What is the fundamental group of $\mathbb{CP}^1$ minus a finite set of points?

Let $\mathbb{C}\mathbb{P}^1$ be the projective space. Let $a_1, \ldots, a_n \in \mathbb{C}$. What is the fundamental group $\pi_1(\mathbb{CP}^1\backslash \{a_1, \ldots, a_n\})$?

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$\mathbb{C}P^1$ is homeomorphic to $S^2$. So, that's just a punctured sphere. –  Joe Johnson 126 Mar 6 '12 at 18:50
To add to Joe's comment: once you remove a point from $S^2$, you get something homeomorphic to the plane via the stereographic projection. –  Dylan Moreland Mar 6 '12 at 18:58

Expanding on the comments, since $\mathbb CP^1$ is homeomorphic to $S^2$, we find $\pi_1(\mathbb CP^1\setminus\{a_1\})\cong0$ since $\mathbb CP^1\setminus\{a_1\}\cong S^2\setminus\{a_1\}\cong \mathbb R^2$ which is contractible.
If we remove $n>1$ points, then this is the same as removing $n-1$ points from $\mathbb R^2$, so we find for $n>1$ that $\mathbb CP^1\setminus\{a_1,\ldots,a_n\}\simeq\bigvee_{i=1}^{n-1} S^1$, and so $\pi_1(\mathbb CP^1\setminus\{a_1,\ldots,a_n\})\cong \mathbb Z*\cdots*\mathbb Z$ (n-1 copies).