In Ahlfors' complex analysis text, page 264, he writes:
When $\Omega$ is the whole plane $F(\zeta)$ has isolated singularities at $\zeta = 0$ and $\zeta=\infty $.
Reading the previous sections, I assume that $\Omega \subset \mathbb C$ is a region invariant under the translations $z \mapsto z+ \omega,z \mapsto z- \omega$, where $\omega \in \mathbb C$ is some constant. $F(\zeta)=f(z)$ where $\zeta=\exp(2 \pi i z/ \omega)$, and $f$ is some meromorphic function in $\Omega$. Also we denote that image of $\Omega$ under $\zeta(z)$ by $\Omega'$.
Here is (what appears to me as) a counterexample to Ahlfors' statement:
Construct a function $f(z)$ meromorphic in $\Omega=\mathbb C$ with periods $\omega, i \omega$, with $\omega \in \mathbb R$, whose poles form the lattice $\{m \omega+i n \omega:m,n \in \mathbb Z \}$. The sequence of poles $\{i \omega n\}_1^\infty$ in the $z$-plane corresponds to the sequence of poles $\{e^{-2 \pi n} \}$ in the $\zeta$-plane, which tends to the point $\zeta=0$.
This shows that $\zeta=0$ is not an isolated singularity of $F(\zeta)$, (the situation at $\infty$ is identical).
Is my reasoning correct, or is Ahlfors right after all? If I'm wrong, please show me why.
Thanks!
You are right: $0$ and $\infty$ need not be isolated. As Ahlfors himself says in the last sentence of 1.1, every meromorphic map on $F:\mathbb C\setminus\{0\}$ arises from some $f$, namely from $f(z)=F(\exp(2\pi i z/\omega))$. A meromorphic function on $\mathbb C\setminus\{0\}$ can easily have poles accumulating to either $0$ or $\infty$, or both.
As a fix, omit "isolated" from the quoted sentence, and perhaps soften it to "may have singularities" because the statement "has singularities" reads as a definite indication of some singular behavior.