# How can a covering map (as defined in Hatcher) fail to be a surjection?

Hatcher defined a covering space as follows:

$\textbf{Defn:}$ A covering space of a space $X$ is a space $\tilde{X}$ together with a map $p: \tilde{X} \to X$ satisfying the following conditions: There exists an open cover $\{ U_{\alpha}\}$ of $X$ such that for each $\alpha$, $p^{-1}(U_{\alpha})$is a disjoint union of open sets in $\tilde{X}$ each of which is mapped by $p$ homeomorphically onto $U_{\alpha}$

I don't understand how such a map can fail to be surjective. Any concrete examples??

For instance, if $\tilde{X}$ is the empty set, and $p$ is the empty map. Then each $p^{-1}(X)$ is empty, so it is an (empty) disjoint union of open sets, "each" of which are mapped homeomorphically onto $X$ (vacuously, as there are no opens to check the condition on).
• And there are intermediate situations: if $X$ is not connected, one component can be completely outside of the image for this reason. – Mariano Suárez-Álvarez Oct 16 '15 at 1:26
• This is not correct. The definitions asks that $p$ maps each disjoint open set onto $U_\alpha$. This cannot happen in your case, since the image of an empty set is empty. – Pedro Tamaroff Oct 22 '15 at 17:31
• @PedroTamaroff I think it is correct. The open cover of $\emptyset$ is not the empty set $\{ \emptyset \}$, it is the empty collection of sets $\{ \}$. Each open $U \in \{ \}$ satisfies any condition you want, since there are no opens to check. Do you disagree? – Lorenzo Najt Oct 24 '15 at 15:31
• @PedroTamaroff I was being imprecise. You take the covering $\{X\}$ of the base space. What I mean is that $p^{-1}(X)$ is an empty disjoint union - it is $\cup_{U \in \emptyset} U$. Since there are no opens in this disjoint union to check, vacuously each maps onto $X$. – Lorenzo Najt Oct 24 '15 at 17:06