Projection from discrete set in covering space

Question

In Kosniowski's book it is said that a trivial example of covering space is consider the canonical projection $\pi : X \times Y \to X$, being $Y$ a discrete space.

My question is: why $Y$ must be discrete?

Thank you very much for the support.

Answer 1

Discrete space does not mean finite.

A space $X$ is said to be discrete, if all subsets are open (and thus also closed).

Answer 2

Discrete does not mean finite. Depending on your definition of a covering space, usually you do allow infinite, discrete fibers. Think for instance of the standard cover of $S^1$ by $\mathbb{R}$ (whose fiber is $\mathbb{Z}$) or of the Riemann surface of the complex logarithm, which is a cover of $\mathbb{C} \backslash\{0\}$ with fiber $\mathbb{Z}$.

There are generalizations with continuous fibers (for instance, fibered manifolds), but they are usually more complicated than covering spaces.