Let $M$ be an orientable surface of genus $g>1$, I can assume compact. Let $f$ be a continuous map from $M$ to $M$. I want to prove that the degree of $f$ is $1$, $0$ or $-1$.
For a surface of genus $1$, a torus $S^1\times S^1$, it is easy to construct a map of any degree. So I have to use topological properties of the genus.
Any help is appreciated
If $M$ is a closed, connected, oriented manifold, let $\|M\|$ denote its Gromov norm (also known as simplicial volume). One of the key properties of the Gromov norm is that if $N$ is another closed, connected, oriented manifold of the same dimension, then for any continuous map $f : M \to N$, the degree of $f$ satisfies
$$|\deg f|\|N\| \leq \|M\|.$$
In particular, if $f : M \to M$, then either $\|M\| = 0$ in which case we get no information about the degree of $f$, or $\|M\| \neq 0$ and so $|\deg f| \leq 1$, i.e. $\deg f \in \{-1, 0, 1\}$.
Gromov proved that the Gromov norm of a closed oriented hyperbolic manifold $M$ of dimension $n$ satisfies
$$\|M\| = \frac{\operatorname{vol}(M)}{\nu_n}$$
where $\nu_n$ is the supremal volume of all geodesic $n$-simplices in hyperbolic $n$-space (which is finite). In particular, $\|M\| > 0$.
For $g \geq 2$, the surface $\Sigma_g$ is hyperbolic and hence $\|\Sigma_g\| \neq 0$. So by the above, any map $f : \Sigma_g \to \Sigma_g$ has degree $-1$, $0$, or $1$.
In fact, as $\Sigma_g$ is hyperbolic for $g \geq 2$, it admits a metric of constant Gauss curvature $-1$, so by Gauss-Bonnet
$$\operatorname{vol}(\Sigma_g) = \int_{\Sigma_g}1 = -\int_{\Sigma_g}-1 = -2\pi\chi(\Sigma_g) = -2\pi(2 - 2g) = (4g - 4)\pi.$$
As $\nu_2 = \pi$, we see that $\|\Sigma_g\| = 4g - 4$.
In contrast, $S^2$ and $S^1\times S^1$ admit self-maps of any degree, so their Gromov norm is zero.
There is a nice formula for surfaces which is called Riemann-Hurwitz formula. Let $X$ and $Y$ surfaces (now the genus can be less than $2$) and let $f:X\longrightarrow Y$ a continuous map which is surjective and not a branched covering, then:
$$\chi(X)=d\chi(Y),$$
where $d$ is the degree of the map. Suppose now that $M$ is a surfaces of hyperbolic type: $i.e.$ $g(M)\ge2$. Then, by the formula above $\chi(M)=d\chi(M)$; so $d=1$.
The case of $d=-1$ appear when you choose different orientation on $M$. Infact the map $f$ induce a map $f_\sharp$ in cohomology: $$f_\sharp: H^2(M,\mathbb{Z})\longrightarrow H^2(M,\mathbb{Z})$$ such that $f_\sharp([M])=d[M]$, where $[M]$ is a fundamental class. Choosing an orientation means choosing a generator of $H^2$. So with different orientations the degree of the map is $-1$.
Finally the degree $0$ appear when you consider continuous map which are homotopic to constant map.
