I'm studying de Rham cohomology, but there is something not quite clear. Consider a manifold $M$ and $U_i \subset M$ for $i\in I$ a collection of open sets.
I can't figure out why when computing the cohomology ring of a disjoint union I get the direct product of the cohomology rings, whereas with cohomology with compact support I get the direct direct sum of the cohomology rings?
What I mean is this: For the "standard" cohomology, this holds: \begin{equation} H^*_{dR}\big(\coprod_{i \in I} U_i\big)=\prod_{i\in I} H^*_{dR}(U_i) \end{equation}
for cohomology with compact support this holds: \begin{equation} H^*_c\big(\coprod_{i \in I} U_i\big)=\bigoplus_{i\in I} H^*_c(U_i) \end{equation}
So my question is: why in one case I get the direct product whereas in the other I get the direct sum? What is the difference between the two? Be as thorough as possible in answering :)
thank you
This is easier than it looks. For vector spaces, the direct sum is just the subspace in the direct product consisting of all those "vectors" for which only finitely many components are non-zero. (More formally the product consists of all functions from $I$ to the disjoint union of the space $H^*(U_i)$ and the sum is the subspace of those functions which have only finitely many non-zero values.) For standard cohomology, the isomorphism is induced by sending a form $\varphi$ to the function that associates to each $i\in I$ the restriction $\varphi|_{U_i}$. For compactly supported cohomology, you use the same map, but observe that any compact subset of $M$ can only intersect finitely many $U_i$ non-trivially. Conversely, given a forms $\varphi_i$ on each $U_i$ you can just view them as a form on the disjoint union of the $U_i$ and the resulting form has compact support if each of the $\varphi_i$ has compact support and only finitely many of the $\varphi_i$ are non-zero.