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What is the fundamental group of the following space in $\mathbf C^n$?

This is the topological space given by $$\{(x_1,\ldots,x_n)\in \mathbf C^n-\{0\} : \vert x_1\vert < 1, \ldots, \vert x_n\vert <1\} - \{x_1\cdot \ldots \cdot x_n =0\}.$$

For $n=1$, I know it's $\mathbf Z$. I'm guessing it should be $\mathbf Z^n$ in general. Is this true? And why?

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Are you sure it's $\mathbb{Z}$ for $n=1$? The way you defined the space it's contractible, so the fundamental group should vanish. –  Piotr Pstrągowski Mar 17 '13 at 14:08
    
Whoops. I was thinking about the punctured polydisc! My apologies. –  Tom Mar 17 '13 at 14:09
    
Can you describe the unit disc (not punctured!) up to homeomorphism? This is a reasonable first step. –  Piotr Pstrągowski Mar 17 '13 at 14:11
    
Ow I think I see what you're getting at. When $n=1$, this is going to be like $\mathbf C$ minus the origin topologically, right? When $n>1$, this is probably a simply connected space, no? It might even be contractable because of all the extra space you get to move around. –  Tom Mar 17 '13 at 14:15
    
@PiotrPstragowski I changed the question again. This should be a less trivial question. –  Tom Mar 17 '13 at 14:23

1 Answer 1

up vote 1 down vote accepted

Your space deformation retracts to the $n$-torus $\prod_i\{\vert x_i\vert=\frac {1}{2} \}$, so has the same fundamental group : $$\mathbb Z^n $$

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