TITLE: De Rham-Weil theorem
QUESTION [6 upvotes]: I am having trouble understanding a couple of points with regard to the De Rham-Weil theorem and was hoping that someone might be able to shed some light.
Let $X$ be a smooth manifold and $\mathcal{F}$ the sheaf of abelian groups on $X$. The theorem states that $H^q(X,\mathcal{F}) \cong H^q((A^{\bullet}(X),d_x))$, where $(A^{\bullet}(X),d_x)$ is the complex of global sections for an acyclic resolution $0 \rightarrow \mathcal{F} \rightarrow A^{\bullet}$. But, if I understand the definition correctly (I suppose I don't), an acyclic resolution has trivial higher cohomology groups. Then the theorem would imply that for for any sheaf the higher cohomology groups are trivial?
Further, I have a proposition in my notes saying that any soft resolution of $\mathcal{F}$ is acyclic. But, for example, the Canonical resolution is soft, but higher cohomology groups are zero only if $\mathcal{F}$ is soft itself, correct?
At the risk of sounding dumb, what is wrong with this picture?
Thanks.

REPLY [2 votes]: 1) Even if the complex of sheaves  $(A^{\bullet}(X),d^\bullet)$ is exact in some degree, this does not imply that the corresponding complex of global sections is exact in the same degree.
For example on the punctured plane $\mathbb C^*$ the complex of sheaves  $\mathcal O \stackrel {d}{\to}  \mathcal O\to 0$ (where $df=f'$) is exact but the corresponding complex of global sections $\mathcal O (\mathbb C^*)\stackrel {d}{\to} \mathcal O(\mathbb C^*)\to 0$ is not exact since no holomorphic function on $\mathbb C^*$ has derivative $\frac 1z$: the function $\log$ cannot be defined globally on $\mathbb C^*$.  
2) This explains that a sheaf that has a soft resolution can very well have non-vanishing cohomology in any degree.
 However your assertion that a sheaf can have vanishing cohomology only if it is soft is wrong.
For example the sheaf $\mathcal O$ of holomorphic functions on $\mathbb C$ is definitely not soft: you cannot extend the function $\frac {1}{z-2}$ defined on the closed unit disc $\mid z \mid \leq 1$ to $\mathbb C$.
But nevertheless $H^i(\mathbb C,\mathcal O)=0$ for all $i\geq 1$ .