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Let $f$ be a meromorphic function with $\text{dom} (f)=\mathcal{M}$, where $\mathcal{M}$ is a non-compact Riemann surface.

If $\mathcal{M}'= \mathcal{M} \cup \{\infty \}$ is the one-point compactification of $\mathcal{M}$, then is $f$ with domain $\mathcal{M}'$ still meromorphic? Does this depend on the compactification method?

Many thanks

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1 Answer 1

up vote 1 down vote accepted

No.

You can see it with the case $M = \mathbb{C}$, $f=\sin$ (here $f$ is even holomorphic).

sinus is bounded in the direction corresponding to the real line, but tends to infinity along the direction of pure imaginary.

So on the riemann sphere $S^2 = \mathbb{C} \cup \{ \infty \}$, $\sin$ admits no continuous continuation and hence no meromorphic continuation.

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