TOPOLOGY HW 8 CLAY SHONKWILER 55.1 Show that if A is a retract of B2, then every continuous map f : A → A has a fixed point. 2 Proof. Suppose r : B → A is a retraction. Thenr|A is the identity map on A. Let f : A → A be continuous and let i : A → B2 be the inclusion map. Then, since i, r and f are continuous, so is F = i ◦ f ◦ r. Furthermore, by the Brouwer fixed-point theorem, F has a fixed point x0. In other words, 2 x0 = F (x0) = (i ◦ f ◦ r)(x0) = i(f(r(x0))) ∈ A ⊆ B . Since x0 ∈ A, r(x0) = x0 and so x0F (x0) = i(f(r(x0))) = i(f(x0)) = f(x0) since i(x) = x for all x ∈ A. Hence, x0 is a fixed point of f. 55.4 Suppose that you are given the fact that for each n, there is no retraction f : Bn+1 → Sn. Prove the following: (a) The identity map i : Sn → Sn is not nulhomotopic. Proof. Suppose i is nulhomotopic. Let H : Sn × I → Sn be a homotopy between i and a constant map. Let π : Sn × I → Bn+1 be the map π(x, t) = (1 − t)x. π is certainly continuous and closed. To see that it is surjective, let x ∈ Bn+1. Then x x π , 1 − ||x|| = (1 − (1 − ||x||)) = x. ||x|| ||x|| Hence, since it is continuous, closed and surjective, π is a quotient map that collapses Sn×1 to 0 and is otherwise injective. Since H is constant on Sn×1, it induces, by the quotient map π, a continuous map k : Bn+1 → Sn that is an extension of i. Since it is an extension of i, k(x) = x for all x ∈ Sn, so k is a retraction of Bn+1 onto Sn. From this contradiction, then, we conclude that i is not nulhomotopic. (b) The inclusion map j : Sn → Rn+1\{0} is not nulhomotopic. 1 2 CLAY SHONKWILER n n+1 Proof. Suppose j is nulhomotopic. Let H : S × I → R \{0} be a ho- motopy between j and a constant map which maps everything to x0 ∈ n+1 n+1 n R \{0}. Let r : R \{0} → S be given by x x 7→ . ||x|| n+1 n Then r is a retraction of R \{0} onto S , as we’ve shown in class. Now, define H0 : Sn × I → Sn by H0(x, t) = (r ◦ H)(x, t). Then, since H is a homotopy between j and a constant map, H0 is continuous and, for all x ∈ Sn, H0(x, 0) = (r ◦ H)(x, 0) = r(H(x, 0)) = r(j(x)) and 0 H (x, 1) = (r ◦ H)(x, 1) = r(H(x, 1)) = r(x0). Since j is merely the inclusion map, j(x) ∈ Sn, so r(j(x)) = i(x). Hence, H0 is a homotopy between the identity map i and the constant map which n maps everything to the point r(x0) ∈ S . In other words, i is nulhomotopic. However, as we showed in part (a) above, i is not nulhomotopic. From this contradiction, then, we conclude that j is not nulhomotopic. (c) Every nonvanishing vector field on Bn+1 points directly outward at some point of Sn, and directly inward at some point of Sn. n n+1 Lemma 0.1. If h : S → R \{0} extends to a continuous map k : n+1 n+1 B → R \{0}, then h is nulhomotopic. n n+1 Proof. Let H : S × I → R \{0} be given by (x, t) 7→ k((1 − t)x). Then H is certainly continuous, since k is. Furthermore, for x ∈ Sn, H(x, 0) = k((1 − 0)x) = k(x) = h(x) and H(x, 1) = k((1 − 1)x) = k(0). Therefore, h is nulhomotopic. Now, we are ready to prove the desired result Proof. Let (x, v(x)) be a nonvanishing vector field on Bn+1. Since this vector n+1 n+1 field is non-vanishing, v is a map from B into R \{0}. Suppose that v(x) does not point directly inward at any point x ∈ Sn in order to derive n+1 n+1 a contradiction. Consider the map v : B → R \{0} and let w be its n n+1 n+1 restriction to S . w extends to a continuous map of B into R \{0} (namely v) so, by the above lemma, w is nulhomotopic. TOPOLOGY HW 8 3 On the other hand, we want to show that w is homotopic to the inclusion n n+1 n n+1 map j : S → R \{0}. Let F : S × I → R \{0} be given by (x, t) 7→ tx + (1 − t)w(x) for x ∈ Sn. In order to be sure that F is well-defined, we must make sure that F (x, t) 6= 0 for all x ∈ Sn, t ∈ I. Since x is always non-zero, F (x, 0) 6= 0 n+1 and, since the image of w is contained in R \{0}, F (x, 1) 6= 0. Now, if F (x, t) = 0 for some 0 < t < 1, then tx + (1 − t)w(x) = 0 or −t w(x) = x. 1 − t However, this means w(x) (and hence v(x)) points directly inward at x. From this contradiction, we see that F (x, t) 6= 0, so F really maps Sn × I → n+1 R \{0}. Now, F is certainly continuous and F (x, 0) = x = j(x) and F (x, 1) = w(x), so j is homotopic to w. Since w is nulhomotopic, so is j. This, however, contradicts part (b) above. From this contradiction we conclude that, in fact, v points directly inward at some point. To show that v points directly outward at some point of Sn, we simply notice that this is equivalent to saying that −v points directly inward at some point. Since (x, −v(x)) defines a vector field, we know by the above proof that, in fact, −v does point directly inward at some point, so v points directly outward at that same point. (d) Every continuous map f : Bn+1 → Bn+1 has a fixed point. Proof. By contradiction. Suppose f(x) 6= x for all x ∈ Bn+1. Then we define v(x) = f(x) − x, which gives a non-vanishing vector field (x, v(x)) on Bn+1. This vector field v cannot point directly outward at any point x ∈ Sn, as that would mean f(x) − x = ax for some positive real a. This in turn would imply that f(x) = (1 + a)x∈ / Bn+1. However, this contradicts the fact proved in part (c) above that, since it is non-vanishing, the vector field v must point directly outward at some point of Sn. From this contradiction, then, we conclude that f has a fixed point. (e) Every n + 1 by n + 1 matrix A with positive real entries has a positive eigenvalue. 4 CLAY SHONKWILER n+1 n+1 Proof. Let T : R → R be the linear transformation whose matrix is n A. Let B = S ∩ {(x1, . , xn+1): x1, . , xn+1 ≥ 0}. Then this B is surely homeomorphic to Bn, so that the result from part (d) holds for continuous maps from B to itself. Let (x1, . , xn+1) ∈ B. Then each component of x is non-negative and at least one is positive. Also, since all entries of A are positive, all the components of the vector T (x) are positive. Hence, the map x 7→ T (x)/||T (x)|| is a continuous map of B into itself, so it has a fixed point x0. Then T (x0) = ||T (x0)||x0, so T (and, hence, A) has the positive real eigenvalue ||T (x)||. (f) If h : Sn → Sn is nulhomotopic, then h has a fixed point and h maps some point x to its antipode −x. Proof. Since h is nulhomotopic, we can, just as we did in part (a), extend h to a continuous map f : Bn+1 → Sn ⊂ Bn+1. By part (d), then, h has n n a fixed point x0. Moreover, x0 ∈ S , since Im(f) = S . Therefore, since f n and h agree on S , h(x0) = x0. To prove that h maps some point to its antipode, first we prove the following parallel of part (d) above: Lemma 0.2. If f : Bn+1 → Bn+1 is continuous, then there exists a point x ∈ Bn+1 such that f(x) = −x. Proof. By contradiction. Suppose f(x) 6= −x for all x ∈ Bn+1. Then v(x) := f(x) + x gives a non-vanishing vector field (x, v(x)) on Bn+1. But the vector field v cannot point directly inward at any point x ∈ Sn, for that would mean −ax = v(x) = f(x) + x for some positive real a. If this were the case, f(x) = −(1 + a)x would lie outside the unit ball. However, we proved in part (c) above that every nonvanishing vector field on Bn+1 points directly inward at some point of Sn. From this contradiction, then, we conclude that there exists a point x ∈ Bn+1 such that f(x) = −x. Now, to finish off the proof, we simply note that by this lemma, our n+1 extension f must have some point x1 ∈ B at which f(x1) = −x1. Again, n n n since Im(f) = S , x1 ∈ S and therefore, since f and h coincide on S , h(x1) = −x1.