# Projection is a covering map iff the topology is discrete

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?

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?