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

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:


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*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.



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