Let $U$ be a topological space, let $I$ and $J$ be discrete sets, and let $f\colon U \times I \rightarrow U \times J$ be a continuous map that commutes with the projection onto the first factor. In other words, for all $u \in U$ and $i \in I$ there exists some $j \in J$ such that $f(u,i) =(u,j)$.
Question: Must $f$ be a covering space map? I allow covering space maps to be non-surjective, so in particular to me $\emptyset \rightarrow X$ is a covering space for all $X$.
If $U$ is connected, then the answer is "yes" since then the $U \times j$ are the connected components of $U \times J$, which implies that there exists some set map $g\colon I \rightarrow J$ such that $f(u,i)=(u,g(i))$ for all $u \in U$ and $i \in I$. But for general $U$ stranger $f$ exist, and I have no idea.
Motivation: In my math stackexchange question here, I explain that this is equivalent to asking whether in complete generality covering space homomorphisms must be covering space maps (note that this question is about the case of covering space homomorphisms between trivial covering spaces). This holds for reasonable spaces, say if there is a universal cover. I wanted to know if it held in general.
I originally asked on math stackexchange since I thought it probably had a trivial answer that I was just missing, but I've convinced myself that it is actually a non-trivial question, so I decided to move it here (while asking it in a different way to bring out the main difficulty better).
