Let $m 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.
Fundamental group of the complement of a linear subspace
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algebraic-topology
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0Do you know that the fundamental group is a homotopy invariant? – 2011-10-25
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0Yes, I do. Could that help finding the solution? – 2011-10-25
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0Think 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. – 2011-10-25
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0This 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. – 2011-10-25
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0I 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. – 2011-10-25
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0I 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! – 2011-10-25
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2Glad to be of assistance! If you complete a proof, you should post it as an answer to your question! – 2011-10-25