I'm trying to prove that de Rham cohomology computes the cohomology of the constant sheaf $\mathbb{R}$ of real valued smooth functions on a manifold.

I would like to do this without using Čech cohomology, if possible.

Fix a smooth manifold $M$. Let $\mathbb{R}$ denote the contant sheaf, $\Omega^k$ the sheaf of $k$-forms on $M$. The Poincaré lemma tells us, that if $U$ (open) is contractible then the sequence $C^{\bullet}(U)$:

$\mathbb{R}(U) \to \Omega^0(U) \to \Omega^1(U) \to \dots$

is exact.

If $\mathcal{C}$ is the site of open sets of $M$, for each $x \in M$ define $\mathcal{C}_x$ to be the subcategory of contractible open neighbourhoods of $x$. The stalk $F_x$ of a sheaf $F$ on $\mathcal{C}$ is the colimit of the composition $\mathcal{C}_x^\mathrm{op} \xrightarrow{\subseteq} \mathcal{C}^\mathrm{op} \xrightarrow{F} \mathcal{Ab}$ as $M$ locally contractible.

Now consider the diagram $D_i: \mathcal{C}_x^\mathrm{op} \xrightarrow{\subseteq} \mathcal{C}^\mathrm{op} \xrightarrow{C^i} \mathcal{Ab}$. The sequence

$D_{-1} \to D_{0} \to D_{1} \to \dots$

of presheaves is exact as its sections are $C^\bullet(U)$.

As $\mathcal{C}_x^\mathrm{op}$ is filtered, the assignment $D \to \mathrm{colim} D$ is exact, so we obtain an exact sequence on stalks

$\mathbb{R}_x \to \Omega^0_x \to \Omega^1_x \to \dots$

for all $x \in M$. This is equivalent to the sequence of sheaves

$\mathbb{R} \to \Omega^0 \to \Omega^1 \to \dots$

being exact.

My questions are:

  1. Is this argument correct?
  2. Do you know a reference that gives a similar proof? http://www3.nd.edu/~lnicolae/sheaves_coh.pdf
  3. Is it possible to give a similarly elementary proof without using stalks?
  • $\begingroup$ Your argument is correct. However, it only shows that you have an exact sequence – still a long way from calculating sheaf cohomology. $\endgroup$ – Zhen Lin Oct 5 '14 at 17:48
  • $\begingroup$ Why a long way? Surely I show $\Omega^k$ acyclic and I'm done? $\endgroup$ – user40167 Oct 5 '14 at 18:12
  • 1
    $\begingroup$ Yes. You don't mean to say it's trivial, do you? $\endgroup$ – Zhen Lin Oct 5 '14 at 20:06

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