I know that the following is true:
Let $Y$ have the discrete topology. Show that if $p:X\times Y\to X$ is the projection, then $p$ is a covering map.
But is the opposite direction also true? So does "$p$ is a covering map" imply that the topology on $Y$ is discrete?
I would argue: Yes it is. We have $p^{-1}(U) = U \times Y$ and we require it to be homeomorphic to $U \times F$ for some non-empty discrete set $F$. So if $Y$ wouldn't be discrete via it's topology, then the product topology of $U \times Y$ would be different to the one of $U \times F$.
Is this correct?