I encountered a theorem from the Book Elementary Topology: Problem Book:
Theorem. Let $p:X\to Z$ be a closed quotient map where $X$ is compact Hausdorff. Then for every topological space $Y$, the function $\Phi:C(Z,Y)\to C(X,Y)$ defined by $f\mapsto f\circ p$ is a topological embedding.
Here $C(Z,Y)$ and $C(X,Y)$ are assumed to be given with the compact-open topology.
It is easy to prove thet $\Phi$ is continuous and injective:
Proof of Continuity. Let $S(C,U)$ be a subbase element of $C(X,Y)$ where $C$ is compact in $X$ and $U$ is open in $Y$. Since the quotient map $p$ is continuous, $p(C)$ is compact in $Z$. Then we claim that
$$\Phi^{-1}(S(C,U))=S(p(C),U)\subseteq C(Z,Y).$$
On the one hand, for each $f\in\Phi^{-1}(S(C,U))$, we have
$$f(p(C))=(f\circ p)(C)=\Phi(f)(C)\subseteq U.$$
On the other hand, for each $f\in S(p(C),U)$, we have
$$\Phi(f)(C)=(f\circ p)(C)=f(p(C))\subseteq U.$$
The equality thus follows. Note that $S(p(C),U)$ is a subbase element of $C(Z,Y)$, hence is open. Thus $\Phi$ is continuous.
Proof of Injectivity. Moreover, for every $f,g\in C(Z,Y)$, if $f\neq g$, then $f(z)\neq g(z)$ for some $z\in Z$. Since $p$ is surjective, we have $z=p(x)$ for some $x\in X$. Note that
\begin{equation*}
\Phi(f)(x)=(f\circ p)(x)=f(p(x))=f(z)
\end{equation*}
and similarly, $\Phi(g)(x)=g(z)$, so $\Phi(f)\neq\Phi(g)$, implying that $\Phi$ is injective.
Finally, it suffices to prove that $\Phi$ is an open map. I followed the hint of the book:
Hint. Let $K\subseteq Z$ be a compact set, $U$ open in $Y$. The image of the open subbase set $S(K,U)\subseteq C(Z,Y)$ is the set of all maps $g:X\to Y$ constant on all $p^{-1}(z)$ where $z\in K$ and such that $g(p^{-1}(K))\subseteq U$. It remains to show that the set $S(p^{-1}(K),U)$ is open in $C(X,Y)$. Since the space $Z$ is Hausdorff, it follows that the set $K$ is closed. Therefore,
the preimage $p^{-1}(K)$ is closed, and hence also compact. Consequently, $S(p^{-1}(K),U)$ is a subbase set in $C(X,Y)$.
The hint is too redundant, so I add more details about that:
First, since $X$ is compact Hausdorff, it is clear that $X$ is normal Hausdorff (both $T_1$ and normal).
The quotient map $p$ is closed, so $Z$ is also normal Hausdorff.
Note that $K$ is compact in $Z$, so $K$ is closed. As a result, the set $p^{-1}(C)\subseteq X$ is closed and hence compact.
The set $S(p^{-1}(K),U)$ is open in $C(X,Y)$ because it is a subbase set.
But here is a problem: We can only obtain that
$$\Phi(S(K,U))\subseteq S(p^{-1}(K),U)$$
The hint also mentioned that
The image of the open subbase set $S(K,U)\subseteq C(Z,Y)$ is the set of all maps $g:X\to Y$ constant on all $p^{-1}(z)$ where $z\in K$ and such that $g(p^{-1}(K))\subseteq U$.
It seems hopeless to deduce $\Phi(S(K,U))$ is open in $C(X,Y)$ by these arguments.
I am not sure whether I have missed something, but I still do not think such argument would go. Hope anyone had good ideas on this.
Update and Comments. Thanks a lot to all the answers and replys. I misunderstood the definition of topological embeddings. Suppose a function $\varphi:X\to Y$ is continuous and injective. If it is also an open map, then it is definitely a topological embedding. But the converse need not be true. The definition of topological embeddings requires that the restriction $\varphi':X\to\varphi(X)$ is a homeomorphism; that is, the image of any open set in $X$ only needs to be open in the subspace $\varphi(X)$, but not necessarily in $Y$. Of course if it is open in $Y$, then it must be open in $\varphi(X)$. This is an easy mistake. I hope any of you who read this could avoid such mistake in the future.
Best Answer
You do not have to prove that $\Phi(S(K,U))$ is open in $C(X,Y)$, but that it is open in $\Phi(C(Z,Y))$.
Let us first observe that $g : X \to Y$ having the form $\Phi(f) = f \circ p$ with $f \in C(Z,Y)$ is equivalent to $g$ being constant on all $p^{-1}(z)$ where $ z \in Z$. Thus $$\Phi(C(Z,Y)) = \{ g \in C(X,Y) \mid g \text{ is constant on all }p^{-1}(z) \text{ where } z \in Z \} .$$ That $g \in \Phi(S(K,U))$ therefore means $g = f \circ p$ with $f \in S(K,U)$, i.e. $f(K) \subset U$. The latter is equivalent to $g(p^{-1}(K)) \subset U$ since $K = p(p^{-1}(K))$. Therefore $$\Phi(S(K,U)) = \\ \{ g \in C(X,Y) \mid g \text{ is constant on all }p^{-1}(z) \text{ where } z \in K, \text{ and } g \in S(p^{-1}(K),U) \} = \\ \Phi(C(Z,Y)) \cap S(p^{-1}(K),U) .$$ By the way, the hint is not correct because it says "$g$ is constant on all $p^{−1}(z)$ where $z \in K$".