Are these two 2-manifolds homeomorphic? I have a 2-Sphere with a finite number $k$ of points removed (at least 3), and I want to equip it with a Riemannian metric of constant negative curvature. 
My first thought was to take a free subgroup $A_k$ of rank $k$ of the isometries of the hyperbolic plane $H$, which acts fix point free and discontinuous on $H$ (such a group exists).
If I take the quotient $H/A_k$, I end up with a complete manifold of constant negative curvature, with isomorphic fundamental group to the punctured Sphere, but are those two manifolds homeomorphic? Maybe this is a terribly clumsy way to try to accomplish this task.
 A: That's not necessarily the case.
Take the sphere with 3 punctures: it has fundamental group free on two generators, right?
Now take the torus with one puncture (which has indeed a hyperbolic structure). You can easily see this has the same fundamental group.
So, to answer your question you probably need to know which group you are using (simply because we have just seen that it depends on that).
To endow the thrice punctured sphere with a hyperbolic metric, you should start with two identical copies of an ideal triangle in $\mathbb{H}^2$, and glue them together. If you are careful you can check that this gives you a hyperbolic surface homeomorphic to $S^2-\{p_1,p_2,p_3\}$.
Do you see how you could generalize this to, say, a 4-punctured sphere?
For another approach, non explicit:
A thrice punctured sphere $S$ can be embedded (continuously) in $\mathbb{C}$, as $\mathbb{C}$ with two holes. So, $S$ has a riemann surface structure. By the riemann mapping theorem, it is holomorphically covered by $\mathbb{H}$, the upper half plane (since we removed at least 2 points). This allows you to put a hyperbolic metric on $S$. For details and references, check the wikipedia article, http://en.wikipedia.org/wiki/Uniformization_theorem#Geometric_classification_of_surfaces
A: I think we are saying the same thing but in a slightly different order. Isn't it true that a sphere with at least 3 points removed is homeomorphic to an open set in a compact Riemann surface of genus $\geq2$?
If so, the latter can be endowed with a constant negative curvature because it is uniformized by the hyperbolic plane.
