On the map $\operatorname{Top}(X,Y \times Z) \longrightarrow \operatorname{Top}(X,Y) \times \operatorname{Top}(X,Z)$

Question

Disclaimer: We define a topological space $X$ to be compact if every open cover has a finite subcover, but $X$ is otherwise allowed to be arbitrary.

I have been faced with the following problem:

Let $X, Y, Z$ be arbitrary (nonempty) topological spaces and let $\operatorname{Top}(A, B)= \{ f: A \longrightarrow B \ \mid\ \text{f is continuous}\}$ be equipped with the compact-open topology. The universal property of the product allows us to find a bijective map: $$\varphi: \operatorname{Top}(X,Y \times Z) \longrightarrow \operatorname{Top}(X,Y) \times \operatorname{Top}(X,Z)$$

We are interested in whether $\varphi$ is a homeomorphism.

I could not prove that $\varphi^{-1} =: \psi : \operatorname{Top}(X,Y) \times \operatorname{Top}(X,Z) \longrightarrow \operatorname{Top}(X,Y \times Z)$ is continuous, or equivalently that $\varphi$ is an open map.

The question is discussed here, here and here, but provided references and proofs rely on defining a compact space as a Hausdorff space whose open covers have finite subcovers.

I am quite sure that in the above generality we do not have a homeomorphism. If we make additional assumptions on $X$, e.g., $X$ locally compact or Hausdorff, $\varphi$ can be proven to be a homeomorphism. Seen here in the case of $X$ being Hausdorff.

I am greatly interested in finding a counterexample where $\varphi$ isn't a homeomorphism. The same goes for a potential proof of $\varphi$ being a homeomorphism under the given definition of compactness.

If you decide to reference anything stating that this is indeed a homeomorphism, please check the definition of compactness beforehand or whether it's used as a Hausdorff space in the proof.

Answer 1

[Edit: This does not work, see the comments. Leaving it here in case anyone can get some ideas from it.]

Isn't this true in general? On the level of continuous functions we have a bijection $$ \phi : \operatorname{Hom}(X, Y \times Z) \to \operatorname{Hom}(X,Y) \times \operatorname{Hom}(X,Z) $$ given by $\phi(f) = (\pi_Y \circ f, \pi_Z \circ f)$, where $\pi_Y$ and $\pi_Z$ are the projections. Let's first prove that this is a continuous function in the compact-open topology. For this it's enough to prove that the function $\operatorname{Hom}(X, Y \times Z) \to \operatorname{Hom}(X,Y)$ given by $\psi(f) = \pi_Y \circ f$ is continuous; we can then conclude by symmetry and the product of continuous functions being continuous.

Let then $K \subset X$ be compact and $U \subset Y$ be open, and let $V(K,U) = \{ f : X \to Y \mid f(K) \subset U\}$ be an element of the subbase of the compact-open topology on $\operatorname{Hom}(X,Y)$. The inverse image of this under $\psi$ is $$ \begin{align*} \psi^{-1}(V(K,U)) &= \{ g : X \to Y \times Z \mid \pi_Y \circ g(K) \subset U \} \\ &= \{ g : X \to Y \times Z \mid g(K) \subset U \times Z \} \\ &= V(K, U \times Z), \end{align*} $$ which is an element of the subbase of the compact-open topology on $\operatorname{Hom}(X,Y\times Z)$. As any open set is generated by the subbase, we see that $\psi$ is continuous.

The inverse of $\phi$ is $\phi^{-1}(f,g) = f \times g$. Let again $K \subset X$ be compact and $U \subset Y \times Z$ be open. The topology on $Y \times Z$ is generated by products of open sets in $Y$ and $Z$, so we may assume $U = V \times W$ where $V \subset Y$ and $W \subset Z$ are open. Now $$ V(K, V \times W) = \{ f : X \to Y \times Z \mid f(K) \subset V \times W \}. $$ If $f(K) \subset V \times W$ then $\pi_Y \circ f(K) \subset V$, and similarly for $Z$. Then $$ (\phi^{-1})^{-1}(V(K, V \times W)) \subset V(K,V) \times V(K,W). $$ Now let $f \in V(K,V)$ and $g \in V(K,W)$. Then $(f\times g)(K) \subset V \times W$ and $\phi^{-1}(f \times g) = (f, g)$, so the inclusion above is an equality. But then $\phi^{-1}$ is also continuous and $\phi$ is a homeomorphism.