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

Question

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?

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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).

Answer 1

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.

Answer 2

You want to prove that $p(\tilde X) = X$. The usual technique to prove that something is equal to the whole space, when that space is connected, is to prove that it open, closed, and nonempty. What follows is missing details that you should fill in:

• Assuming $\tilde X \neq \emptyset$ (otherwise the result is of course false), it follows immediately that $p(\tilde X) \neq \emptyset$.
• If $x \in p(\tilde X)$, the existence of a neighborhood entirely contained in $p(\tilde X)$ of $x$ follows immediately from the definition of a covering space.
• Now if $x$ is in the closure of $p(\tilde X)$: $x \in U_\alpha$ for some $\alpha$, but then every neighborhood of $x$ meets $p(\tilde X)$ (because $x$ is in the closure). I'll let you conclude.

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Remark: I never used the fact that $\tilde X$ is path-connected. In fact, that hypothesis is unnecessary.

Answer 3

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