How can I compute the de Rham cohomology of $\mathbb{R}^3$ minus n lines through the origin? I would like to do it with the Mayer-Vietoris sequence (which is the only thing I know to calculate cohomology besides homotopy invariance). For the case n=1 I have homotopy invariance with $\mathbb{R}^2\setminus{0}$ and this is omotopic to the circle so i can compute the cohomology. Is this homotopy still true in the general case?
De Rham cohomology of the euclidean space without n lines
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algebraic-geometry
differential-geometry
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0Consider the map F(t,x)=(1-t)x+tx/|x|. This is a deformation retract to the sphere with points removed, and hence a homotopy equivalence. – 2012-05-30