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?    
 A: 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.
Now, suppose this surface is compactly embedded in some Euclidean space. If you make a puncture in this surface, the metric will be no longer complete at that point (the geodesics just stop there without any reason). In order to save the metric's completeness, we need to move these punctures off to infinity.
To give a very crude illustration, take a circle and make a puncture in it. You get an open interval and the geodesics will hit the interval's boundary in finite time. In order to save the situation you must expand the interval so that it actually becomes the real line (thought of as Riemannian manifold with flat metric) with boundary off at infinity. Now all is good since you only hit the boundary in infinite time and the metric is complete.
For surfaces, this procedure creates spikes -- you can imagine that the manifold is a gum that deforms as you pull out a point (representing the puncture) off towards infinity.

