Let $\mathbb C^* := \mathbb C \cup \{\infty\}$ be the Riemann sphere. I know that every meromorphic function $f \colon \Omega \to \mathbb C^*$ with $\Omega \subseteq \mathbb C$ open can be seen as a holomorphic function of Riemann surfaces ($\Omega$ having only one chart, the identity, and $\mathbb C^*$ the Riemann sphere).
Does the converse also hold, i.e. if $f\colon \Omega \to \mathbb C^*$ is a holomorphic function of Riemann surfaces, is it meromorphic?
My lecture notes claim that.
I could show everything except for that the poles of $f$ are isolated. What I could show is that the set of all poles is closed in $\Omega$.
Best Answer
As already pointed out, you need the hypothesis that $f\colon \Omega\to \mathbb C_{\infty}$ is non-constant in order for $f$ to necessarily be meromorphic. Indeed this is assuming $\Omega$ is supposed to be a connected open subset, otherwise one needs to assume $f$ is non-constant on each connected component.
Now if $Z_0=f^{-1}(0)$ and $Z_{\infty} = f^{-1}(\infty)$, then $Z_0$ and $Z_{\infty}$ are closed subsets of $\Omega$, and so $U_0 = \Omega\backslash Z_{\infty}$ and $U_\infty = \Omega\backslash U_0$ are open subsets of $\Omega$. Now the restriction of $f$ to $U_0$ is a holomorphic function $f_{|U_0}\colon U_0 \to \mathbb C$ while the restriction of $f$ to $U_\infty$ is a holomorphic function $f_{|U_{\infty}}\colon U_{\infty} \to \mathbb C^\times \cup \{\infty\}$, or equivalently, if $g(z)=1/f(z)$ then $g\colon U \to \mathbb C$ is holomorphic.
Now $Z_0$ is a closed subset of $\Omega$, and the identity theorem applied to $f\colon U_0\to \mathbb C$ shows that $Z_0 = V_0 \cup C_0\subseteq U_0$, where $C_0$ is a closed discrete subset of $U_0$ and $V_0$ is an open subset of $U_0$. Now $f^{-1}(\mathbb C^\times) = U_1$ is an open subset of $U_0$ and $U_0$ is the disjoint union $C_0\sqcup V_0\sqcup U_1$. Similarly $Z_{\infty}$ is a closed subset of $\Omega$ and the identity theorem applied to $g\colon U_\infty\to \mathbb C$ shows that $Z_{\infty}= V_{\infty}\cup C_{\infty}$ where $C_{\infty}$ is a closed discrete subset of $U_{\infty}$, $V_{\infty}$ is an open subset of $U_{\infty}$, and $U_{\infty}$ is the disjoint union $C_{\infty} \sqcup U_1\sqcup V_{\infty}$. But then $$ \Omega\backslash (C_0\cup C_{\infty}) = V_0\sqcup U_1 \sqcup V_{\infty}, $$ a disjoint union of open sets. But as $(C_0\cup C_{\infty})$ is a closed discrete set, $\Omega\backslash(C_0\cup C_{\infty})$ remains path-connected, and so connected. Hence we must have $V_0=V_{\infty}=\emptyset$ (as $U_1=\emptyset$ would mean, as $f$ is continuous, that $f$ is constant). Hence $Z_0$ and $Z_\infty$ are both closed discrete subsets of $\Omega$.
Now $g(z_0)=0$ if and only if $f(z_0)=\infty$, and since $g$ is holomorphic and non-constant $g(z-z_0) = (z-z_0)^kg_1(z)$ where $g_1(z_0)\neq 0$ and $k \in \mathbb Z_{\geq 0}$. It follows that $f(z)= (z-z_0)^{-k}.(1/g_1(z))$ where $1/g_1(z)$ is holomorphic at $z=z_0$, so that $f$ has a pole of order $k$ at $z_0$. Thus viewing $f\colon \Omega\to \mathbb C$ $i.e.$ as a function from $\Omega$ to $\mathbb C$ with singularities at the points in $Z_{\infty}$. Since we have shown those singularities are poles and that the set $Z_{\infty}$ is discrete, it follows that $f$ is meromorphic.
Addendum
The argument above uses the fact that $f$ is continuous, which can be seen using the following claim, which essentially says that continuity is a local property:
Claim: If $f\colon X\to Y$ is a map between topological spaces $X$ and $Y$ and $\mathcal U = \{U_i:i\in I\}$ is an open cover of $X$ (so that $X = \bigcup_{i \in I} U_i$) then $f$ is continuous if and only if $f_{|U_i}$ is continuous for each $i \in I$.
Proof: If $f$ is continuous then for any open set $V \subseteq Y$ we see that $f_{|U_i}^{-1}(V) = U_i\cap f^{-1}(V)$ is an open subset of $U_i$, so that $f_{|U_i}$ is continuous for all $i\in I$. Conversely, if $f_{|U_i}\colon U_i \to Y$ is continuous for all $i\in I$ then $$ f^{-1}(V) = \bigcup_{i\in I}U_i\cap f^{-1}(V) = \bigcup_{i \in I} f_{|U_i}^{-1}(V). $$ Thus $f^{-1}(V)$ is open i $f^{-1}_{|U_i}(V)$ is open for each $i\in I$. But the continuity of $f_{|U_i}$ ensures that $f_{|U_i}^{-1}(V)$ is open in $U_i$, and since $U_i$ is open in $X$ it follows that $f^{-1}_{|U_i}(V)$ is also open in $X$ as required.
Now for $f\colon \Omega \to \mathbb C_{\infty}$ as above, we can apply the claim to the open cover $\Omega = U_0\cup U_\infty$: $f_{|U_0}$ is a holomorphic function which is therefore continuous, and similarly $f_{|U_{\infty}} = 1/g$ where $g$ is a nonvanishing holomorphic function, and hence it is also continuous. By the criterion above it follows that $f$ is continuous. (Note that the chart maps identifying $U_0$ and $U_\infty$ with open subsets of $\mathbb C$ are by definition homeomorphisms, so nothing is being covered up by suppressing them in the notation, and including them just makes the notation cumbersome I think.)