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](https://en.wikipedia.org/wiki/Riemann%E2%80%93Hurwitz_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.



Construction of self-covering map of any surface
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 $84g$. 


