Projection is a covering map iff the topology is discrete

Question

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?

Answer 1

The answer is yes but the argument you provide doesn’t seem bullet-proof to me: in general you cannot say: $X\times Y \cong X \times Z$ and $Y$ has property $p$, therefore $Z$ also has property $p$. You can’t "cancel out the $X$".

Instead you can think of it like this: $Y$ is homeomorphic to the fiber of some point $x\in X$. Can you show that the fiber of a point of a covering map is discrete?