I was trying to solve the following exercise. I wanted to check if my solution was correct/rigorous enough, and ask a question at the end. (The general direction is given in here: holomorphic map between compact Riemann surfaces, and I seem to have expanded out the steps in a detailed way)
Let $h\colon X \rightarrow Y$ be a morphism of Riemann surfaces. Assume that $X$ is connected and $h$ is not constant. Show that the image of $h$ is open in $Y .$ Deduce that $h$ is surjective if we moreover assume that $X$ is non-empty and compact, and $Y$ is connected. This implies that a morphism from a connected compact Riemann surface to a connected non-compact Riemann surface is always constant. In particular, every holomorphic function on a connected compact Riemann surface is constant.
Open mapping theorem: If $U$ is a connected open subset of $\mathbb{C}$, then the image of every non-constant holomorphic function $f: U \rightarrow \mathbb{C}$ is open.
Proof: Let $h\colon X \rightarrow Y$ be such a morphism. Let $O \subset X$ be an open subset of $X$, and we will show $h(O)$ is open. For any $x \in O$, let $(U,\phi)$ be a chart of $O$ containing $x$, and $(V,\psi)$ a chart of $Y$ containing $h(x)$. Then consider the holomorphic map $\psi \circ h \circ \phi^{-1}$ defined on the open set $\phi(U \cap h^{-1}(V))\ni\phi(x)$. Draw an open neighbourhood around $\phi(x)$, and apply $\psi \circ h \circ \phi^{-1}$ to it. The result is an open in $\mathbb{C}$ around $\psi \circ h(x)$, using the open mapping theorem. Composing from the left with $\psi^{-1}$, we obtain an open neighbourhood around $h(x)$. Since this holds for any point $x \in O$, $h(O)$ is open in $Y$.
Now assume that $X$ is connected, non-empty, compact, and $Y$ connected. Let the notation be as above. Then $h(X)$ is compact, non-empty, and connected. As $h$ is an open map, $h(X)$ is open in $Y$. As $h(X)$ is compact, it is closed in Hausdorff space $Y$. As $Y$ is connected, we conclude $h(X) = Y$. Now, if $Y$ is non-compact, a continuous map cannot map a compact space to a non-compact space, so $h$ must be a constant (otherwise we proved it is surjective).
Is my solution correct? Also, I do not understand why this implies that every holomorphic function on a connected compact Riemann surface is constant.
Best Answer
Your solution is correct. For the last part, if you do not understand the explanation given in comments, you could try to prove it by contradiction.
Suppose $f : X \to \mathbb{C}$ is holomorphic with $X$ compact, $f$ non-constant. By open mapping theorem, as you said, $f(X)$ is open in $\mathbb{C}$. But as $f$ is continuous and $X$ compact, $f(X)$ is compact. Hence $f(X)$ closed in $\mathbb{C}$ (since compact subspaces of Hausdorff spaces are closed). By connectedness of $\mathbb{C}$, $f(X) = \mathbb{C}$, and $\mathbb{C}$ is then compact: this is the contradiction.
A more direct way to prove this is using the maximum principle. Suppose $f : X \to \mathbb{C}$ is continuous with $X$ compact. Then $|f| : X \to \mathbb{C}$ is continuous and achieves its maximum at a point $p \in X$. Pick a holomorphic coordinate $z$ around $p$, say $z : U \to X$, with $z(0) = p$, and $U$ connected. Then $f\circ z : U\subset \mathbb{C} \to \mathbb{C}$ is holomorphic, and $|f\circ z| $ has a local maximum at $0$: by the maximum principle, $f\circ z$ is constant. It follows that $f$ is locally constant around $p$. Now, one can easily show that $f^{-1}(f(p))\subset X$ is open and closed, and by connectedness of $X$, $f$ is constant.