TOPOLOGY HW 8 55.1 Show That If a Is a Retract of B 2, Then Every
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.
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