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).

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.
