This is an argument used in Kodaira's Complex Manifolds and Deformation of Complex Structures, Chpt 2, Sec 2, Example 2.5
The equation $z_0^m+\dots+z_3^m=0$ defines an algebraic surface $S$ in $P^3$. Consider $[z_0,z_1,z_2,z_3]\to [z_0,z_1,z_2]$ as projection map restricted to $S$. Then clearly $S$ is a m-fold branched covering of $P^2$ by moving away from $z_0^m+z_1^m+z_2^m=0$ and branching order of $z_0^m+z_1^m+z_2^m$ is $m-1$. Denote $\chi(M)$ euler characteristic of manifold $M$.
"$\chi(S)=m\chi(P^2)-(m-1)\chi(C)$".
$\textbf{Q:}$ How does one arrive at the formula $\chi(S)=m\chi(P^2)-(m-1)\chi(C)$? It looks like this is a minick of Riemann Hurewitz formula in terms of euler characteristics by noting $\chi(pt)=1$. It seems this should be obvious from topologial part but I could not arrive at conclusion easily. It seems that Example 2.3 does need use cohomology argument to compute arithmetic genus and implicitly assumes arithmetic genus same as topological genus.(The Riemann surface side story is clear to me but this is higher dimensional.) The proof of Riemann Hurewitz for Riemann surface case I knew requires pullback of meromorphic forms and knowledge of degree(i.e. Riemann-Roch and Serre duality need to be assumed).
Update:I have found the following link. https://mathoverflow.net/q/71960 Why it is critical to require $Z$ codimension even real codimension here to get $\chi(X-Z)=\chi(X)-\chi(Z)$? Should I naively apply Mayer-Vietoris argument to get this formula as I need to wonder why $Z$ is homototpic to an open set? Even so, I also need to cut off $X-Z$ part a bit away to make sure intersection empty to get additive formula of euler characteristics.
$\textbf{Q':}$ Should Riemann-Hurewitz formula purely topological and thus there should be a topological proof which generalizes naturally to higher dimensional case? It would be nice if someone can offer a nice easy-going reference.