For a positive integer $g$, let us denote the connected closed orientable surface with genus $g$ by $M_g$. Suppose we have an $n$-sheeted covering map $p:M_{g_1} \to M_{g_2}$. I want to show that $p_* : H_2(M_{g_1}) \to H_2 (M_{g_2})$ is multiplication by $n$. (We know that $H_2(M_g)$ is $\Bbb Z$. Actually there is ambiguity of sign, but here I mean "multiplication by $n$ with appropriate generators".)
I tried to use the local degree argument. Although degree is defined only for spheres, but we can define degrees for a map $M_{g_1} \to M_{g_2}$ in an obvious way, and also we can use the same argument to show that the degree of $p_*$ is the sum of the local degrees.
Since $p$ is a covering map, it is a local homeomorphism, so at every point of $M_{g_1}$ the local degree must be $+1$ or $-1$. Fix a point $y$ in $M_{g_2}$. Since $p$ is $n$-to-$1$, $p^{-1}(y)$ consists of $n$ distinct points. Now if I show that the sign of the local degree at each of these $n$ points are equal, then I would be done, but I'm not sure about this. Intuitively it seems true, but how do I have to show this? I don't think local degree is locally constant in general.
Best Answer
Hint: Show that the local degree is constant in a neighborhood (using local orientations as @ConnorMalin suggested). Then since $M_{g_1}$ is connected, the local degree is constant.