Riemann-Hurwitz formula and degree of a mapping

Question

I'm working over my exercise sheet in my course on Riemann surfaces and I have to verify the Riemann-Hurwitz formula $$\chi(X)=n\chi(Y)-\sum_{x\in X}(ord_x(f)-1)$$ for the mapping $f(z)=\frac{(z+1)^2}{z^4},f(\infty)=0$ between $\mathbb{P}^1\to \mathbb{P}^1$. So far only $z=0,z=-1$ contribute to the sum and give multiplicity $4$ and $2$.This would imply that the degree of the map $f$ is -1 since $\chi(\mathbb{P}^1)=2$ but I'm actually not quite convinced on how to compute the degree of this map. There's no definition in my notes, any help is welcome :)

Answer 1

You did not consider all critical points of the mapping. We have $$ f'(z) = -\frac{(z+1)(z+2)}{z^5} $$ so that the critical points are

• $f(0) = \infty$ with multiplicity $4$,
• $f(-1) = 0$ with multiplicity $2$,
• $f(-2) = 1/16$ with multiplicity $2$,
• $f(\infty) = 0$ with multiplicity $2$.

This gives $\sum_{x\in X}(ord_x(f)-1) = 3+1+1+1 = 6$ and $n=4$.