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Just a quick question to verify whether I'm right.

Claim: The fundamental group of the complement of $n$ lines through the origin in $\mathbb{R}^3$ is $F_n$, the free group on $n$ generators.

Proof: remove a line from $\mathbb{R}^3$. We may deformation retract the remaining space onto a cylinder radius $\epsilon$ about the line, and thence to a circle $S^1$. There is no trouble repeating this process with a second distinct line, except that then we will be a wedge union $S^1 \vee S^1$. Continue inductively, and recall that the wedge union of $n$ circles has the stated fundamental group.

I'm only just starting to really get my head around this stuff, so any feedback would be really useful!

Thanks!

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    @MarianoSuárez-Alvarez, I just mean to say that the title of the thread seems to be misleading, as it doesn't match the question in the main post, and I think there needs to be some clarification as to which question should be answered here.2012-04-08

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There is a deformation retraction of ($\mathbb{R}^3$ minus $n$ lines through the origin) to (the unit sphere with $2n$ points removed). The $2n$ points are the intersections of the lines with the sphere, the deformation retraction is along the rays from the origin.

As a result, the fundamental group is actually $F_{2n-1}$, not $F_n$.

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    I was asking this question myself too ; I guess the right way to do it is to use one of the $2n$ points as a "north pole" for the stereographic projection, which leaves us with $\mathbb R^2$ with $2n-1$ points removed. Using van Kampen's theorem, we get $F_{2n-1}$. :)2013-09-01