Morphism induced in cohomology of a covering space It is a basic question but I'm stuck.
If $p:M\rightarrow N$ is a $m$-fold unramified covering between surfaces, why the morphism induced by $p$ in cohomology at level 2 with coefficients in $\mathbb{Z}/m$ is zero?
 A: An $m$-fold covering map $p$ has degree $m$, so $p^* : H^2(N; \mathbb{Z}) \to H^2(M; \mathbb{Z})$ is given by multiplication by $m$. There is a commutative diagram
$$\require{AMScd}
\begin{CD}
H^2(N; \mathbb{Z}) @>{p^*}>> H^2(M;\mathbb{Z})\\
@VVV @VVV \\
H^2(N; \mathbb{Z}_m) @>>> H^2(M;\mathbb{Z}_m)
\end{CD}$$
where the vertical maps are coefficient reduction modulo $m$, and the second horizontal arrow is the one you asked about. 
If $N$ and $M$ are not closed, then $H^2(N; \mathbb{Z}_m) = H^2(M; \mathbb{Z}_m) = 0$ (and hence the map between them is zero). From now on, suppose $N$ and $M$ are closed.
If $N$ and $M$ are orientable, then each vertical map is the quotient map $\mathbb{Z} \to \mathbb{Z}_m$. So the map $H^2(N; \mathbb{Z}) \to H^2(M; \mathbb{Z}_m)$ obtained by going clockwise is zero, and therefore so is the counterclockwise map. As the map $H^2(N; \mathbb{Z}) \to H^2(N; \mathbb{Z}_m)$ is surjective, the map $H^2(N; \mathbb{Z}_2) \to H^2(M; \mathbb{Z}_2)$ must be zero.
If $M$ is not orientable, then $H^2(M; \mathbb{Z}) = \mathbb{Z}_2$ and $H^2(M; \mathbb{Z}_m) = \mathbb{Z}_2$ if $m$ is even or $0$ if $m$ is odd. In the first case, the map $H^2(M; \mathbb{Z}) \to H^2(M; \mathbb{Z}_m)$ is an isomorphism, and in the second case, it is the zero map. Either way, the clockwise map $H^2(N; \mathbb{Z}) \to H^2(M; \mathbb{Z}_m)$ is zero, and therefore so is the counterclockwise map. Whether $N$ is orientable or not, the map $H^2(N; \mathbb{Z}) \to H^2(M; \mathbb{Z}_m)$ is surjective, so as in the previous paragraph, it follows that he map $H^2(N; \mathbb{Z}_m) \to H^2(M; \mathbb{Z}_m)$ is the zero map.

Note, if $M$ and $N$ were manifolds of dimension $n$, the analogous statement with $2$ replaced by $n$ would also be true.
