I am trying to solve the exercise 2.4 chapter III in Hartshorne's "Algebraic Geometry". For this I would like to prove for a sheaf $F$ of Abelian groups on a topological space $X$ and $U$ open subset of $X$ with complement $Z$ the following $$ H^{i}_{Z}(X,j_{.}(F\mid_{U})) = 0$$ where $j:U \longrightarrow X$ the inclusion and $j_{.}$ meaning the extension of a sheaf on U to a sheaf on X by zero. I have proven exercise 2.3 chapter III, so I can use this.

Does anyone know a nice proof for this? Or maybe a counterexample if this doesn't hold? I was able to reduce/adapt this case to proving $H^{i}(X,j_{.}(F\mid_{U})) \cong H^{i}(U,F\mid_{U})$ but I also haven't a rigorous proof for this. Also for it is quite clear because $H^{0}_{Z}(X,j_{.}(F\mid_{U})) =0$, it is just for the higher order groups I have't an argument. (Or maybe this all doesn't hold?).

Thanks in advance! (Not sure if this is supposed to be on mathstack, but there I have not yet received an answer on this).


1 Answer 1


This is not true.

Example. Let $X$ be an irreducible topological space with $|X| \geq 2$ that has a closed point $x \in X$. For example, we may take $X = \operatorname{Spec} R$ for $R$ a domain that is not a field, and $x = \mathfrak m$ for $\mathfrak m \subseteq R$ a maximal ideal. Let $Z = \{x\}$, and let $U = X \setminus Z$.

Then the constant sheaf $\mathbb Z_X$ is flabby on $X$ [Hart, Exc. II.1.16(a)]. We have a short exact sequence [Hart, Exc. II.1.19(c)] $$0 \to j_!\mathbb Z_U \to \mathbb Z_X \to i_*\mathbb Z_Z \to 0,$$ since $\mathbb Z_X|_U = \mathbb Z_U$ and $\mathbb Z_X|_Z = \mathbb Z_Z$ are the constant sheaves on $U$ and $Z$ respectively. Taking sections with support on $Z$ gives a long exact sequence $$0 \to \Gamma_Z(X,j_!\mathbb Z_U) \to \Gamma_Z(X,\mathbb Z_X) \to \Gamma_Z(X,i_*\mathbb Z_Z) \to H^1_Z(X,j_!\mathbb Z_U) \to \ldots.$$ But $\Gamma_Z(X,\mathbb Z_X) = 0$, since all nonzero sections of $\mathbb Z_X$ are supported everywhere. On the other hand, $\Gamma_Z(X,i_*\mathbb Z_Z) = \mathbb Z$, since all sections of $i_*\mathbb Z_Z$ are supported on $Z$. Thus, the map $$\Gamma_Z(X,\mathbb Z_X) \to \Gamma_Z(X,i_*\mathbb Z_Z)$$ is not surjective, so $H^1_Z(X,j_!\mathbb Z_U)$ cannot be zero.

Remark. Your statement that $H^i(X,j_!(\mathscr F|_U)) = H^i(U,\mathscr F|_U)$ is also false. This already fails for $H^0$, because $\Gamma(X,j_!(\mathscr F|_U)) = 0$ if $U \subsetneq X$, by the definition of extension by $0$.


[Hart] Hartshorne, Robin, Algebraic geometry, Graduate Texts in Mathematics 52. Springer-Verlag, New York-Heidelberg-Berlin (1977). ZBL0367.14001.

  • $\begingroup$ Thank for your answer! I have received the same counterexample on mathstack, but a good one nonetheless. I do have a remark: I don't think $\Gamma(X,j_{!}(F\mid_{U})) = 0$ for $U \neq X$ by definition. It is true for the presheaf it is associated to, but j_{!}(F\mid_{U}) is the sheafification of this... $\endgroup$
    – Lilolance
    Apr 23, 2018 at 19:09
  • $\begingroup$ @Lilolance: ah, you're right. But I think if $X$ is irreducible and $\mathscr F$ is constant, it should still be true. In this case, the fact that every nonempty open is connected implies that the presheaf $j_{!,\text{pre}}(\mathscr F|_U)$ is already a sheaf. $\endgroup$ Apr 24, 2018 at 3:56
  • $\begingroup$ Is this true if $H^{i}(U, \mathcal{F}_{U}) = 0$, then $H^{i}(X, j_{!}(\mathcal{F}_{U})) =0$?, where $\mathcal{F}_{U}$ is the restriction sheaf to $U$. $\endgroup$
    – Sunny
    Dec 22, 2020 at 12:59

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.