# Riemann surface with punctures corresponds to a hyperbolic surface with cusps

I am reading a paper on Riemann surfaces and the author used the fact that

$\{$Riemann surfaces with genus $g$ and $n$ punctures$\}$ is in one-to-one correspondence with $\{$ hyperbolic surfaces with genus $g$ and $n$ cusps $\}$. The paper says to equip a Riemann surface with a complete hyperbolic metric, we need to move punctures to infinity. How do we do that?

Can someone help me understanding why this is true, please?

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Hey, do you remember which paper it was? – draks ... Feb 28 at 15:04
I don't remember but if I find it, I'll let you know. – yaa09d Feb 29 at 9:45

First suppose that $n=0$ on a genus $g \geq 2$ surface. Then the surface can be endowed with a hyperbolic metric (this is essentially the uniformization theorem). Note that $g \geq 2$ is necessary, e.g. because of Gauss-Bonnet theorem.