Finally, to prove (e), from (cid:98) f we define f = π ◦ (cid:98) f ◦ i and use (b) and (c). (cid:3) Remark 2.7.
We can not prove that the continuous and bijective map π introduce in lemma 2.6 is always a homeomorphism. However, thereare cases when it is. For example, when X is Hausdorff and the disjointunion is compact (see theorem 8.8 page 58 of [7]). Lemma 2.8.
Let X be a Hausdorff topological space, m ∈ N . Then itexists a continuous and bijective map f , (9) f : (cid:101)(cid:98) X → X/G, such that for every open subset V ⊂ X , f ( θ ( π − ( V ))) is an open subsetof X/G , being θ the map defined in (8) and π defined in lemma 2.6 (b).Proof. Consider the next diagram,(10) X ϕ (cid:15) (cid:15) (cid:98) X θ (cid:15) (cid:15) π (cid:111) (cid:111) X/G (cid:101)(cid:98) X f (cid:111) (cid:111) Let’s define,(11) f : (cid:101)(cid:98) X → X/G [ x ] θ (cid:55)→ f ([ x ] θ ) = [ x ] ϕ , and π the continuous and bijective map defined in lemma 2.6 (b). Themap θ is well defined because the subspaces X j of (4) are disjoint, and f is well defined because [ x ] θ ⊂ [ x ] ϕ ⊂ X for all x ∈ X . From diagram(10), lemma 2.6 (b) and definitions (8) and (11) we get ϕ ◦ π = f ◦ θ . Weknow that ϕ is continuous and π is continuous. With the last equality,we just need to know that θ is continuous to conclude the same for f . But θ is continuous by the universal property of the disjoint uniontopology (see [12] or [5] or [4]) so it is f .Let’s see now f is bijective. By definition of (cid:98) X , like ϕ is onto andby lemma 2.6 for every [ x ] ϕ ∈ X/G it exists x (cid:48) ∈ [ x ] ϕ ∩ (cid:98) X such that( ϕ ◦ π )( x (cid:48) ) = [ x ] ϕ . Thus f ([ x (cid:48) ] θ ) = ( f ◦ θ )( x (cid:48) ) = ( ϕ ◦ π )( x (cid:48) ) = [ x ] ϕ andwe obtain that f is onto. Suppose now [ x ] ϕ = [ x (cid:48) ] ϕ for some x, x (cid:48) ∈ (cid:98) X ;then it exists g ∈ G such that x (cid:48) = gx . Like G is abelian then G x (cid:48) = G x and both x and x (cid:48) are in the same X j . Thus,[ x ] θ = [ x ] θ j = [ gx ] θ j = [ x (cid:48) ] θ j = [ x (cid:48) ] θ , LP FOR ACTIONS OF FINITE ABELIAN GROUPS 7 by definition of the action θ j from lemma 2.3 and by definition (8) of θ . We conclude that f is injective.Finally, let’s prove that for every open subset V ⊂ X , f ( θ ( π − ( V )))is an open subset of X/G . But by the previous paragraphs ϕ ◦ π = f ◦ θ .Thus f ( θ ( π − ( V ))) = ϕ ( V ) and ϕ is open (see theorem 5.12 page 40[7]). (cid:3) With lemma 2.8 and with the next diagram,(12) Y g (cid:47) (cid:47) (cid:101) g (cid:32) (cid:32) g (cid:30) (cid:30) X ϕ (cid:15) (cid:15) (cid:98) X θ (cid:15) (cid:15) π (cid:111) (cid:111) X/G (cid:101)(cid:98) X f (cid:111) (cid:111) , one can reduce the proof of the homotopy lifting property for X/G to the proof of the homotopy lifting property for (cid:101)(cid:98) X . More precisely,we just need to lift a path in (cid:101)(cid:98) X to a path in (cid:98) X to get the lift of apath in X/G to a path in X . In the last diagram, Y is a topologicalspace, g is a continuous map in X , g is a continuous map in (cid:98) X , (cid:101) g is acontinuous map in X/G , ϕ is the map defined in (2), π the continuousand bijective map defined in lemma 2.6 (b), θ is the map defined in(8), f is the continuous and bijective map defined in (11). Lemma 2.9.
Let X and Y be Hausdorff topological spaces and m ∈ N .Let (cid:101) g : Y → X/G a continuous map in
X/G . Then,It exists a continuous map g : Y → X such that (cid:101) g = ϕ ◦ g ⇔ It exists a continuous map g : Y → (cid:98) X such that (cid:101) g = f ◦ θ ◦ g Proof.
Remembering that the next diagram is conmutative,(13) X ϕ (cid:15) (cid:15) (cid:98) X θ (cid:15) (cid:15) π (cid:111) (cid:111) X/G (cid:101)(cid:98) X f (cid:111) (cid:111) , we obtain, (cid:101) g = f ◦ θ ◦ g ⇔ (cid:101) g = ϕ ◦ i ◦ π ◦ g ⇔ (cid:101) g = ϕ ◦ g . (cid:3) So now, our efforts will be focused to the proof of the homotopylifting property for (cid:101)(cid:98) X . We are going to work with the next conmutative EDUARDO BLANCO-G ´OMEZ diagram, for n ∈ N ,(14) [0 , n γ (cid:47) (cid:47) (cid:101) γ (cid:33) (cid:33) (cid:98) X θ (cid:15) (cid:15) (cid:101)(cid:98) X. Lemma 2.9 gives us an advantage we didn’t have before: instead ofworking with the map ϕ , we are going to work with the map θ . Bothare continuous, open and surjective maps ( θ is open by its definitionand by lemma 2.4) but θ is also what we call a covering-by-partsmap . This is a direct consequence of the definition (8) of theta because θ | X j = θ j and θ j is a covering map as stated in lemma (2.3).Our aim is to ”lift by parts” the n-region (cid:101) γ : [0 , n → (cid:101)(cid:98) X and thenglue carefully the lifted pieces. Remark 2.10.
Let X and Y be topological spaces and m ∈ N . Let (cid:101) γ : Y → (cid:101)(cid:98) X be a continuous function. Let θ be the map defined in (8).Take y ∈ Y . Then for every p ∈ (cid:101) γ ( y ) ⊂ (cid:98) X the set { ( θ − ◦ (cid:101) γ )( y ) : y ∈ U } is dense in p for all U ⊂ Y an open neighborhood of y . To provethis, suppose not; then it exists y and U ⊂ Y an open neighborhoodof y such that for some p ∈ (cid:101) γ ( y ) ⊂ (cid:98) X and for some V ⊂ (cid:98) X an openneighborhood of p , the set { ( θ − ◦ (cid:101) γ )( y ) : y ∈ U } is not dense in V ,i.e.,(15) { ( θ − ◦ (cid:101) γ )( y ) : y ∈ U } ∩ V = { p } . Having account that θ − ( (cid:101) γ ( U )) is a saturated set (see [9] page 155)and from the last equality (15) we have, U ∩ ( (cid:101) γ − ◦ θ )( V ) = ( (cid:101) γ − ◦ θ )( { p } ) , but this is impossible because the left member is an open set and theright one is a closed set as: U is open, ( (cid:101) γ − ◦ θ )( V ) is open like V is open, (cid:101) γ continuous and θ open (as stated after diagram (14)), and( (cid:101) γ − ◦ θ )( { p } ) is closed as θ ( p ) = [ p ] θ is a point in (cid:101)(cid:98) X and (cid:101) γ is continuous.Now let Y ⊂ Y a subspace of Y . Suppose γ : Y → (cid:98) X is a continuousfunction such that (cid:101) γ | Y = θ ◦ γ . Take y ∈ ∂Y . Then the set { γ ( y ) : y ∈ U ∩ Y } , with U ⊂ Y an open neighborhood of y , is dense insome p ∈ (cid:101) γ ( y ) ⊂ (cid:98) X . Take any U ⊂ Y an open neighborhood of y .Denote Y = Y ∪ { y } endowed with the relative topology. By theprevious paragraph we have that the set { ( θ − ◦ (cid:101) γ | Y )( y ) : y ∈ U ∩ Y } is dense in every p ∈ (cid:101) γ ( y ) , so, for every V p ∈ (cid:98) X an open neighborhood of p it exists y p ∈ Y suchthat ( θ − ◦ (cid:101) γ | Y )( y p ) ∈ V p , i.e.,(16) ( θ − ◦ (cid:101) γ | Y )( y p ) ∈ V p . LP FOR ACTIONS OF FINITE ABELIAN GROUPS 9 If γ ( y p ) / ∈ V p for every p ∈ (cid:101) γ ( y ), then ( θ ◦ γ )( y p ) / ∈ θ ( V p ), i.e., (cid:101) γ | Y ( y p ) / ∈ θ ( V p ), so, ( θ − ◦ (cid:101) γ | Y )( y p ) / ∈ ( θ − ◦ θ )( V p ). But as V p ⊂ ( θ − ◦ θ )( V p ) then( θ − ◦ (cid:101) γ | Y )( y p ) / ∈ V p and this is a contradiction with (16). Furthermore,in the case that the set θ − ( (cid:101) γ ( y )) is finite, then the set { γ ( y ) : y ∈ U ∩ Y } , with U ⊂ Y an open neighborhood of y , is dense in exactlyone and only one p ∈ θ − ( (cid:101) γ ( y )) ⊂ (cid:98) X when X is Hausdorff. Remark 2.11.
Let X be a Hausdorff topological space, U ⊂ X anopen subset and x ∈ U . Then U \ { x } is open. To prove that, itis enough to see that every x (cid:48) ∈ U \ { x } is an interior point, i.e., itexists U (cid:48) ⊂ U \ { x } an open neighborhood of x (cid:48) . But this is a directconsequence of the existence of U (cid:48) ⊂ U an open neighborhood of x (cid:48) (that exists because U is an open set) and the existence of U (cid:48) ⊂ X an open neighborhood of x (cid:48) that does not contain x (because X isHausdorff). Taking U (cid:48) = U (cid:48) ∩ U (cid:48) we finish.3. Homotopy lifting property for actions of finiteabelian groups on Hausdorff topological spaces
We want to prove now the analogue of theorem 5.22 of [1].
Theorem 3.1.
Let X be a Hausdorff topological space and n ∈ N . Let (cid:101) γ : [0 , n → (cid:101)(cid:98) X be a continuous function. Then it exists γ : [0 , n → (cid:98) X continuous such that diagram (14) conmutes.Proof. First of all, we will split every piece X j . Denote, (cid:101) γ j = (cid:101) γ | (cid:101) γ − ( X j / (cid:101) G j ) . Consider the next diagram,(17) (cid:101) γ − ( X j / (cid:101) G j ) (cid:47) (cid:47) (cid:101) γ j (cid:38) (cid:38) X jθ j (cid:15) (cid:15) X j / (cid:101) G j . Take,(18) (cid:101) γ − ( X j / (cid:101) G j ) = (cid:91) λ j ∈ Λ j Γ λ j j , the decomposition of (cid:101) γ − ( X j / (cid:101) G j ) in connected and locally path-connected(at the same time) components. To make simpler the notation, whenpossible, we will denote Γ j a connected and locally path-connectedcomponent. Denoting as, (cid:101) Γ λ j j = (cid:101) γ (Γ λ j j ) , and , Θ λ j j = θ − j ( (cid:101) γ (Γ λ j j )) , we can rewrite diagram (17),(19) Γ λ j j (cid:47) (cid:47) (cid:101) γ j (cid:32) (cid:32) Θ λ j jθ j (cid:15) (cid:15) (cid:101) Γ λ j j . As Γ λ j j is connected and (cid:101) γ j is continuous therefore using 6.1.3. theorempage 352 of [5], (cid:101) Γ λ j j is connected. In fact, (cid:101) Γ λ j j is a connected componentof X j / (cid:101) G j because so it is Γ λ j j of (cid:101) γ − ( X j / (cid:101) G j ). As a consequence, every y ∈ (cid:101) Γ λ j j has got a neighborhood that is disjoint with every (cid:101) Γ λ (cid:48) j j for every λ (cid:48) j (cid:54) = λ j .From now til the end of the proof, we will work with the sets Θ λ j j .Therefore, we will denote as Λ j the set of indexes associated to λ j ,and as µ j an index of Λ j ; finally, we will denote (cid:98) Θ µ j j a path-connectedcomponent of Θ λ j j or just (cid:98) Θ j when possible.First of all, let’s prove a property of the sets Θ λ j j ,(20) There exists no x ∈ Θ λ j j ∪ Θ λ (cid:48) j j such that, x ∈ Θ λ j j ∩ Θ λ (cid:48) j j . Suppose not. Then (cid:101) γ − ( θ ( x )) ∩ (Γ λ j j ∩ Γ λ (cid:48) j j )(Γ λ j j ∪ Γ λ (cid:48) j j ) (cid:54) = ∅ . But that isimpossible because Γ λ j j and Γ λ (cid:48) j j are locally path-connected components.Take now x ∈ Θ j and (cid:98) Θ j the path-connected component including x . Let (cid:101) x = θ j ( x ) ∈ (cid:101) Γ j and (cid:101) x ∈ (cid:101) Γ j . As (cid:101) Γ j is connected andlocally path connected, then by 21.1 lemma page 175 of [7], (cid:101) Γ j is pathconnected. Therefore take a path (cid:101) δ : [0 , → (cid:101) Γ j which holds (cid:101) δ (0) = (cid:101) x and (cid:101) δ (1) = (cid:101) x ; like θ j is a covering map as stated in lemma 2.3, wecan apply 17.6 theorem page 157 of [7] to obtain a path δ : [0 , → Θ j which holds (cid:101) δ = θ j ◦ δ and δ (0) = x ; moreover, it exists x ∈ θ − j ( (cid:101) x ) sothat δ (1) = x . As the last deduction can be done with every (cid:101) x ∈ (cid:101) Γ j ,we conclude that (cid:98) Θ j holds θ j ( (cid:98) Θ j ) = (cid:101) Γ j . We obtain the next diagram,(21) Γ j γ j (cid:47) (cid:47) (cid:101) γ j (cid:36) (cid:36) (cid:98) Θ jθ j (cid:15) (cid:15) (cid:101) Γ j = (cid:98) Θ j / (cid:101) G j . Our aim with diagram (21) is to apply 21.2 theorem page 176 of [7] toobtain a lift of (cid:101) γ j . To do that, like Γ j is connected and locally path LP FOR ACTIONS OF FINITE ABELIAN GROUPS 11 connected by its definition, we also need the next condition,(22) (cid:101) γ ∗ j ( π (Γ j )) ⊆ θ ∗ j ( π ( (cid:98) Θ j )) , where (cid:101) γ ∗ j and θ ∗ j are the induced maps between the fundamental groups.Let y ∈ (cid:98) Θ j / (cid:101) G j and, ω : π ( (cid:98) Θ j / (cid:101) G j , y ) → (cid:101) G j [ δ (cid:48) ] H Θ / Σ (cid:55)→ σ δ (cid:48) , defined in page 165 of [7] (there is defined as ϕ ). For one hand, like (cid:98) Θ j is path connected, we can apply 19.2 theorem page 166 of [7] to obtainthat θ ∗ j ( π ( (cid:98) Θ j )) = ker ω . On the other hand, looking at the definition of (cid:101) γ ∗ j , we deduce that every loop in Γ j goes to a loop in (cid:101) Γ j by the actionof (cid:101) γ j , so, by definition of ω , we conclude that, (cid:101) γ ∗ j ( π (Γ j )) ⊆ ker ω, that is (22).All in all, we use 21.2 theorem page 176 of [7] to obtain a continuouslift γ j of (cid:101) γ j which holds,(23) (cid:101) γ j = θ j ◦ γ j . From γ j , we want to build a continuous extension γ j : Γ j → (cid:98) Θ j inthis way: denote P Γ j ( (cid:101) γ ) the set of passings-through of (cid:101) γ which changefrom a quotient piece to another one from or towards (cid:101) Γ j (from now tilthe end of the proof we will use the notation P ( γ j ) = P Γ j ( (cid:101) γ )). Taking t ∈ P Γ j ( (cid:101) γ ) and applying remark 2.10, theorem 3.11 page 7 of [1] and5.1 theorem page 215 of [4] we get a continuous extension of γ j on t .Let’s make it with every t ∈ P Γ j ( (cid:101) γ ) so that we obtain a function,(24) γ j : Γ j ∪ P Γ j ( (cid:101) γ ) → (cid:98) Θ j , such that its restrictions to Γ j ∪ { t } are continuous functions, for every t ∈ P Γ j ( (cid:101) γ ). That function can be defined as γ j : Γ j → (cid:98) Θ j becauseevery t ∈ ∂ Γ j on which γ j is not defined, has to be a passing-throughby definition 4.1 page 10 of [1] (if Γ j includes no passing-through in itsboundary, then by lemma 4.8 page 12 of [1] γ j would be the lifting ofthe whole continuous function (cid:101) γ ). Let’s prove that γ j is a continuousfunction. For one hand, γ j is well defined; this is true due to the nexttwo facts: the first one is that there is no p ∈ Im γ j ( P Γ j ( (cid:101) γ )) such that p is in another (cid:98) Θ j (cid:48) ...j (cid:48) k (cid:48) by definition. The second fact is that there isno t ∈ P Γ j ( (cid:101) γ ) such that γ j ( t ) can be associated to two o more differentpoints of the same equivalent class. Suppose not; then γ j ( t ) would notbe well defined in the set Γ j ∪ { t } , but this is a contradiction with the previous lines. On the other hand, we want to prove that γ j iscontinuous; take the next notation, (cid:101) γ j = (cid:101) γ | Γ j and θ j = θ | (cid:98) Θ j . Let’s prove the following equality,(25) γ − j ( V ) = (cid:101) γ − j ( θ j ( V )) ∀ V ⊂ (cid:98) Θ j . But identity (25) is true by construction of γ j ; thus, like (cid:101) γ is continuousand θ is open by lemma 2.4, then for every open V ⊂ (cid:98) Θ j we have that γ − j ( V ) is open concluding that γ j is continuous.Our aim now is to glue carefully the liftings obtained in the previousparagraphs. Let’s define a new concept: a shire . We will say that aset S = { (Γ λ j j , (cid:98) Θ µ j j ) } µ j ∈ Λ j j is a shire if it holds (cid:101) Γ λ j j = θ j ( (cid:98) Θ µ j j ) and thenext two conditions,C1 ∀ (Γ λ j j , (cid:98) Θ µ j j ) ∈ S there exists (Γ λ j (cid:48) j (cid:48) , (cid:98) Θ µ j (cid:48) j (cid:48) ) ∈ S such that it existsa lifting γ i which holds,Im γ i ( P ( γ i )) ∩ ∂ (cid:98) Θ µ j j ∩ ∂ (cid:98) Θ µ j (cid:48) j (cid:48) (cid:54) = ∅ . C2 The set of couples S cannot be split in a disjoint way withrespect to condition C1 .We will say that S is a complete shire if,(26) (cid:91) ∀ (Γ λjj , (cid:98) Θ µjj ) ∈S Γ λ j j = [0 , n . In other case, we will say that S is an incomplete shire . We will saythat S is an univalent shire if,(27) ∀ Γ λ j j ∃ ! µ j such that (Γ λ j j , (cid:98) Θ µ j j ) ∈ S . We want now to prove the next result,(28) If S = { (Γ λ j j , (cid:98) Θ µ j j ) } µ j ∈ Λ j j is an univalent shire ⇒ it exists a continuous function γ S : ∪ Γ λ j j → ∪ (cid:98) Θ µ j j . First of all, take account that the family { Γ λ j j } j , of ’first coordinates’ ofthe shire, is locally finite; like θ is open and (cid:101) γ continuous, it is enoughto prove that the family { (cid:98) Θ µ j j } j , of ’second coordinates’ of the shire,is locally finite; let’s prove it: take x ∈ (cid:98) Θ µ j j ⊂ Θ λ j j and U x ⊂ (cid:98) X anopen neighborhood of x ; take another (cid:98) Θ µ (cid:48) j j ⊂ Θ λ j j ; it is impossible that x ∈ ∂ (cid:98) Θ µ (cid:48) j j because (cid:98) Θ µ j j and (cid:98) Θ µ (cid:48) j j are path-connected components of Θ λ j j .Suppose now that x ∈ ∂ Θ λ (cid:48) j j with λ (cid:48) j (cid:54) = λ j ; in that case ∃ t ∈ ∂ Γ λ j j ∩ ∂ Γ λ (cid:48) j j ,with t ∈ Γ λ j j , being Γ λ j j associated to a path-connected component ofΘ λ j j and Γ λ (cid:48) j j associated to a path-connected component of Θ λ (cid:48) j j ; but LP FOR ACTIONS OF FINITE ABELIAN GROUPS 13 that is impossible because Γ λ j j and Γ λ (cid:48) j j are connected and locally-pathconnected components. All in all, every (cid:98) Θ µ j j can share its boundarywith just one (cid:98) Θ µ j (cid:48) j (cid:48) of every piece X j (cid:48) different from X j . As the numberof pieces of the puzzle (cid:98) X is finite, we conclude that the family { (cid:98) Θ µ j j } j is locally finite, thus, so it is the family { Γ λ j j } j . Therefore, the family { Γ λ j j } j is locally finite. Applying now exercise 9(c) page 127 of [9], thatis a generalization of Pasting lemma (theorem 18.3 page 123 of [9]), andusing continuous functions γ j defined in (24), we obtain the continuousfunction γ S predicted in (28), that is a function because the shire isunivalent, and continuous by construction.Finally, we need to prove that a maximal univalent shire associatedto an x ∈ (cid:98) X is, in fact, a complete shire. Let’s begin to prove the nextstatement,(29) Take S an incomplete shire. Then it exists a shire S (cid:48) such that S ⊂ S (cid:48) . Furthermore, if S is univalent, S (cid:48) can be built as univalent.As S is an incomplete shire, then,[0 , n \ (cid:91) ∀ (Γ λjj , (cid:98) Θ µjj ) ∈S Γ λ j j (cid:54) = ∅ . Like, [0 , n = (cid:101) γ − ( (cid:101)(cid:98) X ) = (cid:91) j ∈ J (cid:101) γ − ( X j / (cid:101) G j )remembering (18) we can take a connected and locally path-connectedcomponent, Γ (cid:48) ⊂ [0 , n \ (cid:91) ∀ (Γ λjj , (cid:98) Θ µjj ) ∈S Γ λ j j that holds ∂ Γ (cid:48) ∩ ∂ ( (cid:83) ∀ (Γ λjj , (cid:98) Θ µjj ) ∈S Γ λ j j ) (cid:54) = ∅ . Take now (cid:98) Θ (cid:48) a path-connectedcomponent of (cid:98) X associated to Γ (cid:48) (in the sense of diagram (21)), thatholds, ∂ (cid:98) Θ (cid:48) ∩ ∂ ( (cid:83) ∀ (Γ λjj , (cid:98) Θ µjj ) ∈S (cid:98) Θ µ j j ) (cid:54) = ∅ . That path-connected componentexists by construction. Define now the new shire like this, S (cid:48) = S ∪ { (Γ (cid:48) , (cid:98) Θ (cid:48) ) } . Therefore S (cid:48) is a (univalent if so it is S ) shire by construction andbecause so it is S . At this point, we have proved statement (29).To finish the proof of the theorem, take x ∈ (cid:98) X and S a maximal uni-valent shire containing x (in its second coordinates). Using statement(29) we conclude that S is a complete univalent shire and applyingstatement (28) we finish the proof. (cid:3) Theorem 3.2.
Let X be a Hausdorff topological space and n ∈ N . Let G be a finite abelian group acting on X. Let (cid:101) γ be an n -region over X/G .Then it exists γ an n -region over X such that (cid:101) γ = ϕ ◦ γ .Proof. This theorem is a direct consequence of lemma 2.9 and theorem3.1. (cid:3)
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