2 Proof for discreteness of the mapping space added; added 99 characters in body

Here is an argumentwhy the space of maps is compact. It works well for proving is a generalization of the argument that shows the finiteness of automorphism groups. I have not thought through the details, so there might be a bug.

First you we show that the mapping space is compact. This is an analytic argument.You can pick, by uniformization, universal covers $E \to \Sigma_i$ ($E\subset C$ is the unit disc). Then assume that $f_n: \Sigma_1 \to \Sigma_2$ is a sequence of maps. The goal is to find a convergent subsequence. You can find lifts of these maps $g_n:E \to E$, such that $g_n(0)$ remains in a bounded ball in the Poincare metric. By Montel's theorem, you find a convergent subsequence of $g_n$; the limit is a function $g:E \to C$ (not yet to $E$). If $|g(z)|=1$ for some $z \in E$, by the maximum principle, $g$ is constant. You can exclude this case, because $g_n (0)$ was to stay in a ball, but the circle as $\infty$ away from $0$. So you can assume that $g_n$ converges to some map $g: E \to E$. $g$ will be invariant under the Deck transformation groups group of $\Sigma_1$ and so it descends to a map $f: \Sigma_1 \to \Sigma_2$. This shows that the space of holomorphic maps is compact. In particular, only finitely many homotopy classes can be realized by holomorphic maps.

Next we show that any homotopy class contains at most one holomorphic map (if it is not constant). I use basic facts from algebraic topology, but no algebraic geometry for that. Let $f: \Sigma_1 \to \Sigma_2$ be a holomorphic map and let $\Gamma \subset \Sigma_1 \times \Sigma_2$ be the graph; it is a complex submanifold. The normal bundle to $\Gamma$ can be identified with $f^{\ast} T\Sigma_2$. This bundle has negative degree if $f$ is not constant, namely $deg (f) \chi (\Sigma_2)$. The algebraic self-intersection number of (the homology class of) $\Gamma$ is thus negative.

If $g$ were another holomorphic map in the same homotopy class as $f$, then the graph $\Delta$ of $g$ has the same homology class as $\Gamma$. Thus the algebraic intersection number of the two graphs is, as argued above, negative. If $g$ and $f$ were different, then $\Gamma$ and $\Delta$ intersect in a finite number of point and the intersection index at each intersection point is positive (because $f$ and $g$ are both holomorphic), so the sum of the intersection indices is nonnegative. On the other hand, these intersection indices should add to the algebraic intersection number of $\Gamma$ with itself, which is negative. Contradiction.

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Here is an argument why the space of maps is compact. It works well for proving finiteness of automorphism groups. I have not thought through the details, so there might be a bug.

First you pick, by uniformization, universal covers $E \to \Sigma_i$ ($E\subset C$ is the unit disc). Then assume that $f_n: \Sigma_1 \to \Sigma_2$ is a sequence of maps. The goal is to find a convergent subsequence. You can find lifts of these maps $g_n:E \to E$, such that $g_n(0)$ remains in a bounded ball in the Poincare metric. By Montel's theorem, you find a convergent subsequence of $g_n$; the limit is a function $g:E \to C$ (not yet to $E$). If $|g(z)|=1$ for some $z \in E$, by the maximum principle, $g$ is constant. You can exclude this case, because $g_n (0)$ was to stay in a ball, but the circle as $\infty$ away from $0$. So you can assume that $g_n$ converges to some map $g: E \to E$. $g$ will be invariant under the Deck transformation groups and so it descends to a map $f: \Sigma_1 \to \Sigma_2$. This shows that the space of holomorphic maps is compact. In particular, only finitely many homotopy classes can be realized by holomorphic maps.