Is a path connected covering space of a path connected space always surjective?

Solution 1:

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 $\tilde{X} \neq \emptyset$.

Solution 2:

This is my attempt to prove it.

Let's assume there is some $x\in X$ such that $x\not\in p(\tilde{X})$. Then given the open cover of the definition, $\{U_a\}_{a\in A }$, there exist some $a$ such that $x\in U_a$.

Let's say $U\subseteq \tilde{X}$ is a preimage of $U_a$, $p(U)\cong U_a$ (pick any such $U$). Since $p|_U$ is an homeomorphism between $U_a$ and $U$, in particular $p|_U$ is surjective in $U_a$ so $p(p^{-1}(x)) = p|_U(p^{-1}(x)) = x$ which is a contradiction.