Question Regarding Riemann-Hurwitz Formula Proof Does someone knows of any good reference for a proof of the Riemann-Hurwitz Formula (of Riemann Surfaces) that uses Spectral-Sequences and Homology ?
Thanks in advance ! 
[ I think I know how to start this kind of proof, but have no idea on how to finish it...If no one has a good reference, I might ask for the specialists' help in finishing my idea]
 A: I'll try to sketch a proof of Riemann-Hurwitz using the Leray spectral sequence. It has the feel of a fun exercise.
To fix notation, let $X$ and $Y$ be compact Riemann surfaces and let $f : X \to Y$ be a finite surjective morphism of degree $d$. We want to compare the topological Euler numbers of these surfaces; these are the numbers defined as
$$
\chi(X) = h^0(X,\mathbb C) - h^1(X,\mathbb C) + h^2(X,\mathbb C)
$$
and similarly for $Y$. We'll use arbitrarily fancy facts of sheaf cohomology to do this.
If $\mathcal F$ is a sheaf on $X$, then the first terms of the Leray spectral sequence read
$$
E_2^{p,q} = H^q(Y, \mathcal R^p f_* \mathcal F) \Rightarrow H^{p+q}(X,\mathcal F).
$$
As the fibers of $f$ are 0-dimensional, we have $\mathcal R^p f_* \mathcal F = 0$ for any $p \geq 1$. Combined with the annihilation of cohomology on $Y$ for dimension reasons, we find that $E^{p,q}_2 = 0$ for any $p \geq 1$ and $q \geq 3$. The second page of the Leray spectral sequence is thus just
$$
E_2^{0,0} \qquad E_2^{0,1} \qquad E_2^{0,2}
$$
and all other entries are zero, so the sequence degenerates at the $E_2$-level. It follows that $H^k(Y,f_*\mathcal F) = H^k(X,\mathcal F)$ for any $k$.
Consider now a point $y$ on $Y$ that is not in the image of the ramification locus of $f$, in other words the preimage $f^{-1}(y)$ consists of $d$ distinct points. Then we see that $f_{\ast} {\mathbb C} = {\mathbb C}^{\oplus d}$. This line of though yields a short exact sequence
$$
0 \longrightarrow f_* \mathbb C \longrightarrow \mathbb C^{\oplus d} \longrightarrow
\mathcal G \longrightarrow 0
$$
where $\mathcal G$ is a skyskraper sheaf supported on the image of the ramification divisor of the morphism $f$. Taking Euler characteristics we get
$$
d\,\chi(Y) = \chi(f_*\mathbb C) + \chi(\mathcal G)
= \chi(X) + h^0(Y,\mathcal G).$$ 
Expressing $h^0(Y,\mathcal G)$ in terms of the degrees of $f$ at its ramification points, and thus showing that it has the expected form, should not be a source of great trouble.
The thing that makes this proof relatively painless is that the Leray spectral sequence degenerates straight away (at least without recourse to heavy machinery) and that calculating the cohomology of a sheaf supported on a finite number of points is easy. The spectral sequence will again degenerate at the $E_2$-level in the case of a morphism between surfaces, but there a finer analysis is needed to calculate the cohomology of the corresponding sheaf $\mathcal G$. In any case the proof points the way to a similar statement for finite surjective morphisms between higher dimensional varieties, though it also seems to indicate that this is not a path one wants to take unless one really needs to.
A: Just in case you don't know how overkill this machinery really is (especially in the case of Riemann surfaces which is what you asked for) I have posted a sketch of the basic argument.  Since it doesn't really answer your question, I have made the answer CW.
Let $f: X \to Y$ be a holomorphic map of Riemann surfaces of degree $d$, i.e. the preimage of all but finitely many points in $Y$ has $d$ points in it.  Let $x_i$ be the branch points of $f$ and $y_j$ be the images of these branch points.
Triangulate $Y$ so that each $y_i$ is a vertex. Lift this triangulation of $Y$ to a triangulation of $X$ via $f$.  Let $F,E,V$ be the number of faces, edges, and vertexes in the triangulation of $Y$.  Then there are $dF$ faces and  $dE$ edges in $X$.  There are almost $dV$ vertexes, but I get less vertexes because at a ramification point in $Y$ there are less than the full $d$ preimages.  How many less?  Exactly the sum of one less than the degrees of the $x_i$!
Feel free to edit this post if it is unclear.
