33.2a) Suppose p^{−1}(U) is open in X. Then f^{−1}(p^{−1}(U)) is open in Y. Since f ∘ p = id, f^{−1}(p^{−1}(U)) = U. Thus U is open in Y. Hence U is open in Y if and only if p^{−1}(U) is open in X. Thus p is a quotient map IF p is surjective. Note p : R → R^{2}, p(x) = (x, 0) is continuous. Note π_{1} : R^{2} → R is continuous and π_{1}(p(x)) = π_{1}(x, 0) = x. Thus p satisfies the hypothesis of this problem, but is NOT a quotient map since it is not surjective. b.) Define f : A → X by f(a) = a. Since f is obtained from the identity map i : A → A by expanding the co-domain, f is continuous. r(f(a)) = r(a) = a for all a ∈ A. Hence r ∘ f is the identity map of A. Since r is continuous and onto, r is a quotient map by part a. 43.8.) e : X × C(X, Y) → Y, e(x, f) = f(x). If d is a metric for Y and C(X, Y) has the uniform topology, then e is continuous. Since d and \overline{d} generate the same topology on Y and continuity is a topological property, we will use \overline{d} to prove the continuity of e. Hence, we will use the following notation: B_{Y} (y, ϵ) = \{z ∈ Y | \overline{d}(y, z) < ϵ\} and B\overline{ρ}(h, ϵ) = \{g ∈ C(C, Y) | \overline{ρ}(h, g) < ϵ\} Take U open in Y and (x, f) ∈ e^{−1}(U). Then e(x, f) = f(x) ∈ U. Hence there exists an r > 0 such that B_{Y} (f(x), r) ⊂ U. f ∈ C(X, Y) implies f^{−1}(B_{Y} (f(x), \frac{r}{2})) is open in X Thus f^{−1}(B_{Y} (f(x), \frac{r}{2}) × B\overline{ρ}(f, \frac{r}{2}) is open in X × C(X, Y). Suppose (z, g) ∈ f^{−1}(B_{Y} (f(x), \frac{r}{2})) × B\overline{ρ}(f, \frac{r}{2}). z ∈ f^{−1}(B_{Y} (f(x), \frac{r}{2})) implies f(z) ∈ B_{Y} (f(x), \frac{r}{2}). Thus d(f(z), f(x)) < \frac{r}{2} g ∈ B\overline{ρ}(f, \frac{r}{2}) implies \overline{ρ}(f, g) < \frac{r}{2}. Hence d(g(z), f(z)) < \frac{r}{2} d(g(z), f(x)) ≤ d(g(z), f(z))+d(f(z), f(x)) < \frac{r}{2} + \frac{r}{2} = r. Thus e(z, g) = g(z) ∈ B_{Y} (f(x), r. Thus (z, g) ∈ e^{−1}(B_{Y} (f(x), r)) ⊂ e^{−1}(U). Hence e is continuous.

43.10) If h : X → Y h′ : X → Y′ are isometric imbedding of metric space (X, d) into complete metric spaces (Y, D) and (Y′, D′), respectively then there is an isometry of (\overline{h(X)}, D) with (\overline{h′(X)}, D′) that equals h′h^{−1} on h(X). Proof: h′h^{−1} : h(X) → h′(X) Claim: If f : A → B is continuous, and A ⊂ X where X is first countable and Y is metrizable, then f can be extended to F : \overline{A} → \overline{B} such that F (a) = f(a) for all a ∈ A. Proof: Suppose f : A → B is continuous. Then if x ∈ \overline{A}, there exists x_{n} ∈ A such that x_{n} → x. Define F : \overline{A} → \overline{B} by F (x) = limf(x_{n}) where x_{n} is any sequence in A such that x_{n} → x. Claim: F is well-defined. Suppose that x_{n} → x and w_{n} → x. Then d(limf(x_{n}), limf(w_{n})) = limd(f(x_{n}), f(w_{n})) = 0 (Can you prove this last equality?). Hence limf(x_{n}) = limf(w_{n}). Since the constant sequence a, a, a, ... converges to a, F (a) = limf(a) = f(a). Hence F extends f. Hence h′h^{−1} : h(X) → h′(X) can be extended to a function F : (\overline{h(X)}, D) → (\overline{h′(X)}, D′) where F (y) = lim[h′h^{−1}(y_{n})] where y_{n} ∈ h(X) and y_{n} → y. Claim F is an isometry: Let y_{n} → z, y_{n} ∈ h(X) and z_{n} → z, z_{n} ∈ h(X) Then D′(F (y), F (z)) = D′(limh′h^{−1}(y_{n}), limh′h^{−1}(z_{n})) = limD′(h′h^{−1}(y_{n}), h′h^{−1}(z_{n})) = lim(D(h^{−1}(y_{n}), h^{−1}(z_{n}))) = limD′(y_{n}, z_{n}) = D′(y, z) since D′ is continuous and h′ and h^{−1} are isometries.