Degree theory and Branched covers. Pekka Pankka

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1 Degree theory and Branched covers Pekka Pankka November 2, 2015

2 Preface These are lecture notes for the course Degree theory and branched covers lectured at the University of Jyväskylä Fall The purpose of these lecture notes is to introduce the necessary theory for the proof of the Chernavskii Väisälä s theorem, see [Väi66]. Theorem (Väisälä 1966). Let f : M N be a discrete and open map between n-manifolds M and N for n 3. Then the branch set B f of f has topological dimension at most n 2 and dim B f = dim fb f = dim f 1 fb f. A discrete and open map 1 is called a branched cover. Recall that a map is discrete if preimage of a point is a discrete set, and a map is open if image of an open set is open. A point x M in the domain of a mapping f : M N is a branch point if f is not a local homeomorphism at x. The branch set B f is the set of all branched points of f. Väisälä s theorem is fundamental in the theory of these mappings. It yields as a corollary that the branch set does not locally separate the domain of the map; the same holds of course for the image of the branch set. As a corollary we obtain Corollary. Let f : M N be a branched cover between n-manifolds. Then f is either orientation preserving or reversing. The proof of Väisälä s theorem requires a substantial amount of preliminary material. The main argument uses local degree theory of proper maps. In order to define the local degree we discuss first (compactly supported) Alexander Spanier cohomology, which we then use to define (local) orientation and the local degree of a proper map. The classical expositions on this theory are Spanier [Spa66] and Massey [Mas78], which we mainly follow. To motivate the abstract theory, we begin with a PL version of Väisälä s theorem. map. 1 In these notes map and mapping are synonymous and typically refer a continuous 1

3 Bibliography [Bor60] Armand Borel. Seminar on transformation groups. With contributions by G. Bredon, E. E. Floyd, D. Montgomery, R. Palais. Annals of Mathematics Studies, No. 46. Princeton University Press, Princeton, N.J., [Dug78] James Dugundji. Topology. Allyn and Bacon, Inc., Boston, Mass.- London-Sydney, Reprinting of the 1966 original, Allyn and Bacon Series in Advanced Mathematics. [Eng78] Ryszard Engelking. Dimension theory. North-Holland Publishing Co., Amsterdam-Oxford-New York; PWN Polish Scientific Publishers, Warsaw, Translated from the Polish and revised by the author, North-Holland Mathematical Library, 19. [Mas78] William S. Massey. Homology and cohomology theory. Marcel Dekker, Inc., New York-Basel, An approach based on Alexander-Spanier cochains, Monographs and Textbooks in Pure and Applied Mathematics, Vol. 46. [Mas91] William S. Massey. A basic course in algebraic topology, volume 127 of Graduate Texts in Mathematics. Springer-Verlag, New York, [RS82] Colin Patrick Rourke and Brian Joseph Sanderson. Introduction to piecewise-linear topology. Springer Study Edition. Springer-Verlag, Berlin-New York, Reprint. [Spa66] Edwin H. Spanier. Algebraic topology. McGraw-Hill Book Co., New York-Toronto, Ont.-London, [Väi66] Jussi Väisälä. Discrete open mappings on manifolds. Ann. Acad. Sci. Fenn. Ser. A I No., 392:10,

4 Contents 1 PL version of Väisälä s theorem 4 2 Alexander Spanier cohomology Spaces Φ k (X) of k-functions Algebraic intermission Cochains C k (X) Cohomologies H ( ) and Hc ( ) Exact sequence of a pair in H ( ) Push-forward in Hc ( ) Exact sequence of a pair in Hc ( ) Homotopy property Mayer Vietoris sequence Open sets in R n and S n Goals H c (R n ) and H c (S n ) Triviality of H k c ( ) in R n for k > n H n c ( ) for open sets in R n and S n Open and closed sets on n-manifolds Orientation and degree From R to Z Orientation Global degree of proper maps Open maps and admissible domains Local degree of open maps Local index of discrete and open maps Väisälä s theorem Statement The branch set has no interior The branch set does not separate locally

5 Chapter 1 PL version of Väisälä s theorem In this section we prove Väisälä s theorem for piece-wise linear branched covers. The purpose of this section is to show that in that context the proof is essentially trivial due to the local rigidity of these maps. The purpose of the section is to also invite the reader to isolate the steps in the proof which require more advanced methods in the general case. To avoid unnecessary terminology, which is not needed in later sections, we discuss the result on the level of simplicial complexes and simplicial maps, and not on the level of PL topology. The gist of the proof is already captured in this context. We begin with some definitions. The convex hull of points x 1,..., x k in R n is { k } k [x 1,..., x k ] = t i x i R n : 0 t i 1, t i = 1. i=1 Let (e 1,..., e n+1 ) be the standard basis (from linear algebra) of R n+1. The convex set n = [e 1,..., e n+1 ] R n+1 is the standard n-simplex. A convex set τ R n+1 is a k-face of n if there exists indices 1 j 1 < < j k+1 n + 1 for which τ = [e j1,..., e jk+1 ]. From now we use the identify R k with its standard copy in R n for 0 k n, that is, R k = R k {0} n k R n. With this identification, we have k n for k n. Let X be a topological space and k N. An embedding σ : k X is called a (singular) k-simplex. Given an l-face τ of n, the restriction σ τ : τ X is called a l-face of σ. Since we are not really interested in the actual parametrization σ but rather the image σ k, we tacitly assume that the mapping for each singular simplex is fixed and we call the set σ( k ) a k-simplex. Similarly, we denote the image σ(τ) simply by τ and call it the l-face σ. We say that k is the dimension on the simplex σ. As a convention, we call the empty set an ( 1)-dimensional simplex. 4 i=1

6 A collection S is a simplicial complex in X if each element σ X is a simplex, S contains all faces of its simplices, and the intersection σ σ of two simplices σ, σ S is a simplex in S (possibly of dimension 1). The complex S is n-dimensional if there exists an n-dimensional simplex in S and all simplices in S have dimension at most n. Given k N, the subcomplex of S consisting of all l-simplices for l k is called the k-skeleton of S and denoted S (k). Let X be a simplicial complex in space X. We say that S triangulates the space X if X = S = σ S σ and S is locally finite, that is, for each x X there exists a neighborhood U X of x so that #{σ S : σ U } <. Let S and S be simplicial complexes triangulating spaces X and X, respectively. A continuous map f : X Y is simplicial from S to S if (a) for each σ S the image fσ is a simplex in S and (b) the map f σ : σ fσ has affine representation, that is, there exists homeomorphisms θ σ : k σ and θ σ : l fσ so that the map θ σ f σ θσ 1 : k l is a restriction of a linear map R k+1 R l+1 ; σ θ σ k f σ fσ l θ σ Remark In particular, f σ maps faces of σ to faces of fσ. We have now all the terminology for the statement of a simplicial version of Väisälä s theorem. For the statement, recall that the branch set B f X of a mapping f : X Y is the set of points x X for which f is not a local homeomorphism at x; f is a local homeomorphism at x if there exists a neighborhood U of x so that f U : U fu is a homeomorphism. Theorem Let n 2 and let S and S be n-dimensional simplicial complexes triangulating spaces M and N, respectively, so that (a) M and N are unions of n-simplices of S and S, respectively, and (b) each (n 1)-simplex τ in S (in S ) is a face of exactly two n-simplices in S (in S ). Let f : M N be a continuous simplicial map from S to S, which is either discrete or open. Then the branch set B f M is contained in the (n 2)-skeleton S (n 2) (i.e. B f S (n 2) ) and its image fb f in (S ) (n 2). Remark Note that a triangulation of an n-manifold always satisfies the additional condition that each (n 1)-simplex is a face of exactly two n- simplices. This can be seen for example by using the Invariance of Domain. 5

7 There are two simple observations which explain why it suffices to consider maps which only discrete or open. Lemma Let S and S be triangulations of spaces M and N, respectively, as in Theorem 1.0.2, and f : M N a simplicial map from S to S. Suppose f is either discrete or open. Then simplices fσ and σ have the same dimension for each σ S. Idea of the proof. Let σ S be a k-simplex. Then fσ S has at most k vertices. Thus fσ is an l-simplex for l k. Thus the dimension of fσ is at most the dimension of σ. Suppose now that there exists a k-simplex σ S so that fσ is an l- simplex for l < k. Let ϕ = θ σ f σ θ σ : k l be the affine representation of f σ and L: R k+1 R l+1 a linear map so that ϕ = L k. Clearly, ϕ is not discrete. Thus f σ is not discrete. To show that f is not open, let ξ be an n-simplex containing σ. Since fσ is not a k-simplex, we observe that fξ is not an n-simplex. Thus fξ has no interior in N and fintξ is not an open set in N. Thus the map f is not open. Remark In fact, it is also easy to see that, under the assumptions of Lemma 1.0.4, the map f σ : σ fσ is a homeomorphism for each simplex σ S. It is important to keep in mind that this fact does not imply that the map f is a local homeomorphism. (Draw a picture.) Proof of Theorem Suppose first that B f is not contained in the (n 1)-skeleton S (n 1), that is, there exists x B f \ S (n 1). Then there exists an n-simplex σ S so that x intσ. Thus the map f σ : σ fσ is not a homeomorphism. This contradicts Lemma Thus B f S (n 1). We show next that actually B f S (n 2). Suppose this is not the case, and let x B f \ S (n 2). Then there exists an (n 1)-simplex τ S (n 1) which contains x in its interior, i.e. x τ \ S (n 2). Let σ + and σ be n-simplices in S so that σ + σ = τ. Let U = int(σ + σ ). Then U is a neighborhood of x in M. Since f τ : τ fτ is a homeomorphism by Lemma and f is not a local homeomorphism, there exists x ± σ ± so that f(x + ) = f(x ). Since fσ + and fσ are n-simplices in S, we conclude that fσ + = fσ. Since fτ is an (n 1)-simplex which a common face of two n-simplices in S, we conclude that f(x) fu. Thus fu is not open in N, which is a contradiction. Hence B f S (n 2). Since fb f f S (n 2) (S ) (n 2), the proof is complete. Remark In this section we merely introduced few of the basic definitions having only Theorem in mind. PL-theory is a vast topic in topology and our discussion (naturally) does not even scratch the surface. 6

8 For more detailed discussion on PL-theory, we refer to [RS82]. Simplicial complexes are a main object in (singular) homology and they are discussed in many text books on algebraic topology, see e.g. Massey [Mas91]. Triangulability problems have a long history going back to Poincaré, see e.g. Manolescu s article on this history at cm/tm.pdf. 7

9 Chapter 2 Alexander Spanier cohomology In this chapter we discuss the definition and basic properties of Alexander Spanier cohomology. The use of Alexander Spanier cohomology in the proof of Väisälä s theorem stems from the good properties of this cohomology with respect to closed sets. To emphasize this aspect we compare it to the more familiar singular cohomology to highlight the differences. The Alexander Spanier cohomology H (X) is the homology of a (co)chain complex (C k (X), d k ) which is a quotient complex of a (co)chain complex (Φ k (X), δ k ) of k-functions in X. More precisely, we have a commutative diagram 0 Φ 0 (X) δ 0 Φ 1 (X) δ 1 Φ 2 (X) δ 2 Φ 3 (X) δ3 0 C 0 (X) d 0 C 1 (X) d 1 C 2 (X) d 2 C 3 (X) d3 where the vertical arrows are quotient maps, and H k (X) = ker d k /imd k Spaces Φ k (X) of k-functions Definition Let X be a space. For k N, a k-function on X is a function X k+1 R. We denote by Φ k (X) the vector space of all k-functions on X. For completeness, we define Φ k (X) = {0} for k < 0. Remark In some sources (e.g. Massey [Mas78]), k-functions are assumed to have finitely many values. Note that a finitely valued k-function φ: X k+1 R has a unique representation φ = λ 1 χ A1 + + λ m χ Am with λ 1,..., λ m R and the pair-wise disjoint sets A 1,..., A m X k+1. 8

10 Definition Let f : X Y be a map. The pull-back homomorphism f! : Φ k (Y ) Φ k (X) is defined by φ f! (φ), where f! (φ): X k+1 R is the function f! (φ)(x 1,..., x k+1 ) = φ(f(x 1 ),..., f(x k+1 )). Example One of the most important maps in the theory is the inclusion map ι AX : A X for A X. The pull-back ι! AX : Φk (X) Φ k (A) is given by formula φ φ A k+1, i.e. pull-back induced by inclusion is just a restriction operator. Note that, for A B X, ι AX = ι BX ι AB and ι! AX = ι! AB ι! BX. It is crucial to notice that there is no assumption on continuity of k- functions or for the map f : X Y. The topology of X comes forth in the notion of support of a k-function, which is a key concept in the theory. Definition Let k N. A k-function φ: X k R is not supported at x X if there exists a neighborhood U X of x for which φ U k = 0. The set null(φ) = {x X : φ is not supported at x} the nullset of φ. The complement of null(φ) in X is the support spt φ of φ in X. For k < 0, we set null(0) = X and spt(0) = 0. Remark We make the some trivial observations. Let φ Φ k (X). 1. The nullset null(φ) is an open set and spt(φ) is closed. 2. If k = 0, the support spt(φ) is the usual support of a function, that is, spt(φ) = {x X : φ(x) 0}. Thus spt(φ) = only for φ = 0 in this case. 3. The support spt(φ) is empty if and only if there exists a covering U of X so that φ U k+1 = 0 for each U U. It is important to notice that the nullset and the support of a k-function are subsets of the space X and not of X k+1. Moreover, it is easy find examples of non-zero k-functions having empty support. Example Let X be any locally compact space Hausdorff space having at least two points x 0 y 0. Let φ: X 2 R be a 1-function be the characteristic function of the point {(x 0, y 0 )}. Then φ 0 but null(φ) = X, since there exists a neighborhood U of x 0 which does not contain y 0. Before continuing, we make some easy observations in sprit Remark 2.1.6(c). Let D X,k = {(x,..., x) X k+1 : x X} be the diagonal of X k+1. Lemma Let φ and ψ be k-functions in X. Then spt(φ) = if and only if there exists a neighborhood W of the diagonal D X,k of X k+1 so that φ W = 0. 9

11 Proof. Suppose first that spt(φ) =. Then null(φ) = X and there exists a covering U of X so that φ U k+1 = 0 for each U U. Let W = U. Then φ W = 0. Suppose now that such neighborhood W of D X,k exists and let x X. Then, by the definition of product topology, there exists a neigborhood U of x in X so that U k+1 W. Then x null(φ). Lemma Let φ and ψ be k-functions in X. spt(φ) spt(ψ) and spt(φψ) spt(φ) spt(ψ). Then spt(φ + ψ) Proof. Let x null(φ) null(psi). Then there exists a neighborhood U of x so that φ U k+1 = 0 and ψ U k+1 = 0. Thus (φ + ψ) U k+1 = 0. Hence x null(φ + ψ). For the second claim, let x null(φ) null(ψ). Then there exists a neighborhood U of x so that either φ U k+1 = 0 or ψ U k+1 = 0. Hence (φψ) U k+1 = 0. Thus x null(φψ). Corollary Let φ and ψ be k-functions in X. Then spt(φ ψ) = if and only if there exists a neighborhood U of the diagonal D X,k so that φ U = ψ U. For the first glimpse of the connection of the support and topology of X, we note the following. It is easy to find a counterexample if the map X Y is not continuous. Lemma Let f : X Y be a continuous map and φ Φ k (Y ). Then spt f! (φ) f 1 spt(φ). Proof. Let y null(φ) and x f 1 (y). Since f is continuous, there exists a neighborhood V of x so that fv is contained in a neighborhood U of y so that φ U k+1 = 0. Hence f! (φ) V k+1 = 0 and x null(f! (φ)). Thus f 1 (null(φ)) null(f! (φ)). Hence spt(f! (φ)) = X \ null(f! (φ)) X \ f 1 (null(φ)) = f 1 spt(φ). This concludes the proof. Remark Note that we could also argue using the observation that, by continuity, given a covering U of Y the collection f 1 U = {f 1 U : U U} is a covering of X and f! φ V k+1 = 0 for each V f 1 U. For inclusions of subsets, we have the following observation. Here and in what follows, we denote the map A X, which is the inclusion of the subset A X to X, by ι AX : A X. Lemma Let φ Φ k (X) and A X a subset. Then spt(ι! AX (φ)) spt(φ) A. Moreover, if U X is an open set, then spt(ι! UXφ) = spt(φ) U. 10

12 Proof. Let x null(φ) A. Then there exists a neighborhood U of x in X so that φ U k+1 = 0. Then (ι! φ) (U A) k+1 = φ (U A) k+1 = 0. Thus null(φ) A null(ι! φ), which is equivalent to the claim. Let now U X be an open subset. Let x null(ι! φ). Then there exists a neighborhood V of x in U so that (ι! φ) V k+1 = 0. Since V is open in X, we have that φ V k+1 = (ι! φ) V k+1 = 0. Thus x U null(φ). Remark The equality for supports need not hold even for inclusions. This is easily seen by considering 0-functions and the inclusion R {0} R 2. Indeed, let X = R 2, A = R {0}, and φ: R 2 R be the characteristic function φ = χ R 2 \A. Then φ Φ 0 (R 2 ) and spt(φ) = R 2. On the other hand, ι! AX φ = 0. Spaces Φ k (X) are connected to a chain complex by introducing coboundary operators. Definition The coboundary operator for k-functions is the homomorphism δ k : Φ k (X) Φ k+1 (X) defined by k+2 δ k (φ)(x 1,..., x k+2 ) = ( 1) l+1 φ(x 1,..., x l 1, x l+1,..., x k+2 ), l=1 where x 1,..., x k+2 X. Φ k+1 (X) for k < 0. For completeness, we define δ k = 0: Φ k (X) Convention To simplify notation, we denote the homomorphism δ k simply by δ to unless it is important to emphasize the domain and range. The typical calculation, common to all homology/cohomology theories, shows that δδ = 0. We leave the verification of this fact to the interested reader. Fact For each k Z, δ k+1 δ k = 0, that is, for every x 1,..., x k+2 X. δ k+1 (δ k (φ))(x 1,..., x k+2 ) = 0 The coboundary operator δ and the pull-back f! clearly commute. Fact Let f : X Y be a map and φ Φ k (Y ). Then δ(f! (φ)) = f! (δ(φ)). 11

13 A fundamental observation is that the coboundary of a k-function has smaller support. We formalize this as follows. Lemma Let k Z and φ Φ k (X). Then spt(δ(φ)) spt(φ). Proof. Let x null(φ). Then there exists a neighborhood U X of x for which φ U k+1 = 0. Thus, for x 1,..., x k+2 U, we have δ k (φ)(x 1,..., x k+2 ) = Thus x null(δ k (φ)). = k+2 ( 1) l+1 φ(x 1,..., x l 1, x l+1,..., x k+2 ) l=1 k+2 ( 1) l+1 φ U k+1(x 1,..., x l 1, x l+1,..., x k+2 ) = 0. l=1 Having the coboundary operator and notion of support at our disposal, we have a topological characterization for the kernel of δ 0. Lemma Let φ Φ 0 (X), that is, a function φ: X R. Then spt(δ(φ)) = if and only if φ is locally constant, that is, there exists an open cover U of X so that φ U : U R is a constant function for each U U. Proof. Suppose such an open cover U exists. We show that spt(δ 0 (φ)) =. Let x X and U U a neighborhood of x. Then, for all x 1, x 2 U, (2.1.1) δ 0 (φ)(x 1, x 2 ) = φ(x 2 ) φ(x 1 ) = 0. Thus x null(δ 0 (φ)). Hence spt(δ 0 (φ)) =. Suppose now that spt(δ 0 (φ)) =. Then there exists by Remark a covering U of X so that δ 0 (φ) U k+1 = 0 for each U U. Thus, by (2.1.1), φ is locally constant. Finally, we define two important subspaces of Φ k (X). Definition A k-function φ Φ k (X) is locally trivial if spt(φ) =, and compactly supported if spt(φ) is a compact subset of X. We denote by Φ k 0 (X) and Φk c (X) the subspaces of locally constant and compactly supported k-functions in Φ k (X), respectively. Note that Lemma immediately yields the following corollary. Corollary For each k Z, we have δ(φ k 0 (X)) Φk+1 0 (X). Moreover, δ(φ k c (X)) Φ k+1 c (X). We also have the following corollary of Lemma Corollary Let f : X Y be a continuous map. Then f! Φ k 0 (Y ) Φ k 0 (X). Furthermore, if f is proper1, f! Φ k c (Y ) Φ k c (X). 1 A continuous map f : X Y is proper if f 1 C is compact for each compact set C Y. 12

14 2.2 Algebraic intermission We recall some basic algebraic notions and facts. Let V be a vector space Quotient spaces Given a subspace W V, the coset v + W of v V is the affine subspace {v + w V : w W }. The set V/W of all cosets {v + W : v V } is a partition of V and it induces an equivalence relation W on V ; we define v W v if and only if v v W. Fact The addition and multiplication with scalar in V descend naturally to corresponding operators on V/W, that is, we have well-defined addition and scalar multiplication in V/W given by (v + W ) + (v + W ) = (v + w) + W and a (v + W ) = (av) + W for v + W, w + W V/W and a R. Convention Typically the element v +W of V/W is denoted also by [v] suppressing the subspace W from the notation. We follow this convention in forthcoming sections. Fact Let f : V V be a linear map and W ker f. Then there exists a unique linear map f : V/W V satisfying V f V v [v] V/W f Moreover f is an isomorphism if fv = V and W = ker f Chain complexes Definition A sequence V # = (V k, α k ) k Z of vector spaces and linear maps α k : V k V k+1 is a chain complex if α k+1 α k = 0. The homology H (V # ) of V is the sequence (H k (V # )) k Z where for each k Z. H k (V # ) = ker α k /Imα k 1 Example Sequences (Φ k (X), δ k ) k Z, (Φ k 0 (X), δk ) k Z, and (Φ k c (X), δ k ) k Z are chain complexes. We are not interested in their homologies, see Spanier [Spa66] for a reason. 13

15 Remark Note that Imα k 1 ker α k, since α k α k+1 = 0. Definition Let V # = (V k, α k ) and W # = (W k, β k ) be chain complexes. A sequence f # = (f k : V k W k ) k Z is a chain map f # : V # W # if α k V k commutes. V k+1 f k f k+1 W k W k+1 β k Example Pull-back homomorphisms f! : Φ k (Y ) Φ k (X), and their restrictions f! : Φ k 0 (Y ) Φk 0 (X) and f! : Φ k c (Y ) Φ k c (X), are chain maps. Convention In this context it is common to consider chain complexes of abelian groups and their homomorphisms. For this reason, we will call the chain maps (and virtually all linear maps) as homomorphisms from now on. It should be kept in mind, though, that groups and homomorphisms we consider are actually vector spaces and linear maps! Lemma Let V # and W # be chain complexes and f # : V # W # a chain map. Then, for each k Z, there exists a well-defined linear map f := f k : H k (V # ) H k (W # ) satisfying f k ([v]) = [f k (v)]. Proof. Since f is a chain map, f k (ker α k ) ker β k and Imf k α k 1 Imβ k 1. Thus we have a diagram f k ker α k ker β k v [v] w [w] ker α k /Imα k 1 f k ker β k /Imβ k Exact sequences Definition Let A, B, and C be vector spaces and f : A B and g : B C linear maps. A sequence is exact at B if ker g = Imf. A f B g C A basic result on exact sequences is the Five Lemma 2. 2 Typically the introduction of the Five Lemma is followed by saying Chace the diagram. 14

16 Fact Let f 1 A 1 f 2 A 2 f 3 A 3 f 4 A 4 h 1 = h 2 h 3 = h 4 A 5 B 1 g 1 B 2 g 2 B 3 g 3 B 4 g 4 B 5 be a commutative diagram of vector spaces and linear maps having exact rows. Suppose that h 2 and h 4 are isomorphisms. Suppose also that h 1 is surjective and h 5 is injective. Then h 3 is an isomorphism. A sequence A # f # B # g # C # of chain complexes and chain maps is exact if h 5 is exact. Definition A sequence A k f k B k g k C k 0 A B C 0 of vector spaces and linear maps is a short exact sequence if it is exact at A, B, and C. A sequence A k 1 A k A k+1 of vector spaces and linear maps is a long exact sequence if the sequence is exact at each A k. Example Let f : V W be a linear map. The sequence 0 ker f is a short exact sequence. V f Imf 0 The short and long exact sequences of chain complexes are defined similarly. Fact Let V # = (V k, α k ) be a chain complex. Then H k (V # ) = 0 if and only if the sequence V k 1 α k 1 V k α k V k+1 is exact at V k 15

17 Fact Let 0 A f B 0 be an exact sequence. Then f is an isomorphism. Indeed, since the sequence is exact, ker f = {0} and Imf = B. A beautiful fact, which motivates for us the whole discussion in this section, is that short exact sequence of chain complexes yields a long exact sequence in (co)homology. Fact Let f # g 0 A # # B # C # 0 be a short exact sequence of chain complexes (and chain maps). Then there exists homomorphisms k : H k (C # ) H k+1 (A # ) (so-called connecting homomorphisms) for which H k (A # ) f H k (B # ) g H k (C # ) k H k+1 (A # ) is a long exact sequence. Idea of the proof: Show that the homomorphism H k (C # ) H k+1 (A # ), [c] [f 1 # β kg 1 # c] is well-defined and satisfies the required properties by chasing the commutative diagram g 0 A k # B k C k 0 0 A k+1 f # B k+1 C k+1 0 β k Finally, we remark the construction of the long exact sequence is natural. Fact Let 0 A # f # B # g # C # 0 h 1 # h 2 # h 3 # 0 A # f # B # g # C # 0 be a commutative diagram of chain complexes and chain maps having exact rows. Then the diagram H k (A # ) f H k (B # ) g H k (C # ) k H k+1 (A # ) h 1 h 2 h 3 h 1 H k (A # ) f H k (B # ) g H k (C # ) k H k+1 (A # ) commutes. 16

18 2.2.4 Remarks All the results in Section 2.2 hold also if we replace vector spaces and linear maps with abelian groups, or modules over commutative rings, and their homomorphisms. 2.3 Cochains C k (X) The Alexander Spanier k-cochains are defined as equivalence classes of k- functions modulo locally trivial k-functions. The formal definitions are as follows. Definition The elements of the quotient space C k (X) = Φ k (X)/Φ k 0(X) are called (Alexander Spanier) k-cochains and the space C k (X) as the space of (Alexander Spanier) k-cochains in X. Similarly, the elements of C k c (X) = Φ k (X)/Φ k 0(X) are compactly supported k-cochains and C k c (X) is the space of compactly supported (Alexander Spanier) k-cochains in X. We note first that Φ 0 0 (X) = {0} by Remark Thus we have the following identification. Fact Quotient maps Φ 0 (X) C 0 (X) and Φ 0 c(x) C 0 c (X), φ [φ], are isomorphisms. The second observation is that a k-cochain has a well-defined support. Lemma Let [φ] C k (X), then the set spt([φ]) = spt(φ) is welldefined. Proof. Let φ be a k-function for which [φ] = [φ ]. Then φ φ Φ k 0 (X), that is, spt(φ φ ) =. Thus spt(φ) spt(φ φ ) spt(φ ) spt(φ ) and spt(φ ) spt(φ φ) spt(φ) spt(φ) by Lemma Thus spt(φ) = spt(φ ). It is a direct consequence of Lemma that the coboundary operator δ k : Φ k (X) Φ k+1 (X) descends to a coboundary operator d k : C k (X) C k+1 (X) on cochains. We leave the details of this fact to the interested reader. Fact For each k Z there exists a linear map d k : C k (X) C k+1 (X) for which d k [φ] = [δ k (φ)] for all Φ k (X). In particular, d k+1 d k = 0 for each k. Furthermore, d k has a well-defined restriction d k c = d k Cc k (X): Cc k (X) Cc k+1 (X). 17

19 Corollary The sequences C # (X) = (C k (X), d k ) k Z and C # c (X) = (C k (X), d k c ) k Z are chain complexes. Proof. It suffices to observe that (d d)[φ] = d[δ(φ)] = [δ(δ(φ))] = 0. Similarly, given a continuous map f : X Y, the pull-back f! : Φ k (Y ) Φ k (X) descends to a pull-back f # : C k (Y ) C k (X). More precisely, we have the following. Lemma Let f : X Y be a continuous map. Then there exists a homeomorphism f # : C k (Y ) C k (X) satisfying f # [φ] = [f! φ]. Furthermore, if f is proper, the restriction f # : C k c (Y ) C k c (X) is well-defined. Proof. By Corollary , f! Φ k 0 (Y ) Φk 0 (X). Thus there exists a homomorphism f # : Φ k (Y )/Φ k 0 (Y ) Φk (X)/Φ k 0 (X) satisfying Φ k (Y ) f! Φ k (X) Φ k (Y )/Φ k 0 (Y ) f # Φ k (X)/Φ k 0 (X) where vertical arrows are quotient maps. Similar application of Corollary gives also the other claim. The inclusions of open sets have again special role regarding the support. We record this observation as a corollary. Corollary Let U X be an open set and c Cc k (X). Then spt(ι # UX c) = spt(c) U. In particular, if spt(c) U then spt(ι# UXc) = spt(c). Proof. Let φ Φ k (X) be a representative of c, that is, [φ] = c. By Lemmas and , spt(ι # UX c) = spt([ι! UXφ]) = spt(ι! UXφ) = spt(φ) U = spt(c) U. The second claim follows immediately. It is straightforward to see that the coboundary operator commutes with f #. We record this as a fact. 18

20 Fact Let f : X Y be a continuous map. Then C k (Y ) d k C k+1 (Y ) f # C k (X) d k f # C k+1 (X) commutes. In particular, f # : C # (Y ) C # (X) is a chain map. Similarly, if f is proper, f # : C # c (Y ) C # c (X) is a chain map. We finish this section by introducing more standard terminology. Definition A k-cochain c C k (X) is a k-cocycle, if d k c = 0, and a k-coboundary if there exists c C k 1 (X) for which d k 1 c = c. By Lemma the 0-cocycles are given by locally constant 0-functions. Corollary Let X be a space and φ Φ 0 (X). Then [φ] is a cocycle if and only if φ is locally constant, that is, there exists an open cover U of X so that φ U : U R is a constant function for each U U. Proof. Since d[φ] = [δ(φ)], we have that d[φ] = 0 if and only if spt(φ) =. This is equivalent, by Lemma , to the fact that φ is locally constant 2.4 Cohomologies H ( ) and H c ( ) Since ker d k+1 imd k 1 for each k Z, we have well-defined quotient spaces and H k (X) = ker d k/ imd k 1 = H k c (X) = ker d k c / imd k 1 c = {[φ] C k (X): d k [φ] = 0} {d k 1 [ψ] C k (X): [ψ] C k 1 (X)} {d k 1 c {[φ] Cc k (X): d k c [φ] = 0} [ψ] Cc k (X): [ψ] c C k 1 (X)} for each k Z. Heuristically, H k (X) measures the amount of non-trivial solutions of the equation d k [φ] = d k [φ ] i.e. the solutions [φ] and [φ ] for which the equation [φ] [φ ] = d k 1 [ψ] does not hold. In the terminology of Section 2.2, H k (X) and H k c (X) are homologies of complexes (C k (X), d k ) k Z and (C k c (X), d k c ) k Z, respectively. They are called Alexander Spanier cohomologies of the space X. Definition For k Z, the vector space H k (X) is the kth Alexender Spanier cohomology group of X and H k c (X) the kth compactly supported Alexander Spanier cohomology group of X. c 19

21 Remark If X is compact, H k c (X) = H k (X). Note however that, although C k c (X) C k (X), the quotient space H k c (X) is not a subspace of the quotient space H k (X). Example Let X be a point. Then { H k (X) = Hc k (X) R, if k = 0, = 0, otherwise. Indeed, the claim is a direct consequence of the observations that Φ k (X) = χ X k+1 and δ k (χ X k+1) = (1 ( 1) k )χ X k+2 for each k, where χ X k+1 is the characteristic function of X k+1, that is, x 1. Example Let X be a space and let L X be the space of locally constant 0-functions on X. Then the linear map L X H 0 (X), φ [φ], is an isomorphism. In particular, H 0 (X) = R if X is connected. Indeed, since C 1 (X) = {0}, the quotient homomorphism ker d 0 H 0 (X) is an isomorphism. On the other hand, by Lemma , the homomorphism L X ker d 0, φ [φ], is an isomorphism. For compactly supported cohomology, we have the following observation. Example Let X be a connected space. Then { Hc 0 (X) R, if X is compact, = 0, otherwise Indeed, if X is compact, H 0 c (X) = H 0 (X) = R. Suppose X is not compact. Then ker d 0 c is the space of constant functions having compact support. Since X is connect and not compact, we have ker d 0 c = 0. Thus H 0 c (X) = 0. Remark The interested reader may be interested calculating H 1 (R) and H 1 c (R) by hand at this stage. We return to this later with other tools. A continuous map f : X Y induces a pull-back in cohomology. Corollary Let f : X Y be a continuous map. Then there exists a homomorphism f : H k (Y ) H k (X) satisfying f [c] = [f # c]. Moreover, if X is Hausdorff and f is proper, there exists a homomorphism f : H k c (Y ) H k c (X) satisfying f [c] = [f # c]. Remark The pull-back f is natural in the sense that, id = id and f h = (h f) when ever the composition h f is defined. In particular, if f : X Y is a homeomorphism, both homomorphisms f : H k (Y ) H k (X) and f : H k c (Y ) H k c (X) are isomorphisms. 20

22 2.5 Exact sequence of a pair in H ( ) We begin with an easy observation. Lemma Let A X be a subset. The homomorphism ι # AX : Ck (X) C k (A), induced by inclusion ι AX : A X, is surjective. Proof. Let φ Φ k (A). We define ψ : X k+1 R by { ψ(x), x A k+1 ψ(x) = 0, otherwise Then ι! AX (ψ) = ψ A k+1 = φ and ι# AX [ψ] = [ι! AX (ψ)] = [φ]. Thus ι# AX is surjective. We denote C k (X, A) = ker By Lemma 2.5.1, we have an exact sequence ( ) ι # AX : Ck (X) C k (A) C k (X). 0 C k (X, A) j C k (X) ι# AX C k (A) 0 of chain complexes, where the homomorphism j is just the inclusion. Since ι # AX is a chain map, the sequence C (X, A) = (C k (X, A), d k ) k Z is a welldefined chain complex. Fact We have a long exact sequence in (co)homology H k (C (X, A)) j H k (X) ι AX H k (A) k H k (C (X, A)) 2.6 Push-forward in H c ( ) From now on X is a locally compact and second countable, Hausdorff space, and A X a closed subset. Let also U = X \ A and denote ι: U X the corresponding inclusion. Fact For each k Z, let Q k c (X, U) = {c C k c (X): spt(c) U}. Then Q (X, U) = (Q k (X, U), d k c ) is a chain complex. Theorem The homomorphism ι # UX : Ck (X) C k (U) maps Q k c (X, U) onto C k c (U) isomorphically, that is, the restriction i XU := ι # UX Q k c (X,U) : Q k c (X, U) C k c (U) is a well-defined chain isomorphism, i.e. a chain map which is also an isomorphism on each level. 21

23 Definition Let σ XU : C c (U) C c (X) be the unique injective homomorphism satisfying Q k c (X, U) i XU = C k c (U) σ XU C k c (X) We state it as a lemma that σ XU does not change the support. We formalize this as follows. Lemma Let U X be an open subset and c Cc k (U). spt(σ XU (c)) = spt(c) as subsets of X. Then Proof. First we note that ι # UX (σ XU(c)) = i XU (σ XU (c)) = c. By Corollary 2.3.7, spt(ι # UX (σ XU(c))) = spt(σ XU (c)). Lemma The homomorphism σ XU : C k c (U) C k c (X) is a chain map. In particular, it decends to a homomorphism in cohomology τ XU : H k c (U) H k c (X), [c] [σ XU c]. Proof. Since ι # is a chain map, we have i XU (dσ XU (c)) = di XU σ XU (c) = d(c) = i XU σ XU d(c) for each c C k c (U). Thus dσ XU = σ XU d. Hence σ XU is a chain map and decends to cohomology. Definition The homomorphism τ XU : H k c (U) H k c (X), [c] [σ XU c], is the push-forward induced by U X in compactly supported cohomology. Homomorphisms σ XU : C k c (U) C k c (X) and τ XU : H k c (U) H k c (X) are natural in the following sense. Lemma Let U V X be open sets. Then σ XU = σ XV σ V U. In particular, τ XU = τ XV τ V U. Proof. We observe first that, clearly, Q k c (X, U) Q k c (X, V ). Moreover, the isomorphism i XV : Q k c (X, V ) Cc k (V ) has a well-defined restriction i U XV : Qk c (X, U) Q k c (V, U). Indeed, given [φ] Q k c (X, U), i XV [φ] Cc k (U) and spt(ι # UX [φ]) = spt(ι! UX φ) U. Thus i XV [φ] Q k c (V, U), which proves the claim. We show next that i V U i U XV = i XU. Let [φ] Q k c (X, U). Then (i V U i U XV )[φ] = [ι! V U(ι! V X(φ))] = [(ι! V U ι! V X)(φ)] = [ι! UX(φ)] = i XU [φ]. Thus this claim is proved. 22

24 We conclude that the diagram Q k j 1 c (X, U) Q k j 2 c (X, V ) C k c (X) i XU i U XV i XV Q k c (V, U) j 3 Cc k (V ) σ XV i V U C k c (U) σ V U where homormorphisms j 1, j 2, and j 3 are inclusions, is well-defined and commutes. Hence σ XV σ V U = σ XU by uniqueness Compact supports of classes in H k c (X) Using the push-forward, we formalize the notion of compact support of a cohomology class. This gives a justification for the term compactly supported cohomology. Lemma Let a H k c (X). Then there exists a pre-compact open subset U X so that a Imτ XU. Proof. Let u Cc k (X) so that a = [u]. Since spt u is well-defined and compact, and the space X is locally compact, there exists a pre-compact open set U containing spt u with U compact. Since u Q k c (X, U), there exists v Cc k (U) so that σ XU (v) = u. Thus τ XU ([v]) = [σ XU (v)] = [u]. Lemma Let U X be a pre-compact open subset and a H k c (U) such that a ker τ XU. Then there exists a pre-compact open subset V X containing U so that a ker τ V U. Proof. Let u Cc k (U) be a representative of a, that is, a = [u]. Since [σ XU (a)] = τ XU (a) = 0, there exists w Cc k (X) so that σ XU (u) = δw. Let W be a pre-compact open neighborhood of spt(w) and V = U W. Since w Q k c (X, V ), there exists v Cc k (V ) so that σ XV (v) = w. Thus σ XV (σ V U (u)) = (σ V U σ XV )(y) = σ XU (u) = δw = δσ XV (v) = σ XV (δv). Since σ XV is injective, we conclude that σ V U (u) = δv. Thus τ V U (a) = [σ V U (u)] = [δw] = Proof of Theorem We prove now that i XU = ι UX Q k c (X, U): Q k c (X, U) C k c (U) 23

25 is an isomorphism. i XU is well-defined: Let c Q k (X, U). Then ι # c C k (U) and spt(ι # (c)) = spt(c) U is a compact set. Thus the restriction is well-defined. Injectivity: Let [φ] Q k (X, U) and suppose that [φ] ker ι #, or equivalently that spt(ι! φ) =. Since spt(φ) = spt(ι! (φ)) = and spt(φ) U, we have spt(φ) =. Thus [φ] = 0 and ker ι # is trivial. Hence ι # Q k (X,U) is injective. Surjectivity: It remains to prove that ι # Q k (X, U) = Cc k (U). Let [φ] Cc k (U). We construct a k-function ψ Φ k (X) so that [ψ] Q k c (X, U) and ι # [ψ] = [φ] by extending φ to a k-function ψ Φ k (X). Since spt(φ) is compact, we may fix pair-wise disjoint open sets W and V so that spt(φ) W and X \ U V and W is compact. 3 For each x U \ spt(φ), we fix a neighborhood U x U so that φ Ux = 0. We define a k-function ψ : X k+1 R by { φ(x), if x W k+1 ψ(x) = 0, otherwise. Note that spt(ψ) spt(φ) as subsets of X. Indeed, let x null(φ). Suppose first that x U \ spt(φ). Then ψ U k+1 = φ x U k+1 = 0. Suppose now that x x U. Then x V and we have ψ V k+1 = 0. Thus x null(ψ). Hence spt(ψ) spt(φ). In particular, ψ Φ k c (X). It remains now to show that [φ] = ι # [ψ], or equivalently null(φ ι! ψ) = U. Let x U. If x spt(φ) then x W and φ W k+1 = ψ W k+1 = (ι! ψ) W k+1 = 0. If x U \ spt(φ) then φ Ux = ψ Ux = (ι! φ) Ux. Thus x null(φ ι! ψ). This concludes the proof of Theorem Exact sequence of a pair in H c ( ) The goal in this section is to prove a counterpart of Fact in compactly supported cohomology. Theorem Let X be a locally compact and second countable Hausdorff space and A X a closed subset. Then there exists homomorphisms k : Hc k (A) Hc k+1 (X \ A) for which (2.7.1) H k c (X \ A) τ AX H k c (X) ι AX H k c (A) is a long exact sequence. The sequence (2.7.1) is natural in the following sense. k H k+1 c (X \ A) 3 Recall that a locally compact Hausdorff space X is T 3, i.e. given x X and a closed set E X \ {x} then there exists pair-wise disjoint neighborhoods of x and E. Use now compactness of spt(φ) and local compactness of X to find W and V. 24

26 Theorem Let A and B be closed subsets of X so that A B, and let U = X \ A and V = X \ B. Then the diagram Hc k τ XV (X \ B) Hc k (X) ι BX Hc k (B) τ UV ι AB H k c (X \ A) τ XU H k c (X) ι AX H k c (A) B H k+1 c (X \ B) τ UV A Hc k+1 (X \ A) where the rows are exact sequences of pairs (X, A) and (X, B), commutes. Idea of the proof of Theorem The proof of Theorem spans over following three subsections. In the proof is based on comparisons of four(!) complexes: subcomplexes Q c(x, U) Cc (X, A) of Cc (X) and subcomplexes Rc(X, U) Φ c(x, A) of Φ k c (X), where U = X \ A. The subcomplex Q c(x, U) is familiar from the construction of the pushforward and contains cochains supported away from A. The complex Cc (X, A) is slightly larger subcomplex of Cc k (X) which consists the cochains which belong to the kernel of the restriction ι # AX. More formally, we set Q k c (X, U) = {c C k c (X): spt(c) U} and ( ) Cc k (X, A) = ker ι # AX : Ck c (X) Cc k (A), and define Q c(x, U) = ( ) Q k c (X, U), d k Z and C c (X, A) = ( ) Cc k (X, A), d, k Z where the coboundary operator d is the restriction of d: C c (X) C c (X). We leave it to the reader to check that these complexes are well-defined and that Q c(x, U) is a subcomplex of C c (X, A). Remark It is important to keep in mind that c C k c (X, A) does not imply that spt(c) A =. The subcomplexes R c(x, U) and Φ c(x, A) of Φ c(x) are, corresponding representatives of Q c(x, U) and Φ c(x, A). More formally, we set and and define R c(x, U) = R k c (X, U) = {φ Φ k c (X): spt(φ) U} Φ k c (X, A) = {φ Φ k c (X): spt(ι! AXφ) = 0}, ( ) Rc k (X, U), δ k Z and Φ c(x, A) = 25 ( ) Φ k c (X, A), δ,

27 where δ is a restriction of δ : Φ c(x) Φ c(x). Again, we leave it to reader to verify that these complexes are well-defined and that R c(x, U) is a subcomplex of Φ k c (X, A). The goal of the first subsection (Section 2.7.1) is to show (Lemma that, for each k Z, there exists an isomorphism H k ( Φ c (X, A) / R c(x, U) ) H k ( C c (X, A) / Q c(x, U) ). We then continue using topological arguments to show (Theorem 2.7.5) that H k ( Φ c (X, A) / R c(x, U) ) = 0 for all k Z. Heuristically, the existence of this isomorphism states that a cohomology class of cochain, which trivializes when restricted to A, can be represented with a cochain having support in X \ A. This heuristic statement is made precise in Lemma 2.7.7; the reader may be interested to observe that actually the topological Lemma is the heart of the proof of everything here. These arguments together then give that, for each k Z, there exists an isomorphism (Corollary 2.7.6) H k (Q c(x, U)) H k (C c (X, A)). Using this isomorphism, we then finish the proof of Theorem in Section Quotient complexes Φ c(x, A)/R c(x, U) and C c (X, A)/Q c(x, U) Recall that, given a complex A = (A k, α) k Z and a subcomplex B = (B k, α) k Z, we have the quotient complex A /B = (A k /B k, ᾱ) k Z, where the coboundary operator ᾱ is defined using the diagram a a+b k A k α A k+1 a a+b k+1 A k /B k ᾱ A k+1 /B k+1 The purpose of this section is to prove the following algebraic result, which we formulate as a lemma. First we fix some notation. Let q : Φ k c (X, A) C k c (X, A)/Q k c (X, U) be the composition Φ k c (X, A) φ [φ] C k c (X, A) q c c+q k c (X,U) C k c (X, A)/Q k c (X, U) 26

28 Lemma The homomorphism q : Φ k c (X, A) C k c (X, A)/Q k c (X, U) admits a factorization (2.7.2) Φ k c (X, A) φ φ+r k c (X,U) Φ k c (X, A)/R k c (X, U) φ [φ] q q C k c (X, A) c c+q k c (X,U) C k c (X, A)/Q k c (X, U) where q : Φ k c (X, A)/Rc k (X, U) Cc k (X, A)/Q k c (X, U) is an isomorphism. In particular, ) ( ) q : H k (Φ k c (X, A)/Rc k (X, U) H k Cc k (X, A)/Q k c (X, U) is an isomorphism. Proof. We show first that q is surjective. Let c Cc k (X, A)/Q k c (X, U). Then c = [φ] + Q k c (X, U) for some [φ] Cc k (X, A). Then, by definition of Cc k (X, A), ι # AX [φ] = 0. Thus spt(ι! AX φ) =. Hence φ Φk c (X, A) and q(φ) = [φ] + Q k c (X, U). We show next that ker q = Rc k (X, U). Let φ Rc k (X, U). Then [φ] Q k c (X, U). Thus q(φ) = 0. On the other hand, if q(φ) = 0, then [φ] Q k c (X, U). Thus spt(φ) U, i.e. φ Rc k (X, U). Since ker q = Rc k (X, U), we may factor q through Φ k c (X, A)/Rc k (X, U), i.e. we find a homomorphism q : Φ k c (X, A)/Rc k (X, U) Cc k (X, A)/Q k c (X, U) making the diagram (2.7.2) commutative. Since q is surjective, also q is surjective. Since q is injective by construction, q is an isomorphism. Since q is a chain map, we conclude that also q is a chain map. Thus q is a chain isomorphism and q is an isomorphism Homology of Φ c(x, A)/R c(x, U) In this section we prove the following result. Theorem For each k Z, H k ( Φ c (X, A) / R c(x, U) ) = 0 A simple short exact sequence argument now yields the following corollary. Corollary The inclusion j : Q c(x, U) C c (X, A) induces, for each k Z, an isomorphism j : H k (Q c(x, U)) H k (C c (X, A)). 27

29 Proof. Consider the short exact sequence (2.7.3) 0 Q c(x, U) j Cc (X, A) Cc (X, A)/Q c(x, U) 0 of chain complexes. By Lemma and Theorem 2.7.5, ( H k C c (X, A) / Q c(x, U) ) ( = Hk Φ c (X, A) / Rc(X, U) ) = 0. Hence the sequence 0 H k (Q c(x, U)) j H k (C c (X)) 0 is exact and j is an isomorphism. We may reformulate Theorem as follows. Lemma Let φ Φ k c (X, A) be a cocycle, that is, δφ = 0. Then there exists ψ Rc k (X, U) and ρ Φc k 1 (X, A) for which φ = ψ + δρ. Proof of Theorem assuming Lemma We need to show that ker δ = Im δ, where δ is the coboundary operator of quotient complex Φ c(x, A) / Rc(X, U). More concretely, given a cochain φ Φ k c (X, A) / Rc k (X, U) so that δ φ = 0, we show that there exists φ Φc k 1 (X, A) / Rc k 1 (X, U) so that φ = δ φ. Let φ Φ k c (X, A) / Rc k (X, U) be a cocycle, that is, δ φ = 0. Then φ = φ + Rc k (X, U), where φ Φ k c (X, A) satisfies δφ = 0. By Lemma 2.7.7, there exists ψ Rc k (X, U) and ρ Φ k 1 c (X, A) so that φ = ψ + δρ. Thus φ = (ψ + δρ) + Rc k (X, U) = δρ + Rc k (X, U) = δ(ρ + Rc k 1 (X, U)). This concludes the proof. We are now left with the proof of Lemma The proof is topological and we do in two steps. We begin with an observation on (non-continuous) perturbations of the identity map id: X X. For this, we have a small remark on coverins of space X. Remarks on coverings The space X we consider is locally compact, second countable, and Hausdorff. Thus X is regular [Dug78, Theorem XI.6.4] and X is Lindelöf [Dug78, Theorem XI.7.2]. Regular Lindelöf spaces are paracompact [Dug78, Theorem VIII.6.5]. Recall that a Hausdorff space X is paracompact if each open covering has a locally finite refinement. 28

30 Recall that a covering V refines covering U if for each V V there exists U U for which V U, and that a covering U of X is locally finite if for each x X there exists a neighborhood W so that #{U U : U W } <. An important on paracompact Hausdorff spaces we use is that each open covering has a star refinement, see [Dug78, Theorem VIII.3.5]. Some definitions are in order. Given a covering U of X, a star of U U in U is the set U = {U U : U U }. A star U of U is the covering U = {U : U U}. So, by [Dug78, Theorem VIII.3.5], given a covering U of a paracompact Hausdorff space X, there exists a covering V of X so that V refines U. Perturbation of the identity and chain homotopy in Φ k (X) Lemma Let A X be a closed set and U = X \ A. Let W be a locally finite open covering of X. Then there exists a function 4 f : X X so that fw W for each W W and fw A A. Moreover, f has the following properties: (a) Suppose φ Φ k c (X) satisfies φ (W A) k+1 = 0 for each W W meeting A, then f! φ Rc k (X, X \ A). (b) Let F i : X k X k+1 be the function (x 1,..., x k ) (f(x 1 ),..., f(x i ), x i,..., x k ). Suppose ψ Φ k c (X) satisfies (ψ F i ) (W ) k+1 = 0 for each W W meeting A. Then ψ F i Rc k 1 (X, X \ A). Further, if (ψ F i ) (W A) k+1 = 0 for each W W, then ψ F i Φ k c (X, A). Proof. We define f as follows. For each W W, we fix a point x W W, and if W A, we fix x W W A. Then, for each x X \ W A, we choose W W, and define f(x) = x W. For x W A, we fix W W for which x W and W A, and then define f(x) = x W. Now, by construction, fw A if W W meets A. To show that fw W, let W W and x W. Then f(x) = x W, where W W, x W, and W W. Thus W W and f(x) W. To show (a), let φ Φ k c (X, A). We need to show that spt(f! φ) X \ A, or equivalenly, A null(f! φ). Let x A and let W W be such that x W. Let x 1,..., x k+1 W. Then f(x i ) W A for each i = 1,..., k. Thus f! φ(x 1,..., x k+1 ) = φ(f(x 1 ),..., f(x k+1 )) Thus f! φ W k+1 = 0. Hence x (f! φ). 4 Read: non-continuous function. = φ (W A) k+1(f(x 1),..., f(x k+1 )) = 0. 29

31 Claim (b) is proved with essentially the same argument. Let i {1,..., k} and ψ R k c (X, U). Let x A and W W a neighborhood of x. Let also x 1,..., x k W and observe that F i (x 1,..., x k ) (W ) k+1. Thus (ψ F i )(x 1,..., x k ) = 0. Thus (ψ F i ) W k = 0. Hence x null(ψ F i ). This completes the proof. Finally, we need to show that spt(ι! AX (ψ F i)) = if ψ (W A) k+1 = 0 for each W W which meets A. Let x A and W W a neighborhood of x. Let x 1,..., x k W A. Then f(x j ) W A for each j = 1,..., i. Hence F i (x 1,..., x k ) (W A) k+1. Thus (ψ F i )(x 1,..., x k ) = 0. We conclude that x null(ψ F i ). This concludes the proof. It is interesting observation that the maps f : X X are chain homotopic to the identity id: X X in Φ k (X). We apply this result to subspaces of Φ k c (X) but we formalize this general result as follows. Lemma Let f : X X be a (non-continuous) map. Let D : Φ k (X) Φ k 1 (X) be a homomorphism defined by (Dφ)(x 1,..., x k ) = k ( 1) i+1 φ(f(x 1 ),..., f(x i ), x i,..., x k ) i=1 for φ Φ k (X) and x 1,..., x k X. Then id f! = δd + Dδ Proof. We have (δdφ)(x 1,..., x k ) = and k ( 1) j+1 (Dφ)(x 1,..., x j,..., x k ) j=1 (Dδφ)(x 1,..., x k ) = k ( 1) i+1 (δφ)(f(x 1 ),..., f(x i ), x i,..., x k ). i=1 On the other hand, (Dφ)(x 1,..., x j,..., x k ) j 1 = ( 1) i+1 φ(f(x 1 ),..., f(x i ), x i,..., x j,..., x k ) i=1 + k ( 1) (i+1) 1 φ(f(x 1 ),..., f(x j ),..., f(x i ), x i,..., x k ) i=j+1 30

32 and Thus (δφ)(f(x 1 ),..., f(x i ), x i,..., x k ) i = ( 1) j+1 φ(f(x 1 ),..., f(x j ),..., f(x i ), x i,..., x k ) j=1 k + ( 1) (j+1)+1 φ(f(x 1 ),..., f(x i ), x i,..., x j,..., x k ) j=i (δdφ)(x 1,..., x k ) + (Dδφ)(x 1,..., x k ) k j 1 = ( 1) j+1 ( 1) i+1 φ(f(x 1 ),..., f(x i ), x i,..., x j,..., x k ) j=1 i=1 k k + ( 1) j+1 ( 1) i φ(f(x 1 ),..., f(x j ),..., f(x i ), x i,..., x k ) + + j=1 i=j+1 k i ( 1) i+1 ( 1) j+1 φ(f(x 1 ),..., f(x j ),..., f(x i ), x i,..., x k ) i=1 j=1 k k ( 1) i+1 ( 1) j φ(f(x 1 ),..., f(x i ), x i,..., x j,..., x k ) i=1 j=i By rearranging the double sums, we obtain (δdφ)(x 1,..., x k ) + (Dδφ)(x 1,..., x k ) = ( 1) i+j φ(f(x 1 ),..., f(x i ), x i,..., x j,..., x k ) i<j i>j ( 1) i+j φ(f(x 1 ),..., f(x j ),..., f(x i ), x i,..., x k ) + j i( 1) i+j φ(f(x 1 ),..., f(x j ),..., f(x i ), x i,..., x k ) j i( 1) i+j φ(f(x 1 ),..., f(x i ), x i,..., x j,..., x k ) k = φ(f(x 1 ),..., f(x i 1 ), x i,..., x k ) i=1 k φ(f(x 1 ),..., f(x i ), x i+1,..., x k ) i=1 = φ(x 1,..., x k ) φ(f(x 1 ),..., f(x k )). 31