Let X be a topological space, let $\mathcal{U} = \{U_i\}$ be a cover of X, and let $\mathcal{F}$ be a sheaf of abelian groups on X. If X is separated, each $U_i$ is affine, and $\mathcal{F}$ is quasi-coherent, then Cech cohomology computes derived functor cohomology; in general one only gets a spectral sequence $$ H^p(\mathcal{U},\underline{H}^q(\mathcal{F})) \Rightarrow H^{p+q}(X,\mathcal{F}) $$ where $\underline{H}^q(\mathcal{F})$ is the presheaf $U \mapsto H^q(U,\mathcal{F}|_U)$.

Question: For q > 0, $\underline{H}^q(\mathcal{F})$ sheafifies to 0.

For a quasi-coherent sheaf $\mathcal{F}$ this is clear because cohomology vanishes on affines. Is this really true in general? Brian Conrad states this in the introduction to his notes on cohomological descent.


Yes, this is true in general.

It suffices to show the stalks vanish. Pick $x \in X$ and take an injective resolution $0 \to {\cal F} \to I^0 \to \cdots$. For any open $U$ containing $x$, we get a chain complex

$$0 \to I^0(U) \to I^1(U) \to \cdots$$

whose cohomology groups are $H^p(U,{\cal F}|_U)$.

Taking direct limits of these sections gives the chain complex

$$0 \to I^0_x \to I^1_x \to \cdots$$

of stalks, which has zero cohomology in positive degrees because the original complex was a resolution. However, direct limits are exact and so we find

$$0 = {\rm colim}_{x \in U} H^p(U,{\cal F}|_U) = {\underline H}^p({\cal F})_x$$

as desired.

Generally, cohomology tells you the obstructions to patching local solutions into global solutions, and this says that locally those obstructions vanish.


Here's another short proof: denote by $I,J$ the inclusion of sheaves on X into presheaves and sheafification respectively, then

$$ \underline{H}^p(\mathcal{F})\cong R^pI(\mathcal{F}). $$

Since $J$ is exact,

$$ J\circ R^pI\cong R^p(J\circ I) $$

and the later vanishes for $p>0$ as $J\circ I=id_{\mathfrak{Ab}(X)}$. So $\underline{H}^p(\mathcal{F})$ sheafifies to 0 for $p>0$.

  • $\begingroup$ This is an interesting approach, but it relies a lot on understanding derived functors. The longer proof reveals more, so to say. $\endgroup$ – Bombyx mori Apr 3 '17 at 19:04
  • 2
    $\begingroup$ @Bombyxmori Depends on point of view. For me this proof reveals the conceptual reason behind the fact, which I could not really see in the longer one. $\endgroup$ – მამუკა ჯიბლაძე Apr 3 '17 at 20:14
  • $\begingroup$ @მამუკაჯიბლაძე: I agree. Upvoted. $\endgroup$ – Bombyx mori Oct 31 '17 at 22:36

I do not think that sheaves of abelian groups need to be locally acyclic. Let me say what I mean in an example. Take $X=\mathbb{C}^{2}$ with the classical (metric) topology. Let $\mathcal{F} = \mathbb{Z}_{D}$ where $D=\mathbb{C}^{\times} \times \{0\}$.

Then for any arbitrarily small polydisk $U$ containing $(0,0)$ we have $H^{1}(U,\mathbb{Z}_{D}) = H^{1}(U \cap D, \mathbb{Z})$

is not vanishing. It seems in this example that $\underline{H}^{1}(\mathcal{F})_{(0,0)}$ does not vanish.

  • $\begingroup$ Oh, I could have just used $\mathbb{C}^{\times}$ in $\mathbb{C}$ $\endgroup$ – Oren Ben-Bassat Mar 28 '11 at 10:11
  • $\begingroup$ Just to clarify, the assumptions of locally affine and quasi coherence are crucial. The simple example of the push forward of the constant sheaf $\mathbb{Z}$ under the inclusion of $\mathbb{C}^{\times}$ into $\mathbb{C}$ gives a sheaf of abelian groups which is not locally acyclic. $\endgroup$ – Oren Ben-Bassat Nov 1 '11 at 17:33
  • $\begingroup$ Why should we have $H^{1}(U,\mathbb{Z}_{D}) = H^{1}(U \cap D, \mathbb{Z})$ ? If $X=\mathbb{C}$ and $D=\mathbb{C}^\times$, then $\mathbb{Z}_{D}$ fits into an exact sequence of sheaves on $\mathbb{C}$: $0\rightarrow \mathbb{Z}_{D}\rightarrow \mathbb{Z}\rightarrow \mathbb{Z}_{0}\rightarrow 0$, where the latter two are the constant sheaf resp. the skyscraper sheaf at $0$ in $\mathbb{C}$. The sequence remains exact, when computing global sections on any open disk $U$ containing $0$, and since $H^1(U,\mathbb{Z})=0$, the long exact sequence implies $H^1(U,\mathbb{Z}_D)=0$. Or not? $\endgroup$ – user_1789 Dec 30 '16 at 13:26

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