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?
Best Answer
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 .