# Fundamental group of the complement of a linear subspace

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.

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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? – user18233 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
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