I encountered a theorem from the Book *Elementary Topology: Problem Book*:

**Theorem.** Let $p:Xto Z$ 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 $fmapsto fcirc p$ 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 $finPhi^{-1}(S(C,U))$, we have

$$f(p(C))=(fcirc p)(C)=Phi(f)(C)subseteq U.$$

On the other hand, for each $fin S(p(C),U)$, we have

$$Phi(f)(C)=(fcirc p)(C)=f(p(C))subseteq U.$$

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 $f,gin C(Z,Y)$, if $fneq g$, then $f(z)neq g(z)$ for some $zin Z$. Since $p$ is surjective, we have $z=p(x)$ for some $xin X$. Note that

begin{equation*}

Phi(f)(x)=(fcirc p)(x)=f(p(x))=f(z)

end{equation*}

and similarly, $Phi(g)(x)=g(z)$, so $Phi(f)neqPhi(g)$, 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 $Ksubseteq Z$ 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 $g:Xto Y$ constant on all $p^{-1}(z)$ where $zin K$ 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 $g:Xto Y$ constant on all $p^{-1}(z)$ where $zin K$ 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.