ALGEBRAIC TOPOLOGY IV. Definition 1.1. Let A, B be abelian groups. The set of homomorphisms ϕ: A B is denoted by

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1 ALGEBRAIC TOPOLOGY IV DIRK SCHÜTZ 1. Cochain complexes and singular cohomology Definition 1.1. Let A, B be abelian groups. The set of homomorphisms ϕ: A B is denoted by Hom(A, B) = {ϕ: A B ϕ homomorphism} Example 1.2. Let A = Z, B any abelian group. Then Hom(Z, B) = B by identifying ϕ with ϕ(1). If A = Z/2, B = Z, any homomorphism ϕ: Z/2 Z satisfies ϕ( 1) + ϕ( 1) = 0 Z which implies ϕ( 1) = 0, so Hom(Z/2, Z) = 0. Similarly Hom(Z/n, Z) = 0. Lemma 1.3. Let A, A, B, B be abelian groups. (1) Hom(A, B) is an abelian group. (2) If F is a field, then Hom(A, F) is a ector space oer F. (3) If I is a set and A i an abelian group for eery i I, then ( ) Hom A i, B = Hom(A i, B) i I i I (4) If f : A A is a homomorphism, there is an induced homomorphism f : Hom(A, B) Hom(A, B) gien by f (ϕ) = ϕ f. (5) If g : B B is a homomorphism, there is an induced homomorphism g : Hom(A, B) Hom(A, B ) gien by g (ϕ) = g ϕ. Example 1.4. Let C be a chain complex of abelian groups, and B an abelian group. Then let C n = Hom(C n, B). Also, for n+1 : C n+1 C n, define δ n = ( n+1 ) : C n C n+1. Furthermore δ n+1 δ n = 0, so we can define H n (C ) = ker δ n /im dδ n 1. Date: March 2,

2 2 DIRK SCHÜTZ Definition 1.5. A sequence of abelian groups (C n ) n Z with homomorphisms δ n : C n C n+1 such that δ n+1 δ n = 0 for all n Z is called a cochain complex, the resulting groups H n (C ) are called cohomology groups. Let X be a topological space, A X a subset. Then the singular cohomology groups of (X, A) with coefficients in the abelian group G are gien by H k (X, A; G) = H k (Hom(C (X, A), G)) for all k 0, where C (X, A) is the singular chain complex of the pair (X, A). If A =, we simply write H k (X; G). Example 1.6. To do calculations, we want to use cellular (co)homology rather than singular. Using the usual cell decompositions for the following spaces, we get { H (S n G = 0, n ; G) = 0 otherwise for n 1 and for all abelian groups G. For RP 2 we get H (S 0 ; G) = with all other groups zero and { G G = 0 0 otherwise H 0 (RP 2 ; Z) = Z H 1 (RP 2 ; Z) = 0 H 2 (RP 2 ; Z) = Z/2 H 0 (RP 2 ; Z/2) = Z/2 H 1 (RP 2 ; Z/2) = Z/2 H 2 (RP 2 ; Z/2) = Z/2 with all other groups zero. Remark 1.7. Typical properties from homology carry oer to cohomology ia Hom: If f : C D is a chain map, then f : Hom(D, G) Hom(C, G) is a cochain map: f δ = δ f, so f induces a map on cohomology groups. If H : C D +1 is a chain homotopy between two chain maps f, g : C D, then H : Hom(D n, G) Hom(C n 1, G) satisfies H δ + δh = f g, so f and g induce the same map on cohomology groups. Corollary 1.8 (Homotopy Inariance). Let X, Y be topological spaces, and let f, g : X Y maps which are homotopic. Then f, g induce the same homomorphism f = g : H (Y ; G) H (X; G).

3 ALGEBRAIC TOPOLOGY IV 3 Theorem 1.9 (Excision). Let X be a topological space, A X and B A with B A o. Then the inclusion map i: (X B, A B) (X, A) induces an isomorphism i : H = (X, A; G) H (X B, A B; G). f g Lemma Let 0 C D E 0 be a short exact sequence of chain complexes such that each E n for n Z is a free abelian group. Then for any abelian group G there is a short exact sequence of cochain complexes 0 Hom(E, G) g Hom(D, G) f Hom(C, G) 0 with a corresponding long exact sequence H k (C; G) δ H k+1 (E; G) g H k+1 (D; G) f H k+1 (C; G) Remark This does not work in general if E n is not free abelian, but the only thing that can fail is the surjectiity of f. Theorem Let (X, A) be a pair of topological spaces and G an abelian group. Then there is a long exact sequence of cohomology groups which is natural with respect to maps H k (A; G) δ H k+1 (X, A; G) H k+1 (X; G) H k+1 (A; G) Theorem 1.13 (Mayer-Vietoris sequence). Let X be a topological space, A, B X open subsets with X = A B and G an abelian group. Then there is a long exact sequence H k 1 (A B; G) H k (X; G) (i A,i B ) H k (A; G) H k (B; G) i j H k (A B; G) Remark From what we hae done with cohomology groups, it is not clear that the cohomology groups arising from the singular cochain complex gie the same as from the cellular cochain complex as was the case with homology. But we hae all the tools now needed to show this by adapting the homology proof. We will next show that the cohomology groups are in fact determined by the homology groups. 2. Extensions and the Uniersal Coefficient Theorem Remark 2.1. If A groups, then f B g C 0 is an exact sequence of abelian 0 Hom(C, G) g Hom(B, G) f Hom(A, G)

4 4 DIRK SCHÜTZ is an exact sequence. Een if f is injectie, f need not be surjectie. We now gie a construction that measures the failure of f being surjectie. Definition 2.2. Let M be an abelian group. A free presentation of M is gien by a free abelian group F and a surjection p: F M. Denote K = ker p so that we hae a short exact sequence This gies an exact sequence Define 0 K F M 0. 0 Hom(M, G) Hom(F, G) Hom(K, G) Ext p (M, G) = Hom(K, G)/im(Hom(F, G) Hom(K, G)). Lemma 2.3. Let M, G be abelian groups. Then Ext p (M, G) does not depend on the free presentation, and there always exists a free presentation p: F M. We thus drop the p from the notation and write Ext(M, G). Furthermore, if M, M are abelian groups, and f : M M and g : M M are homomorphisms, then there are homomorphisms f : Ext(M, G) Ext(M, G) and g : Ext(M, G) Ext(M, G) with (g f) = f g. Example 2.4. Let F be a free abelian group. Then id: F F is a free presentation, and Ext(F, G) = 0 for all G. In particular, Ext(Z, G) = 0 for all G. Ext(Z/n, Z) = Z/n. Ext(Z/n, Z/n) = Z/n. Ext(Z/9, Z/15) = Z/3. Lemma 2.5. Let M, M, G be abelian groups. Then Ext(M M, G) = Ext(M, G) Ext(M, G) Ext(M, Q) = 0 Ext(M, R) = 0. Theorem 2.6 (Uniersal Coefficient Theorem). Let C be a chain complex such that C n is free abelian for all n Z and G an abelian group. Then there is a short exact sequence which splits 0 Ext(H n 1 (C), G) H n (C; G) Hom(H n (C), G) 0 In particular, H n (C; G) = Hom(Hn (C), G) Ext(H n 1 (C), G) Corollary 2.7. Let (X, A) be a pair of topological spaces, G an abelian group. Then there is a short exact sequence 0 Ext(H n 1 (X, A), G) H n (X, A; G) Hom(H n (X, A), G) 0 which splits, that is, H n (X, A; G) = Hom(Hn (X, A), G) Ext(H n 1 (X, A), G)

5 ALGEBRAIC TOPOLOGY IV 5 Remark 2.8. If f : X Y is a map, there is a ladder 0 Ext(H n 1 (Y ), G) H n (Y ; G) Hom(H n (Y ), G) 0 f 0 Ext(H n 1 (X), G) H n (X; G) Hom(H n (X), G) 0 Howeer, if we write x H n (Y ; G) as x = (a, b) Hom(H n (Y ), G) Ext(H n 1 (Y ), G) it is not always true that f (x) = (f (a), f (b)). Corollary 2.9. Let (X, A) be a pair of topological spaces. Then for all n 0, and H n (X, A; Q) = Hom(Hn (X, A), Q) H n (X, A; R) = Hom(Hn (X, A), R) H 1 (X, A; Z) = Hom(H1 (X, A), Z). Example Let X be a finite CW-complex. Then each homology group H k (X) is finitely generated, that is, f H k (X) = Z r Z/k 1 Z/k l with k 1 k 2,..., k l 1 k l. The Z r produces a Z r in H k (X; Z) ia Hom and nothing ia Ext, and each Z/k i produces a Z/k i in H k+1 (X; Z) ia Ext, and nothing ia Hom. If we use coefficients in Z/2, a Z/2-summand produces Z/2 in H k (X; Z/2) and a Z/2 in H k+1 (X; Z/2). 3. Cup products and the ring structure of cohomology Definition 3.1. Let k be a commutatie ring with identity and X a topological space. If ϕ C k (X; k) and ψ C l (X; k), define their cup product ϕ ψ C k+l (X; k) by ϕ(σ) = ϕ(σ [ 0,..., k ]) ψ(σ [ k,..., k+l ]) where σ : k+l X is a singular k + l-simplex, and k+l = [ 0,..., k+l ]. Lemma 3.2. Let X be a topological space and ϕ C k (X; k) and ψ C l (X; k). Then δ(ϕ ψ) = δϕ ψ + ( 1) k ϕ δψ. Remark 3.3. If ϕ and ψ are both cocyles, then ϕ ψ is a cocycle. If ϕ or ψ is a coboundary, then ϕ ψ is a coboundary. This means there is a well-defined cup product on cohomology gien by H k (X; k) H l (X; k) H k+l (X; k) ([ϕ], [ψ]) [ϕ ψ] f

6 6 DIRK SCHÜTZ This product is associatie and distributie, as this already holds on the cochain leel. Furthermore, define 1 C 0 (X; k) by 1(σ 0 ) = 1 k for eery 0-simplex σ 0 : 0 X. This defines a cohomology class 1 H 0 (X; k) with the property that If we write x 1 = x = 1 x for all x H k (X; k) H (X; k) = H k (X; k) this is a ring with multiplication gien by the cup product and identity 1 H 0 (X; k). It is in fact a graded ring in the sense that it is a direct sum of abelian groups A k such that multiplication sends A k A l to A k+l. Example 3.4. Let X = S n with n 1. There is a generator x H n (S n ; k) and the identity 1 H 0 (S n ; k). There are no interesting products by grading reasons: x x = 0 as the whole group H 2n (S n ; k) = 0. Example 3.5. Look at RP 2 with the triangulation in Figure 1. The red k= Figure 1. The projectie plane circle represents a cochain A by saying that A([ i, j ]) = 1 Z/2 if it intersects [ i, j ], and 0 otherwise. This is in fact a cocycle. Also, a = [ 0, 1 ] + [ 1, 2 ] [ 0, 2 ] generates H 1 (RP 2 ), and since A(a) = 1 Z/2, A represents the generator of H 1 (RP 2 ; Z/2) = Hom(H 1 (RP 2 ), Z/2) = Z/2. For eery simplex σ we write σ C (RP2 ; Z/2) for the dual cochain. That is, σ (σ) = 1 Z/2 and σ (τ) = 0 for all τ σ. It is then easy to check that A A = [ 1, 3, 4 ] + [ 2, 4, 5 ] + [ 1, 3, 5 ].

7 ALGEBRAIC TOPOLOGY IV 7 Now [ i, j, k ] is a cocycle for eery 2-simplex [ i, j, k ] in the triangulation. But as eery 1-simplex [ i, j ] sits in exactly two 2-simplices, we see that the coboundaries are exactly those combinations [ i, j, k ] which inole an een number of 2-simplices. Therefore A A represents the generator of H 2 (RP 2 ; Z/2) = Z/2. For T 2 we use the triangulation in Figure Figure 2. The torus We hae H 1 (T 2 ) = Z 2 = a, b, where a is represented by [ 0, 1 ] + [ 1, 2 ] [ 0, 2 ], and b is represented by [ 0, 3 ] + [ 3, 6 ] [ 0, 6 ]. Also, H 2 (T 2 ) = Z generated by an alternating sum of all the 2-simplices. The torus is orientable, and we choose the counterclockwise orientation. The blue circle with its orientation pointing upward represents a 1-cochain A as follows. If the circle does not meet the 1-simplex [ i, j ], then A anishes on that 1-simplex, and if the circle intersects the 1-simplex [ i, j ] we define A([ i, j ] = ±1 Z, where the sign is determined as follows: First use the direction of the 1-simplex coming from the ordering of the simplicial complex (going from i to j for i < j) followed by the direction of the blue circle. If this represents the orientation of the torus, the sign is +1, and 1 otherwise. This sign conention ensures that A is a cocycle. Note that the only 1-simplex on which A is 1 is [ 2, 7 ]. We can do the same with the red circle, to obtain a cocycle B, and in fact B is +1 on all the 1-simplices it meets. Denote the cohomology class of A by α, and of B by β. As H 1 (T 2 ; Z) = Hom(H 1 (T 2 ), Z) = Z 2, and α(a) = 1, α(b) = 0, β(b) = 1 and β(a) = 0, α and β represent generators of H 1 (T 2 ; Z). A direct calculation shows that A B = [ 4, 5, 8 ], and B A = [ 4, 7, 8 ]. Both represent a generator of H 2 (T 2 ; Z), but note that [ 4, 5, 8 ] has the same orientation as the one chosen for the torus, while [ 4, 7, 8 ] has the

8 8 DIRK SCHÜTZ opposite orientation. Hence α β = β α, and this element generates H 2 (T 2 ; Z). Another calculation shows that A A = ([ 2, 7, 8 ] + [ 1, 2, 7 ] ) = δ( [ 2, 7 ] ), so α α = 0. A similar calculation shows β β = 0. Theorem 3.6. Let X be a topological space, and k a commutatie ring. Then H (X; k) is graded-commutatiein the sense that for a H k (X; k) and b H l (X; k) we hae a b = ( 1) k l b a. Remark 3.7. If a H k (X; k) with k odd, then a a = a a which implies 2 a a = 0. This need not imply that a a = 0 (compare RP 2 ), but if H 2k (X; k) has no 2-torsion, one can conclude that a a = 0 (as in T 2 ). Remark 3.8. If A X, we hae the short exact sequence 0 C (X, A; k) C (X; k) C (A; k) 0 so we can interpret C (X, A; k) as those cochains on X that anish on simplices in A. And if ϕ anishes on C k (A), then ϕ ψ anishes on C k+l (A). We therefore get relatie cup products H k (X, A; k) H l (X; k) H k+l (X, A; k) H k (X; k) H l (X, A; k) H k+l (X, A; k). Also, if ϕ anishes on C k (A) and ψ anishes on C l (B), then ϕ ψ anishes on C k+l (A)+C k+l (B). This is not quite the same as C k+l (A B), but if A, B are open subsets of X, or if X is a CW complex and A, B subcomplexes, we get a cup-product H k (X, A; k) H l (X, B; k) H k+l (X, A B; k). Proposition 3.9 (Naturality). If f : X Y is a map, the induced maps f : H n (Y ; k) H n (X; k) satisfy f (α β) = f α f β for all α H k (Y ; k) and β H l (Y ; k). The same formula holds in the relatie case. Example Let X and Y be CW complexes with chosen basepoint x 0 X and y 0 Y, and denote X Y the one-point union along the basepoints. Then the inclusions i X : X X Y and i Y : Y X Y induce an isomorphism (i X, i Y ): H k (X Y ; k) H k (X; k) H k (Y ; k)

9 ALGEBRAIC TOPOLOGY IV 9 for k > 0 by Mayer-Vietoris, with inerse isomorphism induced by the retractions r X : X Y X and r Y : X Y Y ia r X + r Y : H k (X; k) H k (Y ; k) H k (X Y ; k). To calculate cup-products, let a, b H (X Y ; k). a = rx (a 1) + ry (a 2) and b = rx (b 1) + ry (b 2), and get a b = (r X(a 1 ) + r Y (a 2 )) (r X(b 1 ) + r Y (b 2 )) We can then write = r X(a 1 b 1 ) + r X(a 1 ) r Y (b 2 ) + r Y (a 2 ) r X(b 1 ) + r Y (a 2 b 2 ). Now to understand r X (a 1) r Y (b 2), look at (i X, i Y )(r X (a 1) r Y (b 2)). Then i X(r X(a 1 ) r Y (b 2 )) = (r X i X ) (a 1 ) (r Y i X ) (b 2 ) But r Y i X : X Y is the constant map to y 0, so induces 0 on cohomology. For this and similar reasons, we get that r X (a 1) r Y (b 2) = 0 and r Y (a 2) r X (b 1) = 0, which means that a b = r X(a 1 b 1 ) + r Y (a 2 b 2 ). Now let M be the surface of genus 2, and f : M T 2 T 2 the obious map. Then H 1 (M; Z) = a1, β 1, α 2, β 2 where α i = f (A i ) and β i = f (B i ) for i = 1, 2 and A 1, B 1 generate the first summand of H 1 (T 2 ; Z) in H 1 (T 2 T 2 ; Z) and A 2, B 2. It follows from naturality that α 1 β 1 and α 2 β 2 both generate H 2 (M; Z) = Z, and all other combinations (up to commuting) are The fundamental class of a manifold Definition 4.1. A Hausdorff space X is called a topological manifold of dimension n, if each point x X has a neighborhood U which is homeomorphic to R n. It is called a topological manifold with boundary, if each x X has a neighborhood U homeomorphic to R n or to [0, ) R n 1. Then X = {x X x does not hae a neighborhood homeomorphic to R n }. Example 4.2. A discrete space X is a topological manifold of dimension 0. S n is a topological manifold of dimension n, D n is a topological manifold with boundary of dimension n. Lemma 4.3. Let X n be a topological manifold and x X. Then H k (X, X {x}) = { Z k = n 0 k n

10 10 DIRK SCHÜTZ Lemma 4.4. R n is not homeomorphic to [0, ) R n 1. If X is a topological manifold with boundary of dimension n, then X is a topological manifold of dimension n 1. Remark 4.5. Topological manifolds need not admit triangulations. We will howeer assume that the manifolds we look at admit a triangulation. Lemma 4.6. Let X be a manifold of dimension n with a triangulation, and let σ be an (n 1)-simplex. Then there exist exactly two n-simplices τ 1, τ 2 such that σ is a face of τ 1 and a face of τ 2. Theorem 4.7. Let X be a compact, connected manifold of dimension n, possibly with boundary X. Let k be a commutatie ring. Then H n (X, X; k) is either k or 2 k = {k k 2k = 0}. Note that for k = Z/2, we hae k = 2 k. Definition 4.8. Let M n be a compact, connected manifold and k a commutatie ring. We call M k-orientable, if H n (M, M; k) = k. A generator [M] H n (M, M; k) is called a fundamental class. For k = Z/2 there is a unique generator for any compact, connected manifold. For k = Z there exist two generators [M] and [M], proided that M is Z-orientable. 5. Duality Theorems Definition 5.1. Let X be a topological space, ϕ C p (X; k), σ : n X a singular simplex with p n. Then let σ ϕ = ϕ(σ [0,..., p])σ [p,..., n] C n p (X; k). The induced bilinear map is called the cap-product. Lemma 5.2. We hae C n (X; k) C p (X; k) C n p (X; k) (σ ϕ) = ( 1) p ( σ ϕ σ δϕ). As with the cup-product, we get a cap-product : H n (X; k) H p (X; k) H n p (X; k). Definition 5.3. Let X be a topological space, ϕ C p (X; k) and σ C p (X; k). Then define This induces a bilinear map σ, ϕ = ϕ(σ) k., : H p (X; k) H p (X; k) k which is called the Kronecker product.

11 ALGEBRAIC TOPOLOGY IV 11 Lemma 5.4. Let ε : H 0 (X; k) k be the map induced by ε(σ 0 ) = 1 for eery σ 0 : 0 X. Then ε (x α) = x, α for all x H p (X; k) and α H p (X; k). Example 5.5. Recall the triangulation of the torus from Figure 2, and the cocycle A defined in Example 3.5. Let [T 2 ] H 2 (T 2 ) be the fundamental class corresponding to the counterclockwise orientation on the torus, and α H 1 (T 2 ; Z) the cohomology class represented by A. Then [T 2 ] α = ([ 1, 2, 6 ] [ 1, 5, 6 ] + [ 5, 6, 8 ] [ 5, 7, 8 ] + +[ 2, 7, 8 ] [ 1, 2, 7 ]) A = [ 2, 6 ] + [ 6, 8 ] [ 7, 8 ] [ 2, 7 ] which is easily seen to represent the homology class b H 1 (T 2 ). Lemma 5.6. If f : X Y is a map, then f x α = f (x f (α)) for x H n (X, k) and α H p (Y ; k). Furthermore, if β H k (X; k) and γ H m (X; k), then (x β) γ = x (β γ). Theorem 5.7 (Poincaré Duality). Let k be a commutatie ring with 1. If M is a closed k-orientable n-manifold with fundamental class [M] H n (M; k), then D : H k (M; k) H n k (M; k) gien by is an isomorphism for all k Z. D(α) = [M] α Example 5.8. Let M = RP n and α H 1 (RP n ; Z/2) be the generator. Then α k generates H k (RP n ; Z/2) for all k = 1,..., n. Theorem 5.9 (Lefschetz Duality). Let k be a commutatie ring with 1. If M is a compact k-orientable n-manifold whose boundary M is decomposed as the union of two compact (n 1)-dimensional manifolds A and B with a common boundary A = B = A B, then cap product with a fundamental class [M] H n (M, M; k) gies isomorphisms for all k Z. D : H k (M, A; k) H n k (M, B; k) Example Lefschetz Duality with A = M and B = is and with A = and B = M is H k (M, M) = Hn k (M) H k (M, M) = H n k (M).

12 12 DIRK SCHÜTZ Theorem 5.11 (Alexander Duality). Let K be a finite CW-complex and f : K S n an embedding. Then H i (S n f(k); k) = Hn i 1 (K; k) for all i Z and k a commutatie ring with 1. Corollary Let K be a non-orientable closed surface. Then K cannot be embedded into R The Lefschetz Fixed Point Theorem Definition 6.1. Let K, L be simplicial complexes. A simplicial map from K to L is a function f defined on the ertices of K to the ertices of L such that simplices in K are sent to simplices in L. A simplicial map induces a map f : K L on realizations by ( m ) m f t i i = t i f( i ) i=0 Definition 6.2. Let K be a simplicial complex with ertex set { 0,..., k }. The barycentric subdiision K of K is the simplicial complex defined as follows. Vertices are the simplices of K are of the form [σ 0,..., σ i ], where each σ j is a face of σ j+1. Example 6.3. If K is the 2-simplex on ertices 0, 1, 2, then K has ertices 0, 1, 2, 0 1, 0 2, 1 2, The 1-simplices containing 0 are [ 0, 0 1 ], [ 0, 0 2 ] and [ 0, ]. We can always realize K so that K = K. Definition 6.4. For x K where K is a simplicial complex, the carrier of x, carr(x), is the smallest simplex of K containing x. Let f : K L be continuous. A simplicial approximation to f is a simplicial map g : K L with g(x) carr(f(x)) for each x K i=0 Lemma 6.5. Let K and L be simplicial complexes. (1) If g : K L is simplicial, then g(carr(x)) = carr(g(x)) for eery x K. (2) If g is a simplicial approximation of f : K L, then f g. Definition 6.6. If is a ertex of the simplicial complex K, then the open star of is St k () = {x K carr(x)}. Proposition 6.7. Let K be a simplicial complex. Then St k () is open in K, and {St K () K 0 } coer K. Proposition 6.8. If 0,..., p are ertices of the simplicial complex K, then Stk ( i ) if and only if [ 0,..., p ] span a p-simplex in K

13 ALGEBRAIC TOPOLOGY IV 13 Theorem 6.9 (Simplicial Approximation Theorem). Let K, L be simplicial complexes and f : K L continuous. Then there is an integer r 0 such that there is a simplicial approximation g : K [r] L to f, where K [r] is k- times iterated barycentric subdiision of K. Definition A (co)chain complex C oer a field F is of finite bounded type, if each (co)chain group is a finite dimensional F-ector space and only finitely many of them are non-zero. Let V be a finite dimensional F-ector space and f : V V a linear map. The trace of f, denoted tr f, is the sum of diagonal elements in a matrix representing the linear map. If C is a finite bounded type (co)chain complex oer F, and f : C C a (co)chain map, then the trace of f is defined as where f i is the map in degree i. tr f = ( 1) i tr f i Theorem 6.11 (Hopf Trace Formula). Let C be a (co)chain complex of bounded finite type oer a field F and f : C C a (co)chain map. Then tr f = ( 1) i tr f i : H i (C) H i (C). Definition Let K be a finite simplicial complex and f : K K a map. The Lefschetz number of f is defined as L F (f) = ( 1) i tr f i : H i ( K ; F) H I ( K ; F) F Remark Using simplicial approximation and the Hopf-Trace formula, we can determine the Lefschetz number using the simplicial chain complex of K. Furthermore, we get L F (f) = ε F (tr g ) where ε F : Z F is the unique ring homomorphism sending 1 to 1 F, and g is obtained from f ia simplicial approximation. Theorem 6.14 (Lefschetz Fixed Point Theorem). Let K be a finite simplicial complex and f : K K a map. If L F (f) 0, then f has a fixed point. Corollary Let K be a finite simplicial complex which is contractible. Then eery map f : K K has a fixed point. Let the map f : S n S n hae degree different from ( 1) n+1. Then f has a fixed point. Example The Lefschetz Fixed Point Theorem also holds for finite CW-complexes and compact topological manifolds. (1) Let X = {(1 e t )e it C t 0} S 1 and Y the cone of X. There exists a fixed point free map f : Y Y despite Y being contractible.

14 14 DIRK SCHÜTZ (2) Let f : RP 2n RP 2n be a map. Then L Q (f) = 1, so f has a fixed point. (3) Let f : CP n CP n be a map. Then L Q (f) = 1 + k + k k n for some k Z. For n een, this is non-zero. Definition A space X is said to hae the fixed point property (FPP), if eery map f : X X has a fixed point. Example Eery contractible finite CW-complex X has the FPP. For een n, RP n, CP n and HP n hae the FPP. For odd n, RP n and CP n do not hae the FPP. For n 2, HP n has the FPP.