I just took my smooth manifold theory final and would like to go over one of its most interesting questions.

Let $M$ and $N$ be smooth manifolds and let $f:M\to N$ be a covering projection.

Claim: If $N$ is orientable, then $M$ is orientable.

Proof: Let $p\in N$. Since $N$ is orientable, it is the case that $p$ is in the domain of an oriented local frame $\left(\sigma_i\right)$ defined on an open subset $U$ of $N$. Since $f$ is a covering projection, it is the case that $f^{-1}\left(U\right)$ is an open subset of $M$ for which there exists a countable, open, pairwise disjoint cover $\left\{V_\alpha\right\}_{\alpha\in A}$ such that each $\left.f\right|_{V_\alpha}:V_\alpha\to U$ is a diffeomorphism. Therefore, each $\left(\left.f\right|_{V_\alpha}\right)_*:T_qV_\alpha\to T_{\left.f\right|_{V_\alpha}\left(q\right)}U$ is an isomorphism for all $q\in V_\alpha$, which implies that $\left(\left(\left.f\right|_{V_\alpha}\right)^*\sigma_i\right)$ is an oriented local frame defined on $V_\alpha$. $\blacksquare$

Recall that $\left(\sigma_i\right)$ is an ordered $n$-tuple, taking $N$ to be $n$-dimensional, of sections of the tangent bundle $\pi_N:TN\to N$ of $N$, such that $\left(\left.\sigma_i\right|_p\right)$ is a basis for $E_p=\pi^{-1}_N\left(p\right)$ for all $p\in U$.

Although I am not yet certain that this is a valid proof, I am pretty confident about it.

The converse of the claim does not hold because $f$ need not be surjective, in which case $M$ and $f\left(M\right)$ may be orientable but $N$ might not. For example, let $N=\mathbb S^2/\sim$, let $M=N\setminus\left\{\text{north and south poles}\right\}$, and let $f:M\hookrightarrow N$ be the inclusion map, where $\sim$ is the usual antipodal equivalence. Then $f$ is a local diffeomorphism, $M$ is orientable, and $N$ is not orientable.

I enjoyed this class way too much.