Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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

share|improve this question
3  
$\mathbb{C}P^1$ is homeomorphic to $S^2$. So, that's just a punctured sphere. –  Joe Johnson 126 Mar 6 '12 at 18:50
2  
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

1 Answer 1

up vote 5 down vote accepted

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).

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.