I am misunderstanding something in Theorem 2.1.9 in Dimca’s Sheaves in Topology:
Let $X$ be a real smooth manifold. Then the natural morphism from the constant sheaf to the de Rham complex
$$\mathbb{R}_X \rightarrow \Omega_X^\bullet$$
is an acyclic resolution of $\mathbb{R}_X$.
I am getting confused about in which sense $\Omega_X^\bullet$ is acyclic: if it is acyclic as a complex, then it has no cohomology above $H^0$ by definition. But, of course a smooth real manifold can still have cohomology above $H^0$!
So I would like to confirm that acyclic here means a resolution using $\Gamma$-acyclic sheaves, i.e., something like injective or flasque sheaves.
Edit: Maybe we want hypercohomology of both sides, since Poincar\’e’s lemma says that the de Rham complex $\Omega_X^\bullet$ really is acyclic as a complex. But then the statement
$$ H^k(X,\mathbb{R}) \cong \frac{\text{ker } d: \Omega^k(X) \rightarrow \Omega^{k+1}(X)}{\text{Im }d: \Omega^{k-1}(X) \rightarrow \Omega^k(X)} $$
makes no sense to me, because the right hand side here should be zero for $k>0$ if $\Omega^\bullet(X)$ is acyclic as a complex
Best Answer
The following statements are true:
However, it is not true that either of these statements has anything to do with the exactness of the sequence of global sections $\Omega^0(X) \to \Omega^1(X) \to \dots$, which very frequently has interesting cohomology in all possible degrees (up to $\dim X$). (But why would you expect it to be this? There is much more information in a sheaf than just its global sections.)