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!


  • $\begingroup$ Thanks! I agree the details could be messy, but now at least I know I have the right idea. $\endgroup$ Commented Apr 8, 2012 at 19:41
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    $\begingroup$ 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) $\endgroup$
    – William
    Commented Apr 8, 2012 at 22:48
  • $\begingroup$ @you, and in general the fundamental group will depend on the intersection pattern (but I don't see if it will only depend on it; probably not—and this should be known, by people who work on subspace arrangements) $\endgroup$ Commented Apr 8, 2012 at 23:16
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    $\begingroup$ 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! $\endgroup$ Commented Apr 8, 2012 at 23:19
  • $\begingroup$ @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. $\endgroup$
    – William
    Commented Apr 8, 2012 at 23:25

1 Answer 1


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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    $\begingroup$ How do you show that the unit sphere with 2$n$ points removed has fundamental group $F_{2n-1}$? Thanks! $\endgroup$ Commented Apr 10, 2012 at 23:45
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    $\begingroup$ 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 $\endgroup$ Commented Apr 12, 2012 at 2:14
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    $\begingroup$ 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 $F_{2n-1}$. :) $\endgroup$ Commented Sep 1, 2013 at 12:11
  • $\begingroup$ How do I see the deformation retraction? $\endgroup$ Commented May 5, 2020 at 21:52
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    $\begingroup$ @DanLewis3264 You may be familiar with the deformation retraction of $\mathbb R^3 \setminus \{0\}$ to $S^2$ given by $H(x, t) = \frac{x}{1 + t(\| x\| - 1)}$. Now just restrict the domain to $\mathbb R^3$ minus $n$ lines through the origin. $\endgroup$
    – Alex G.
    Commented Oct 30, 2021 at 3:48

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