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?