Killing cohomology of surfaces in finite covers Let $S$ be a closed orientable surface and $R$ a commutative ring.  Given a nonzero element $\alpha \in H^1(S;R)$ is there a finite cover $p : \tilde{S} \to S$ such that $p^*(\alpha) = 0 \in H^1(\tilde{S}; R)$?  
 A: No. This has nothing to do with $S$ being a surface. In what follows all we assume is that $S$ can run the classification theory of covers (path-connected, locally path-connected, semilocally simply connected). It holds for any CW complex.
There is a natural isomorphism $H^1(S;G) \cong \text{Hom}(\pi_1 S, G)$ for any $S, G$. Given any element $\phi \in \text{Hom}(\pi_1 S, G)$, if $f: S' \to S$ is a covering space, the correspinding map $\phi f_* = 0$ if and only if $f$ factors through the cover corresponding to $\text{ker}(\phi) \subset \pi_1 S$. (We use here the classification of covering spaces over reasonable spaces.)
If $\text{ker}(\phi)$ has infinite index, then the cover corresponding to $\text{ker}(\phi)$ is an infinite cover. For instance, if $G$ is infinite and $\phi$ is surjective (or at least its image has finite index), then you cannot kill $\phi$ in a finite cover. 
Because every subgroup of $\Bbb Z$ is zero or infinite, in particular, you cannot kill any nontrivial class of $H^1(S;\Bbb Z)$ in a finite cover.
A: No; in fact, if $R$ is torsion-free, there never is such a finite cover.  If $p:\tilde{S}\to S$ is a finite cover, then $p_*\pi_1(\tilde{S})$ has finite index in $\pi_1(S)$.  We can naturally identify $H^1(S;R)$ with $\operatorname{Hom}(\pi_1(S),R)$ and similarly for $\tilde{S}$.  If $p^*(\alpha)=0$, that means that $\alpha$ vanishes on $p_*\pi_1(\tilde{S})$.  But that implies $\alpha$ vanishes on all of $\pi_1(S)$ (a homomorphism from a group to $R$ which vanishes on a finite index subgroup must vanish on the whole group, since $R$ is torsion-free).  Thus $p^*(\alpha)=0$ implies $\alpha=0$.
Alternatively, you could use Poincaré duality: if $\alpha\neq 0$, there exists $\beta\in H^1(S;R)$ such that $\alpha\cup\beta\neq 0$, and then $p^*(\alpha\cup\beta)\neq 0$ since $p^*:H^2(S;R)\to H^2(\tilde{S};R)$ is multiplication by the degree of $p$ which is injective if $R$ is torsion-free.
