If $X$ is a path connected topological space, a covering space of $X$ is a space $\tilde{X}$ and a map $p:\tilde{X} \to X$ such that there exists an open cover $\left\{ U_\alpha \right\}$ of $X$ where $p^{-1}(U_\alpha)$ is a disjoint union of homeomorphic open sets ($p$ being homeomorphism between them).
Are there path connected covering spaces of a path connected $X$ which are not surjective? Why?
Allen Hatcher's book 'Algebraic Topology' says the disjoint union of open sets mentioned may be empty/null in some cases (e.g. $X$ not path connected with $p$ the identity on a path component). I'm really asking if the path connectedness of the spaces can be used to show there is no 'folding'. For instance, if $X = \tilde X = [−1,1]$ and the covering map the absolute value. (This example is not a covering space as the preimage of any open set around 0 is bad. I hope this demonstrates my point though).
Best Answer
Let $p: \tilde{X} \rightarrow X$ be a covering map in the broad definition (empty sum allowed, as Hatcher does). First, $p[\tilde{X}]$ is open in $X$: let $y \in X$ with $x \in \tilde{X}$ such that $p(x) = y$. Let $U$ be an evenly covered neighbourhood of $y$, so $p^{-1}[U] = \sum_{i \in I} O_i$ (disjoint sum over index set $I$). As $x$ is in the sum, it is non-empty: some $O_{i_{0}}$ contains $x$ and $p |_{O_{i_0}}$ is a homeomorphism between $O_{i_{0}}$ and $U$. In particular $U \subset p[\tilde{X}]$, so $y$ is an interior point of $p[\tilde{X}]$.
Suppose $y \notin p[\tilde{X}]$. Then again take an evenly covered neighbourhood $U$ of $y$: $p^{-1}[U] = \sum_{i \in I} O_i$, disjoint sum over some index set $I$, where for every $i \in I$, the map $p|_{O_i}$ is a homeomorphism between $O_i$ and $U$. This means (!) that $I = \emptyset$, as otherwise we'd have a preimage for $y$, contradicting how we picked $y$. So in fact $p^{-1}[U] = \emptyset$, and this shows that $U \subset X\setminus p[\tilde{X}]$, so $p[\tilde{X}]$ is closed.
Now if $X$ is connected and $\tilde{X}$ is non-empty, then $p[\tilde{X}]$ is closed, open and non-empty, so equals $X$ by connectedness. So $p$ is surjective.
These are the only conditions we need: $X$ connected and $\tilde{X} \neq \emptyset$.