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Let $X$ be a topological space and $\mathcal F$ is a sheaf of abelian groups on it. Let $Y$ be a closed subspace of $X$. Let $\mathscr{H}^0_Y \mathcal F$ be the subsheaf of $\mathcal F$ with supports in $Y$. Let $H^i _Y (X,\mathcal F)$ denote the $i$-th cohomology groups with supports in $Y$. Then is it true that

$$H^i (X, \mathscr{H}^0_Y \mathcal F) = H^i _Y (X,\mathcal F) $$ ?

I guess that the above is true. Can anyone please give a proof?

Sandip
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    Can you explain the notation? Thank you. – Armando j18eos May 06 '16 at 20:41
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    I am concerned that the subsheaf with supports functor is only left exact,so there are additional terms in the Grothendieck spectral sequence for composition. For example where (I think) exactness fails, let X be the plane and Y the origin, we are over a field. Then consider the sheaf F to be the pushward of the structure sheaf of the residue field at the origin, and G the pushforward of the structure sheaf of the local ring at the origin. Then G surjects onto F, by sending a germ to its value at the origin, but the subsheaf of G supported at the origin is 0, while for F it is all of k. – Elle Najt May 06 '16 at 21:27
  • @AreaMan : It is only left-exact, but it is exact on flasque sheaves (using a bunch of results in Hartshorne). Therefore the Godement resolution allows a proof. See my answer. – Patrick Da Silva May 07 '16 at 20:37
  • @PatrickDaSilva Thanks, that is really cool and good to know. – Elle Najt May 08 '16 at 05:29
  • @Patrick: Unfortunately the proof is wrong. The way the flasque resolution is created, the kernel, cokernels are not flasque. So even though it is an exact functor on short exact sequences, you can't go beyond that and say that it is exact on flasque sheaves. So you do not get a resolution by applying $\mathscr {H}^0 _{Y}( . )$. – Shubhodip Mondal May 12 '16 at 05:45
  • Indeed, think of something more basic. For any sheaf there is a flasque resolution. Now the global section is left exact, and by results in Hartshorne global section is "exact" on flasque sheaves. So by your argument, by applying global section functor you should still get an exact sequence. Does that mean the higher cohomology groups are zero for any sheaf? ! – Shubhodip Mondal May 12 '16 at 05:48
  • ^ Now if the given sheaf happens to be flasque, then the kernel, cokernels remain flasque. So it is coherent with the fact that flasque's have higher cohomology zero. – Shubhodip Mondal May 12 '16 at 06:00
  • Well, apply your logic in the "more basic" case. – Shubhodip Mondal May 12 '16 at 08:21

2 Answers2

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Construct a counterexample as follows.

Let $X = \text{Spec} A$. (Assume $A$ to be Noetherian.) Choose an ideal $\mathfrak a$ such that $ U= X \setminus V(\mathfrak a)$ is NOT an affine scheme.(Later we show this is feasible.) Therefore, by Serre's theorem, there exists a coherent sheaf $\mathscr{F'}$ such that $H^1(U, \mathscr{F'}) \ne 0$. By extension of Coherent sheaves [Hartshorne II Ex. 5.15] property, we get a coherent sheaf $\mathscr{F}$ on $X$ such that $\mathscr{F}|_U = \mathscr{F'}$.

Now take $Y = V(\mathfrak a)$. Then $\mathscr{H}^0_{Y} (\mathscr{F})$ is also a coherent sheaf [Hartshorne II Ex. 5.6 (e)]. Therefore, $ H^i (X, \mathscr{F}) = H^i (X, \mathscr{H}^0_{Y} (\mathscr{F})) = 0$ for $i > 0$ by Serre's theorem.

Now by parts of the long exact sequence in [H, III, Ex. 2.3 (e)], we have the following exact sequence:

$0 =H^1(X, \mathscr{F}) \to H^1(U, \mathscr{F'}) \to H^2_{Y}(X, \mathscr{F})$.

So, $H^1(U, \mathscr{F}')$ injects into $H^2_{Y}(X, \mathscr{F})$. Since $H^1(U, \mathscr{F}') \ne 0$,we must have $ H^2_{Y}(X, \mathscr{F}) \ne 0$.

But we have $H^2(X, \mathscr{H}^0_{Y} (\mathscr{F})) = 0$. Contradicting the desired result.

Now this situation can be easily achieved by taking say $A = k[x,y]$ and $\mathfrak a = (x,y)$. See this for a proof of why $U$ is not affine $\mathbb{A}^{2}$ not isomorphic to affine space minus the origin .

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Shubhodip Mondal
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EDIT : Indeed, a very subtle detail managed to slip into the cracks : the argument in which a functor is exact on short exact sequences is exact on all sequences works only in abelian categories (e.g. not in subcategories of abelian categories), and the category of flasque sheaves is not abelian (it doesn't have kernels).

What always remains true though is the Grothendieck spectral sequence $$ H^i(X, \mathscr H^j_Y \mathcal F) \Rightarrow H^{i+j}_Y(X, \mathcal F). $$

(Wrong proof here for the record) Consider the sheaf of discontinuous sections of $\mathcal F$ and denote it by $\mathcal G$. We show that $\mathscr H^0_Y \mathcal G$ is flasque. Given open subsets $U \subseteq V \subseteq X$ and $s \in \mathscr H^0_Y \mathcal G(U) = \Gamma_{Y \cap U}(U,\mathcal G|_U)$, we have a map $s : U \to \bigcup_{p \in U} \mathcal F_p$ with $s(p) \in \mathcal F_p$ and $s$ is supported on $Y$. We extend this to a section $s_V : V \to \bigcup_{p \in V} \mathcal F_p$ by extending by zero on $V \backslash U$.

It follows that $s_V|_U = s$ and $(s_V)_p \neq 0$ implies $p \in Y$ by the following argument. Let $W = X \backslash Y$, which is open. Since $s|_{U \cap W} = 0$ (because $s|_{U \cap W}$ has zero stalks on $U \cap W$ since $s$ has support on $U \cap Y$, thus $s|_{U \cap W}$ is zero), $s_V|_{V \cap W} = 0$ (because as a function, $s_V|_{(V \backslash U) \cap W} =0$, so together with $s|_{U \cap W} = 0$, $s_V|_{V \cap W} = 0$). So for any $p \in V \cap W$, $(s_V)_p = (s_V|_{V \cap W})_p = 0$. This shows that $(s_V)_p \neq 0$ implies $p \in V \cap Y$. We deduce that $\mathscr H^0_Y \mathcal G$ is flasque.

Therefore if $0 \to \mathcal F \to \mathcal G^{\bullet}$ is a Godement resolution for $\mathcal F$ (i.e. repeatedly taking the sheaf of discontinuous sections of the cokernel of the previous map, defining the resolution inductively), $0 \to \mathscr H^0_Y \mathcal F \to \mathscr H^0_Y \mathcal G^{\bullet}$ is a flasque resolution for $\mathscr H^0_Y \mathcal F$ (the functor $\mathscr H^0_Y(-)$ is exact on flasque short exact sequences by Hartshorne, Chapter III, 2.3(b) and it is obviously left-exact). Taking global sections with supports on the first, we obtain the complex $$ 0 \to \Gamma_Y(X,\mathcal F) \to \Gamma_Y(X,\mathcal G^{\bullet}) $$ and taking global sections on the second, we obtain $$ 0 \to \Gamma(X,\mathscr H^0_Y \mathcal F) \to \Gamma(X,\mathscr H^0_Y \mathcal G^{\bullet}). $$ The cohomology of the first complex is $H^i_Y(X,\mathcal F)$ and the cohomology of the second complex is $H^i(X,\mathscr H^0_Y \mathcal F)$. But the two complexes are equal.

Hope that helps,

Patrick Da Silva
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