A covering with connected total space and finite fibres

Question

I have a covering map $p:X\to Y$ with $X$ connected and the fibres are finite.

What could I conclude?

Could I conclude that $Y$ is connected (or only locally)?

Could I conclude that the fibers have the same number of points (if $Y$ is connected, yes!).

Thank you so much.

Answer 1

Definition: A continuous surjective map $p : X \to Y$ is a covering map if every $y \in Y$ admits an open neighbourhood $U$ such that $p^{-1}(U)$ is a disjoint union of open sets in $X$, each of which is mapped homeomorphically onto $U$ by $p$; such a neighbourhood is called evenly covered.

The function $\Phi : Y \to \mathbb{N}\cup\{\infty\}$, $y \mapsto |p^{-1}(y)|$ is locally constant due to the existence of evenly covered neighbourhoods. Equipping $\mathbb{N}\cup\{\infty\}$ with the discrete topology, the map $\Phi$ is continuous. As the continuous image of a connected set is connected, $\Phi$ is constant on each connected component of $Y$. In particular, if $Y$ is connected, all the fibers of $p$ have the same cardinality. If $Y$ is not connected, this is no longer the case. For example, let $X = \{0, 1, 2\}$ and $Y = \{0, 1\}$, both equipped with the discrete topology, then $p : X \to Y$ given by $p(0) = p(2) = 0$ and $p(1) = 1$ is a covering map with $p^{-1}(0) = \{0, 2\}$ and $p^{-1}(1) = \{1\}$.

With your hypotheses, $Y$ is connected as it is the continuous image of a connected set, i.e. $Y = p(X)$. Therefore, all the preimages of $p$ have the same (finite) cardinality.