Reference request for the fact that if the base space is orientable then the total space is orientable too (for surfaces)?

Question

I need a reference for the fact that if you have a covering $f:\tilde{\Sigma}\longrightarrow\Sigma$ and $\Sigma$ is orientable then $\tilde{\Sigma}$ is also orientable. $\tilde{\Sigma}$ and $\Sigma$ are surfaces. The fact is that I never took an homology theory course at university so I don't know how to prove it by this way, but I need to use this fact for my thesis, so I wanted to give at least a reference. It's quite strange that I didn't find the proof of this seemingly basic fact in any book, if you know where to find it I would be very grateful!

Answer 1

Using the differential topology definition of orientability, i.e. that the top exterior power of the tangent bundle is trivial, we can yield the result without too much work.

I claim that it suffices to show that $f^*T^*\Sigma\cong T^*\tilde{\Sigma}$ for any smooth covering map $f: \tilde{\Sigma}\to \Sigma$. This suffices as if $\bigwedge^n T^*\Sigma\cong \mathbb{R}\times \Sigma$ then $\mathbb{R}\times \tilde{\Sigma}\cong f^*(\mathbb{R}\times \Sigma) \cong f^*\bigwedge^n T^*\Sigma\cong\bigwedge^nf^*T^*\Sigma\cong \bigwedge^nT^*\tilde{\Sigma}$. The fact that $f^*T^*\Sigma\cong T^*\tilde{\Sigma}$ comes from the fact that $f$ is a local diffeomorphism. The dual of the differential of $f$ can be seen as a map $T^*f: f^*T^*\Sigma\to T^*\tilde{\Sigma}$ (the regular differntial is a map $Tf: T\tilde{\Sigma}\to f^*T\Sigma$ to its dual yields a map going the opposite direction). Because $f$ is a local diffeomorphism $Tf$ is an isomorphism and hence $T^*f$ is an isomorphism.

For the characteristic class based approach suggested by Vincent Boelens, the previous work tells us that $f^*T\Sigma\cong T\tilde{\Sigma}$ and so by naturality of the first Stiefel whitney class $w_1(\tilde{\Sigma})=w_1(T\tilde{\Sigma})=w_1(f^*T\Sigma)=f^*w_1(T\Sigma)=f^*w_1(\Sigma)=0$.

How do we see that the first Stiefel Whitney class obstructs orientability? For $E\to X$ a real vector bundle over a CW complex, $w_1(E)=w_1(\bigwedge^n(E))$, so if $\bigwedge^nE\cong \mathbb{R}\times X$ then $w_1(E)=0$. If $w_1(\bigwedge^n(E))=0$ then the classification of real line bundles by $H^1(X,\mathbb{Z}/2)$ tells us that two real line bundles are isomorphic if and only if they have the same first Stiefel Whitney class and hence $\bigwedge^n(E)\cong \mathbb{R}\times X$.