Equality in de Rham cohomology

Question

Let $U_1,U_2,...,U_r$ be open sets in $\mathbb{R}^n$ such that $U_i\cap U_j =\emptyset$ for all $i \neq j$. Then prove, $H^k_{dR}(\bigcup_{i=1}^{r} U_i)=\bigoplus_{i=1}^{r} H^k_{dR} (U_i)$

Answer 1

Well, if those sets are pairwais disjoint, then everything splits as a direct sum: the $k$-forms $$\Omega^k(\cup_i U_i)\simeq\oplus_i\Omega^k(U_i)$$ is given by $\omega\mapsto (\omega|_{U_i})_i$ and the De Rham differential $d$ respects this splitting. So, both the kernel and image of $d$ respect this splitting as well..

Note, however, that for an infinite number of $U_j$'s the cohomology of $\cup_j U_j$ equals the product of the cohomologies, not the direct sum.