Covering map + homotopy equivalence = homeomorphism? How to show that a covering map which is also a homotopy equivalence is a homeomorphism?
 A: Let $p:\tilde X \to X$ be the covering map. As a homotopy equivalence it induces an isomorphism $\pi_1(\tilde X,\tilde x_0)\to\pi_1(X,x_0)$ for any $x_0\in X.$ Now, assume $p(y_0)=p(y_1)=x_0.$ We want to show $y_0=y_1.$ Since a homotopy equivalence induces a bijection between the path-components of $\tilde X$ and the path-components of $X$, the points $y_0,y_1$ must lie in the same path-component. So let $\alpha$ be a path from $y_0$ to $y_1.$ Its image is a loop $p\alpha$at $x_0.$ By the induced isomorphism $p_*$ there is a unique class $[\beta],$ such that $p_*[\beta]=[p\alpha],$ i.e. $p\beta$ is homotopic to $p\alpha,$ and this homotopy lifts to a unique path-homotopy between the loops $\beta$ and the unique lift of $p\alpha,$ which is $\alpha.$ This implies that $\alpha$ is a loop and $y_0=y_1.$
Now injectivity makes $p$ a homeomorphism.
The length of the proof depends on how much you know about covering maps. If you know that each loop $\gamma$ at $x_0$ such that $[\gamma]\in p_*(\pi_1(\tilde X))$ has a lift which is a loop, then it follows immediately that the $\alpha$ in the proof is a loop.
A: We can modify a little Stefan's reasoning. After reducing problem to path component we can firstly note that following lemma holds:

(Conjugacy Theorem). Let $p:\tilde X \to X$ be a covering map. For any $q \in X$, as $\tilde q$ varies over the fiber $p^{-1}(q)$, the set of subgroups $p_*\pi_1(\tilde X, \tilde q) \subset \pi_1(X,q)$ is exactly one conjugacy class.

but homotopy equivalence induces isomorphism of fundamental groups so the conjugacy class must contain one element and so our covering is $1$-sheeted so is injective and hence homeomorphism.
