I'm reading about conformal mappings in Ahlfors' text, and I got to this section:
In the following $\Omega$ denotes a plane region of connectivity $n > 1$. The components of the complement are denoted by $E_1, E_2, \dots, E_n$, and we take $E_n$ to be the unbounded component. Without loss of generality we can and will assume that no $E_k$ reduces to a point, for it is clear that a point component is a removable singularity of any mapping function, and consequently the mappings remain the same if this isolated boundary point is added to the region.
I think this statement is false. For example, consider the region $\Omega=\{0<|z|<1 \} $ which has connectivity 2, with components $E_1=\{0\}$ and $E_2=\{|z| \geq 1\} \cup \{ \infty \}$.
The function $f(z)=\frac{1}{z}$ maps $\Omega$ conformally onto its exterior $\{|z|>1\}$, and although $E_1$ reduces to point, the singularity of $f$ there is of a pole type (which is not removable).
Am I right here? If so, what do you think the author meant to say?
Thanks!
Apparently Ahlfors wrote that with mappings between regions of the Riemann sphere in mind.
If you consider only plane regions, as you noticed, an isolated singularity can be a pole, and thus is not a removable singularity of a holomorphic function $f \colon \Omega \to \mathbb{C}$. When considering regions in the sphere, the mappings are viewed as holomorphic mappings $\Omega \to \widehat{\mathbb{C}}$, and then an isolated singularity of a conformal map is necessarily removable - viewed as a holomorphic function, if the isolated singularity is not removable in that sense, it must be a simple pole (by injectivity; a holomorphic function cannot be injective in a punctured neighbourhood of an essential singularity or multiple pole), and extending $f$ by $f(z_0) = \infty$ into the isolated singularity then produces a conformal mapping between regions in the sphere.