# De Rham cohomology of the euclidean space without n lines

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?

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I think this has been asked here before... Maybe you can find it answered and all? –  Mariano Suárez-Alvarez May 30 '12 at 21:55
In any case: you can deform the complement of a finite set of lines through the origin to the complement if a finite set of points in the sphere. The latter space is diffeomorphic to the complement of a finite set of points in the plane, and you can compute its cohomology using Mayer-Vietoris and an induction. –  Mariano Suárez-Alvarez May 30 '12 at 21:57
Instead of thinking about lines you can think about cylinders meeting around the origin. For the case n=3 you get something homotopy equivalent to a cube with a disc removed at each side. –  M.B. May 30 '12 at 22:04
Also, it should be clear how you can go on from here to general $n$. (On a second note this should be algebraic-topology, not differential-geometry.) –  M.B. May 30 '12 at 22:07
@MarianoSuárez-Alvarez: i searched for the question but i didn't find it. Anyway i already computed the cohomology of the complement of a finite sets of points in the plane. How can I demonstrate the homotopy equivalence? And why isn't it homotopic to the complement of only a point (i imagine to retract all the lines at the origin..) –  balestrav May 30 '12 at 22:15
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