# Fundamental group of complement of $n$ lines through the origin in $\mathbb{R}^3$

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!

-
Sounds right : ) – Rudy the Reindeer Apr 8 '12 at 19:33
This is basically right, though writing down all the details could be messy. – Grumpy Parsnip Apr 8 '12 at 19:39
Thanks! I agree the details could be messy, but now at least I know I have the right idea. – Edward Hughes Apr 8 '12 at 19:41
Do you mean "minus $n$ lines" like in the title, or "minus $n$ lines through the origin"? There is a significant difference: if the $n$ lines are disjoint then the fundamental group is $F_{n}$ (seen by deformation retracting onto ($\mathbb{R}^2$ minus $n$ points)), but if $n\geq 2$ and they all intersect at the same point then the fundamental group is $F_{2n-1}$ (as shown by user8268) – you Apr 8 '12 at 22:48
You say «There is no trouble repeating this process with a second distinct line». What process? WHat you explain in the case of one line cannot be done when there are two of them! – Mariano Suárez-Alvarez Apr 8 '12 at 23:19

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 $F_n$.
How do you show that the unit sphere with 2$n$ points removed has fundamental group $F_{2n-1}$? Thanks! – Edward Hughes Apr 10 '12 at 23:45
Have you heard of stereographic projection? It is a homeomorphism from the complement of a point on $S^n$ to $\mathbb{R}^n$. Relevant link: en.wikipedia.org/wiki/Stereographic_projection – John Stalfos Apr 12 '12 at 2:14
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 $F_{2n-1}$. :) – Patrick Da Silva Sep 1 '13 at 12:11