Algebraic Topology Homework 4 Solutions

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1 Algebraic Topology Homework 4 Solutions Here are a few solutions to some of the trickier problems... Recall: Let X be a topological space, A X a subspace of X. Suppose f, g : X X are maps restricting to the identity on A. Then a homotopy relative to A (or just: a homotopy rel. A) from f to g is a map H : X I X satisfying: (1) H(a, t) = a for all a A and all t I, (2) H(x, 0) = f(x) for all x X, and (3) H(x, 1) = g(x) for all x X. Exercise 1. Let C be the cylinder [ π, π] S 1, π 2 : C S 1 the projection. Consider the map f : C C defined by f(s, x) := (s, A s+π x), where ( ) cos θ sin θ A θ := sin θ cos θ is the map S 1 S 1 given by counter-clockwise rotation through the angle θ. Note that f restricts to the identity map on C = {±π} S 1. Define γ : I C by γ(t) := ( π + 2πt, e 1 ). Let T := C/ be the torus obtained from C by identifying the two boundary circles: ( π, x) (π, x) for x S 1. By the universal property of quotients, f determines a map f : T T and γ determines a map γ : S 1 = I/{±1} T. Define b : S 1 T by b(x) := (0, x). Fix a generator e for H 1 (S 1 ) = Z. Our Mayer-Vietoris computation in class shows that H 1 (T ) is freely generated by α := γ e and β := b e. a) Show that f : C C is not homotopic rel. C to the identity. Hint: If it were, then argue that π 2 fγ and π 2 γ would be homotopic as loops based at e 1 S 1. b) Show that, in the ordered basis (α, β), H 1 (f) : H 1 (T ) H 1 (T ) is given by ( ) Hint: First use the explicit basis of H 1 (T ) so check that the projections π 1 : T [ π, π]/{±π} = S 1 and π 2 : T S 1 induce an isomorphism (π 1, π 2 ) : H 1 (T ) H 1 (S 1 ) H 1 (S 1 ). Conclude that f is not homotopic to Id : T T. c) Observe that b gives an alternative proof of a. Solution: For (a): Suppose H : C I C is a homotopy rel. C from f to the identity. Then one sees immediately from the definition of relative homotopy that the composition I I γ Id C I 1 H C π 2 S 1

2 2 would be a homotopy of paths from π 2 fγ to π 2 γ. But π 2 γ is the constant loop at e 1 S 1, while (π 2 fγ)(t) = π 2 (f( π 2 πt, e 1 )) so that π 2 fγ is a generator of π 1 (S 1 ) = Z. = π 2 ( π + 22πt, A 2πt e 1 ) = (cos(2πt), sin(2πt)), For (b): In fact, π 1 π 2 : T S 1 S 1 is a homeomorphism so the fact that π 1 π 2 is an isomorphism is just the general principle that π 1 is compatible with products. Now one just calculates that π 1 α = [π 1 γ] π 2 α = [π 2 γ] = [1 e1 ] = 0 π 1 β = [1 0 ] π 2 β π 1 fα = [π 1 fγ] π 2 fα π 1 fβ = [π 1 fb] = 0 π 2 fβ = [π 2 fb]. For (c): If H : C I C is a homotopy rel. C from f to the identity, then, in particular, we have H( π, x, t) = ( π, x) (π, x) = H(π, x, t), so H would descend to a homotopy H : T I T from f to Id. Exercise 2. Let S n R 2 be the circle of radius 1/(2n) centered at the point (1/(2n), 0) and let X := n=1 be the Hawaiian earring (with the subspace topology from the inclusion X R 2 ). Let C(X) R 3 be the cone on X, v the vertex of the cone. View X as a subspace of C(X) (the top of the cone). Let x 0 X C(X) be the origin. Show that C(X) does not deformation retract onto x 0 even though the inclusion of x 0 into C(X) is a homotopy equivalence. Hint: Suppose H : C(X) I C(X) is a deformation retract onto x 0. Show that for every n {1, 2,... }, there is a point x n S n C(X) and a t n I such that H(x n, t n ) = v. (If not, then you could somehow use H to get a deformation retract of S n onto x 0.) Solution: The tricky part is to establish the assertion in the hint. If this were not true, then for some n {1,... }, H would restrict to a continuous map H : S n I C(X)\{v}. Composing this map with the projection from the vertex p : C(X) \ {v} X, followed by

3 3 a retract r : X S n of the inclusion S n X, we would get a deformation retract from S n onto x 0, which is absurd. Now that this is known, just pick, for each n {1, 2,... } a point x n S n and a time t n I such that H(x n, t n ) = v. Now use the fact that X I is compact to pass to a subsequence of (x n, t n ) that converges, say to (x, t), in X I. We must have x = x 0 because for any ɛ > 0 we can pick an N big enough that S N, S N+1,... are contained in the ɛ-ball around x 0, hence, in particular x N, x N+1,... are contained in this ball. Now continuity of H implies that H(x 0, t) = v x 0, contradicting the assumption that H is a deformation retract of C(X) onto x 0. Exercise 3. Show that the degree map deg : Hom HotTop (S 1, S 1 ) Z yields a bijection between homotopy classes of maps S 1 S 1 and the integers. It might be helpful here to first show that any map f : S 1 S 1 is homotopic to a map that fixes 0 S 1 = R/Z, then note that the Hurewicz isomorphism is natural so that if f : (S 1, 0) (S 1, 0) is a map of pairs, then the Hurewicz isomorphism identifies H 1 (f) and π 1 (f). Exercise 4. For any x R n, let t x : R n R n be the homeomorphism given by translation along x: t x (y) := x+y. By abuse of notation, we also use t x to denote the homeomorphism t x (R n \ {y}) : R n \ {y} R n \ {x + y} for any y R n. Suppose f : R n R n is a homeomorphism. Then f 0 := t f(0) (f (R n \ {0})) is an automorphism of R n \ {0}. This induces an automorphism (f 0 ) of H n 1 (R n \ {0}) = Z, which is necessarily either the identity (in which case we say that f is orientation preserving, or that the degree deg f of f is 1) or multiplication by 1 (in which ase we say that f is orientation reversing, or that deg f = 1). Show that this notion of orientation preserving has the following expected properties: a) Degree is a group homomorphism from Aut(R n ) to the two element group {±1}. b) Every translation of R n is orientation preserving. c) If f : R n R n is the homeomorphism obtained from an invertible matrix A GL n (R), then deg f is the sign of det A R. (This might be a tiny bit tricky, painful, and/or tedious if you don t know anything about the topology of GL n (R), but if worst comes to worst, you can use part a) to reduce to analysing elementary matrices. For this, it will be convenient to use property (e) of degree for maps of spheres on Page 134 in Hatcher, which I did not discuss in class.) Solution: No one really did a good job with the last part. The point is that any invertible matrix is a product of elementary matrices, so by the first part we reduce the last part to the case where A is an elementary matrix E. If the elementary matrix E is a permutation matrix, then det E is the sign of the corresponding permutation to see that this agrees with the degree, just write the permutation as a product of transpositions and apply Property (e) on Page 134 in Hatcher. If E is the elementary matrix given by adding λ times row i to row j (i j), then det E = 1. On the other hand, a path from 0 to λ in R gives an obvious path (through elementary matrices) from Id to E (a homotopy from Id to E), so the degree of E is also 1. If E is given by multiplying row i by λ R, then

4 4 det E = λ. In this case a path from λ/ λ to λ in R gives a homotopy from a matrix whose degree is λ/ λ (by Property (e)) to E. Exercise 5. Exercise 2 on Page 155 in Hatcher. Exercise 6. Exercise 33 on Page 158 in Hatcher. Exercise 7. (Cf. Exercise 8 on Page 155 in Hatcher) Let f(x) = a 0 +a 1 x+ +a d x d C[x] be a degree d polynomial with complex coefficients (a d 0). We also write f : C C for the map determined by this polynomial. Let f(x, y) := a 0 y d + a 1 xy d a d x d be the homogeneous degree d polynomial in C[x, y] obtained by homogenizing f. (1) Show that [x : y] [ f(x, y) : y d ] is a well-defined map ˆf : CP 1 CP 1. (You have to check that this makes sense on equivalence classes and that ( f(x, y), y d ) (0, 0) when (x, y) (0, 0).) (2) Viewing C as a subset of CP 1 via the embedding x [x : 1], show that ˆf extends f. (3) Note that CP 1 = S 2 so it makes sense to speak of the degree of an endomorphism of CP 1. Show that the degree of ˆf is d and that the local degree of ˆf at a root α C CP 1 of f is the multiplicity of α as a root of f. (4) Observe that all the maps ˆf : CP 1 CP 1 thus-constructed (for d fixed) are homotopic. Solution: The first two parts are straightforward. I would next prove the last part, which is easy: the space of all degree d polynomials is just C C d (look at the coefficients), which is connected the last part is immediate. Now, at least for the first assertion in the third part, it suffices to treat the case f = x d, where ˆf[x : y] = [x d : y d ]. Then ˆf 1 ([0 : 1]) = {[0 : 1]} and, on the standard neighborhood C = U 1 CP 1 of [0 : 1], ˆf coincides with the map x x d from C to C, which certainly has local degree d at the origin. In general, suppose α is a root of f of multiplicity e and let us show that the local degree of f : C C at α is e. We can write f = (x α) e g where g(α) 0. Choose a small closed disc D around α in C such that α is the only root of f in D and let U be the interior of D. By definition of local degree, we want to show that f : H 2 (U, U α) H 2 (C, C ) is multiplication by e. Looking at the LESs of the pairs involved here, we see that it is equivalent to prove that f : H 1 (U α) H 1 (C ) is multiplication by d. Note that g : U α C is homotopic to the constant map 1 because C is homotopy equivalent to S 1 via the argument map z z/ z and g/ g makes sense on the entire disc D. It follows that f : U α C is homotopic to x (x α) e, which is clearly degree e. Exercise 8. Let n be the standard n-simplex, so that Id : n n is a singular n-simplex in n. Write [ n ] for the corresponding element of the complex C( n, n ) of singular chains in n modulo singular chains in n. Obviously [ n ] is a cycle in C( n, n ), so it determines some homology class [ n ] H n ( n, n ).

5 5 On the other hand, we showed in class that H n ( n, n ) = Z. We claim that [ n ] is a generator for H n ( n, n ). This is not so obvious because our computation(s) of H n ( n, n ) involve constructions through which it is difficult to follow the homology class [ n ]. For example, the closely-related statement that the cycle n [ n ] := ( 1) i (Id i n ) i=0 generates H n 1 ( n ) = Z is difficult to see from our Mayer-Vietoris-based calculation of H n 1 ( n ) because the singular chain [ n ] is not actually in the complex C U ( n ) for the usual cover U, V of n = S n 1 used in that argument, so one would have to go into the details of barycentric subdivision to try to follow this cycle through that computation (=painful!). Here is an easy way to prove our claim: (1) Show that if the cycle [ n ] were a boundary in C( n, n ), then any singular n-simplex σ : n n would be a boundary in C( n, n ), hence H n ( n, n ) would be zero, which it is not. (2) Since H n ( n, n ) = Z, we conclude from the previous part that our class [ n ] must be d times a generator of H n ( n, n ) for some positive integer d in particular [ n ] is divisible by d in H n ( n, n ). Show that this implies that any homology class in H n ( n, n ) is divisible by d. Since H n ( n, n ) = Z, we must have d = 1, proving our claim. (3) Conclude, by studying the LES for the pair ( n, n ), that [ n ] is a generator for H n 1 ( n ) = Z.