MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.

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

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.