5

Let $m<n-1$ be two positive integers.

Consider $\mathbb{R}^m$ as a subspace of $\mathbb{R}^n$ via $\mathbb{R}^m\times \{(0,0,...0)\}$.

Any suggestions on how to compute $\pi_1(\mathbb{R}^n\backslash\mathbb{R}^m)$?

I have no idea how to tackle this in the general case, and for the computation of fundamental groups, Van Kampen is the only real method at my disposal so far.

  • Do you know that the fundamental group is a homotopy invariant? – Olivier Bégassat Oct 25 '11 at 17:10
  • Yes, I do. Could that help finding the solution? –  Oct 25 '11 at 17:16
  • Think of $\mathbb{R}^3$. Take away a plane, you are left with two open halfspaces which both contract to a point. So $$\mathbb{R}^3\setminus\mathbb{R}^2\simeq\mathbb{S}^0 \times\mathbb{R}+^*\times\mathbb{R}^2\approx\mathbb{S}^0.$$ Take away a line and you are left with $$\mathbb{R}^3\setminus\mathbb{R}^1\simeq\mathbb{S}^1\times \mathbb{R}+^\times\mathbb{R}\approx \mathbb{S}^1.$$ Take away a point and you are left with $$\mathbb{R}^3\setminus\mathbb{R}^0\simeq\mathbb{S}^2\times\mathbb{R}_+^\approx \mathbb{S}^2$$ Here $\simeq$ stands for isomorphism and $\approx$ for homotopy equivalence. – Olivier Bégassat Oct 25 '11 at 17:56
  • This pattern generalizes... And all you need to know are the fundamental groups of the spheres : $\pi_1(\mathbb{S}^1)\simeq\mathbb{Z}$ and all other fundamental groups are teh trivial groups. – Olivier Bégassat Oct 25 '11 at 17:56
  • I pictured your second example before asking the question and deduced that you'd end up with $\mathbb{Z}$ as the fundamental group, however, I lack the geometric intuition to generalize this to higher dimensions. Anyways, I'll pursue this approach further now that I know that it is supposed to work. –  Oct 25 '11 at 17:59
  • I think I got it figured out, the 'factorization' into a sphere, the positive reals and a linear space makes perfect sense. Now I just have to put it down formally. Thanks! –  Oct 25 '11 at 18:13
  • 2
    Glad to be of assistance! If you complete a proof, you should post it as an answer to your question! – Olivier Bégassat Oct 25 '11 at 18:32

0 Answers0