Construction of self-covering map of any surface

Question

Let $\Sigma(g,n)$ be an $n$-punctured surface of genus $g$.

If we assume that $f:\Sigma(g,n)\rightarrow\Sigma(g,n)$ is a branched self-covering map of degree $d$, then the equality follows from the Riemann-Hurwitz formula $$ \chi(\Sigma(g,n)) = d\cdot\chi(\Sigma(g,n)) - \sum_{p\in\Sigma(g,n)}(e_{p}-1) $$ where $\chi(\Sigma(g,n))$ is the Euler characteristic and $e_{p}$ is the ramification index for $p\in\Sigma(g,n)$.

Hence we have data about ramification points and their indices.

Question. Is it possible to construct such a self-covering map for any surface?

I could construct it for a torus by expressing the map as a matrix. But i have no idea for a general case.

Thank you for your time and effort.

Answer 1

First of all, the Riemann-Hurwitz formula with $\chi<0$ implies that for every self-covering $d=1$ so it is an automorphism. The only punctured surfaces with $\chi\geq 0$ are torus, sphere, and sphere with one or two punctures. For torus and sphere with two punctures self-coverings are easy to describe, for the rest, they are just automorphisms.

On the other hand the group of holomorphic automorphisms when $\chi<0$ is finite, and a generic Riemann surface has none.

If you mean just topological self-coverings, then you can construct them. All compact surfaces of given genus $g$ are homeomorphic. Take one which has holomorphic automorphisms and transplant them to your surface. The number of holomorphic automorphisms of a surface of genus $g>1$ is at most $84g$.