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While teaching a class on de Rham cohomology, I was trying to build an exercise showing that in a space (=smooth manifold) $M$ which is the union of $k$ contractible opens, every $k$-fold product in $H^{*>0}(M)$ is $0$.

When trying to mimic the "classical" proof of this fact, one stumbles on the fact that when $U$ is open in M, a form on $U$ does not always extend to a form on $M$ (even if $U$ is contractible).

[The strategy being to change each form by a cohomologous one which is zero on one open, so that the product is zero everywhere...]

In all the references I found, relative cohomology groups $H^*(X,U)$ are defined only when $U$ is a submanifold, closed as a subset, probably to avoid this problem.

However, there should be a way to correct this strategy to prove the result. Does anyone know how to do this?

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  • $\begingroup$ If $U \subset X$ is a compact codimension 0 submanifold (of course, with boundary), $H^*(X,U)$ is isomorphic to the cohomology with compact support of $H^*(X \setminus U)$. This might be an acceptable substitute for you. In particular, for the inclusion of an open set in the whole space, we don't run into extension problems. $\endgroup$ – user98602 Dec 4 '17 at 16:06
  • $\begingroup$ I don't quite understand your comment: in my situation, $U$ is not compact... $\endgroup$ – Oblomov Dec 5 '17 at 13:29
  • $\begingroup$ Its closure is if you're willing to assume X compact, and if you chose your open set judiciously, the closure of U should still be contractible. $\endgroup$ – user98602 Dec 5 '17 at 13:31
  • $\begingroup$ Well, in the examples I have in mind, $\overline{U}$ is no longer contractible (the affine charts of the projective spaces, namely. I can probably take my opens a little bit smaller here though). Still, the statement is true without extra assumption on $\overline{U}$ (due to comparison with singular cohomology where it is true)... $\endgroup$ – Oblomov Dec 5 '17 at 16:15
  • $\begingroup$ Yes, in all of those cases it would be harmless to choose an ever-so-slightly smaller chart. The cup-length bounds you get from this method will be the same as with contractible open sets with whatever closure, though I don’t really want to prove that... $\endgroup$ – user98602 Dec 5 '17 at 16:21

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