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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 is at most $84g$, where $g$ is the genus.

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 $84g$.

First of all, the Riemann-Hurwitz formula with $\chi<0$ implies that for every self-covering $d=1$ so it is an automorphism.

On the other hand the group of holomorphic automorphisms 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 is at most $84g$, where $g$ is the genus.

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 is at most $84g$, where $g$ is the genus.

No. For hyperbolic surfaces, every self-covering must be an automorphism (because the hyperbolic metric has finite area, and each self-covering is a local isometry, so it must divide the area on its degree). On the other hand the group of automorphisms is finite, and a generic surface has none.

First of all, the Riemann-Hurwitz formula with $\chi<0$ implies that for every self-covering $d=1$ so it is an automorphism. 

On the other hand the group of holomorphic automorphisms 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 is at most $84g$, where $g$ is the genus.