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Let $L=L_1 \cup ... \cup L_n$ be the union of $n$ distinct lines through the origin in $\mathbb{R}^{3}$. I'd like a convincing argument that $\mathbb{R}^{3} \setminus L$ is homotopy equivalent to a wedge of $n$ circles (if that is true). In fact, I especially care about the case $n=2$.

I know this sounds like a homework problem, but I have other purposes in mind (this space naturally showed up as the fibre in a certain fibration) and I don't find typical text-book explanations of such problems very convincing, so I would appreciate a clear answer.

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Thanks. That is convincing. If you post it as an answer, I can accept it. –  A. Pascal Oct 8 '10 at 17:19

1 Answer 1

up vote 4 down vote accepted

First, deformation retract $\mathbb{R}^3$ minus $L$ to $S^2$ minus $2n$ points (you can do this since you've removed the origin). Stereographically project from one of the punctures, and you've got $\mathbb{R}^2$ minus $2n-1$ points. Choose a point away from the punctures and draw disjoint based loops around each of the remaining holes. Now deformation retract to those loops and you've got a wedge of $2n-1$ circles.

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