Show that $ \left( Y,y_0\right)^{(S^1, s_0)} \cong \left(Y,y_0\right)^{(I, \{0,1\})} $

Question

I want to show that $ \left( Y,y_0\right)^{(S^1, s_0)} \cong \left(Y,y_0\right)^{(I, \{0,1\})} $. I've already shown that if $Y$ is a locally compact Hausdorff space, then the composition map $\circ : Y^X\times Z^Y \to Z^X $ is continuous. Then proved that $ S^1 \cong \dfrac{I}{\{0,1\}}$ induces a homeomorphism $ \left( Y,y_0\right)^{(S^1, s_0)} \cong \left(Y,y_0\right)^{(I/\{0,1\}, \ast)}$. My problem is with the last step, that is showing that $ \left(Y,y_0\right)^{(I, \{0,1\})} \cong \left(Y,y_0\right)^{(I/\{0,1\}, \ast)}$.

The problem boils down to:

If I have a Hausdorff locally compact space $X$ and $A\subset X$ is compact, then $ (Y,y_0)^{(X/A, \ast)} \cong (Y,y_0)^{(X,A)}$. I know the map $ \psi : (Y,y_0)^{(X/A, \ast)} \to(Y,y_0)^{(X,A)} $ given by $ \psi(f) = f\circ p$, where $p : X \to X/A $ is the projection, will be a continuous bijection. The issue is showing it is an homeomorphism.

I would really like some references that prove this result.

I think I have a proof:

We have that the projection $\pi: X \to X/A$ is continuous. Therefore, the function \begin{align*} \varphi: (Y,y_0)^{(X/A,*)} &\to (Y,y_0)^{(X,A)} \\ f &\mapsto f \circ \pi \end{align*} is a continuous bijection. We will show that $\varphi$ is an open map. Indeed, let $f \in N(K,U) \cap (Y,y_0)^{(X/A,*)}$, where $K \subset X/A$ is compact and $U \subset Y$ is open. Since $A \subset X$ is compact and $X$ is Hausdorff, we have that $\pi: X \to X/A$ is proper ( $ \pi$ is a closed map and $\pi^{-1}(y)$ is compact for all $y \in A/X$). Thus $\pi^{-1}(K) \subset X$ is compact. Observe that $\varphi(f)(\pi^{-1}(K)) = (f \circ \pi)(\pi^{-1}(K)) = f(K) \subset U$. Hence $\varphi(f) \in N(\pi^{-1}(K), U) \cap (Y,y_0)^{(X, A)}$. We claim that $N(\pi^{-1}(K), U) \cap (Y,y_0)^{(X, A)} \subset \varphi\left(N(K,U) \cap (Y,y_0)^{(X/A,*)}\right)$. Indeed, let $g \in N(\pi^{-1}(K), U) \cap (Y,y_0)^{(X, A)}$. There exists a unique function $h \in (Y,y_0)^{(X/A,*)}$ such that $\varphi(h) = h \circ \pi = g$. Observe that $$ h(K) = h(\pi(\pi^{-1}(K))) = (h \circ \pi)(\pi^{-1}(K)) = g(\pi^{-1}(K)) \subset U. $$ Therefore $h \in N(K,U) \cap (Y,y_0)^{(X/A,*)}$, and consequently $g = \varphi(h) \in \varphi\left(N(K,U) \cap (Y,y_0)^{(X/A,*)}\right)$. Thus $N(\pi^{-1}(K), U) \cap (Y,y_0)^{(X, A)} \subset \varphi\left(N(K,U) \cap (Y,y_0)^{(X/A,*)}\right)$, and $\varphi$ is an open map. Hence it is a homeomorphism.

I think this can be generalized even further (if it is correct): If X is a locally compact Hausdorff space and $\psi : X \to Y$ is a continuous proper map (maybe surjective is necessary), then the map $f \mapsto f\circ \psi$ is an homeomorphism.

Answer 1

I will just rewrite my proof with some remarks.

Some important facts:

Fact 1: If $Y$ is Hausdorff, locally compact, then the composition map $\circ : Y^X \times Z^Y \to Z^X$ is continuous. The proof can be found in here and here.

Fact 2: The Universal Property of Quotient topology is used to show that $f \mapsto f\circ \pi $ is a bijection.

Fact 3: If $A\subset X$ is closed, then $ \pi : X\to X/A$ is a closed map. Indeeed, if $F\subset X$ is a closed subset, then $F\cap A = \emptyset$ or $F\cap A \neq \emptyset$, therefore $\pi^{-1}(\pi(F)) = F$ or $\pi^{-1}(\pi(F)) = F\cup A$. Either case $\pi(F)$ is closed.

Fact 4: If $f: X\to Y$ is a closed map with $f^{-1}(y)\subset X$ compact for every $y\in Y$, then $f$ is proper. The proof can be found here.

Fact 5. Using facts 3 and 4 and the fact that $X$ is Hausdorff, we get that if $A \subset X $ is compact (therefore closed), then $\pi : X\to X/A$ is proper.

Now the proof: We have that the projection $\pi: X \to X/A$ is continuous. Therefore, the function \begin{align*} \varphi: (Y,y_0)^{(X/A,*)} &\to (Y,y_0)^{(X,A)} \\ f &\mapsto f \circ \pi \end{align*} is a continuous bijection. We will show that $\varphi$ is an open map. Indeed, let $f \in N(K,U) \cap (Y,y_0)^{(X/A,*)}$, where $K \subset X/A$ is compact and $U \subset Y$ is open. Since $A \subset X$ is compact and $X$ is Hausdorff, we have that $\pi: X \to X/A$ is proper ( $ \pi$ is a closed map and $\pi^{-1}(y)$ is compact for all $y \in A/X$). Thus $\pi^{-1}(K) \subset X$ is compact. Observe that $\varphi(f)(\pi^{-1}(K)) = (f \circ \pi)(\pi^{-1}(K)) = f(K) \subset U$. Hence $\varphi(f) \in N(\pi^{-1}(K), U) \cap (Y,y_0)^{(X, A)}$. We claim that $N(\pi^{-1}(K), U) \cap (Y,y_0)^{(X, A)} \subset \varphi\left(N(K,U) \cap (Y,y_0)^{(X/A,*)}\right)$. Indeed, let $g \in N(\pi^{-1}(K), U) \cap (Y,y_0)^{(X, A)}$. There exists a unique function $h \in (Y,y_0)^{(X/A,*)}$ such that $\varphi(h) = h \circ \pi = g$. Observe that $$ h(K) = h(\pi(\pi^{-1}(K))) = (h \circ \pi)(\pi^{-1}(K)) = g(\pi^{-1}(K)) \subset U. $$ Therefore $h \in N(K,U) \cap (Y,y_0)^{(X/A,*)}$, and consequently $g = \varphi(h) \in \varphi\left(N(K,U) \cap (Y,y_0)^{(X/A,*)}\right)$. Thus $N(\pi^{-1}(K), U) \cap (Y,y_0)^{(X, A)} \subset \varphi\left(N(K,U) \cap (Y,y_0)^{(X/A,*)}\right)$, and $\varphi$ is an open map. Hence it is a homeomorphism.