I'm trying to do Hartshorne's exercises on local cohomology at the moment and seem to be stuck in Exercise III 2.5. The problem goes as follows:

$X$ is supposed to be a Zariski space (i.e a Noetherian and sober topological space) and $P\in X$ a closed point. Let $X_P$ denote the subset of $X$ consisting of all points specializing to $P$, and give it the induced topology. Let $j:X_P\rightarrow X$ be the inclusion map and write $\mathcal F_P=j^{-1}\mathcal F$ for any sheaf (of abelian groups) $\mathcal F$ on $X$. Then for all $i\in\mathbb Z$ and sheaves $\mathcal F$ we have \begin{equation} H_P^i(X,\mathcal F) = H^i_P(X_P,\mathcal F_P) \end{equation}

I succeeded in showing that the identity holds for $i=0$, so to prove the claim it would suffice to know that the $H^i_P(X_P,(~\cdot~)_P)$ define a universal $\delta$-functor. I tried to do this with the standard procedure by showing effaceability of these functors for $i>0$. So first I tried to show that pulling back along $j$ preserves flsaqueness, but meanwhile I'm quite convinced that this isn't true. I do not see why injectives would go to injectives, either. Then I thought maybe for special sheaves like sheaves of discontinuous sections flasqueness is preserved, but, as before, it didn't lead anywhere. Does anyone have some good advice?


  • $\begingroup$ Did you ever confirm whether the claim is false that pulling back along $j$ preserves flasqueness? I came up with a proof that seems to work, but of which I'm a little suspicious. Thanks for any comment. $\endgroup$ – Mr. Chip Aug 25 '16 at 14:32
  • $\begingroup$ $j$ does in fact preserve flasqueness. A proof is given in the second half of Takumi's answer. $\endgroup$ – Andreas Aug 29 '16 at 12:11

By the excision isomorphism (Exercise III.2.3(f)), we have already $$ H^i_P(X,\mathscr{F}) \cong H^i_P(V,\mathscr{F}\rvert_V) \cong \varinjlim_{V \ni P} H^i_P(V,\mathscr{F}\rvert_V), $$ where the direct limit is over the restriction maps induced by the inclusions $P \in W \subset V$. So it remains to show $\varinjlim_{V \ni P} H^i_P(V,\mathscr{F}\rvert_V) \cong H^i_P(X_P,\mathscr{F}_P)$. Note that for any inclusion $P \in W \subset V$, we have the morphism of long exact sequences from Exercise III.2.3(e) $$\require{AMScd}\begin{CD} \cdots @>>> H^{i-1}(V,\mathscr{F}\rvert_V) @>>> H^{i-1}(V^P,\mathscr{F}\rvert_{V^P}) @>>> H^{i}_P(V,\mathscr{F}\rvert_V) @>>> H^{i}(V,\mathscr{F}\rvert_V) @>>> H^{i}(V^P,\mathscr{F}\rvert_{V^P}) @>>> \cdots\\ @. @VVV @VVV @VVV @VVV @VVV\\ \cdots @>>> H^{i-1}(W,\mathscr{F}\rvert_W) @>>> H^{i-1}(W^P,\mathscr{F}\rvert_{W^P}) @>>> H^{i}_P(W,\mathscr{F}\rvert_W) @>>> H^{i}(W,\mathscr{F}\rvert_W) @>>> H^{i}(W^P,\mathscr{F}\rvert_{W^P}) @>>> \cdots \end{CD} $$ where $V^P$ denotes the punctured open set $V \setminus \{P\}$, and similarly for $W$. Note that every open set $V$ containing $P$ contains $X_P$ as well by the fact that it is closed under generization by Exercise II.3.17(e), and so the direct limit taken over all $V \ni P$ is the same as the direct limit taken over all $V \supset X_P$. Using the universal property of the direct limit, and the fact that direct limits are exact in the category of abelian groups [Atiyah–Macdonald, Exc. 1.19], we therefore get the corresponding morphism of long exact sequences $$\begin{CD} \cdots @>>> \varinjlim_{V \ni P}H^{i-1}(V,\mathscr{F}\rvert_V) @>>> \varinjlim_{V \ni P}H^{i-1}(V^P,\mathscr{F}\rvert_{V^P}) @>>> \varinjlim_{V \ni P}H^{i}_P(V,\mathscr{F}\rvert_V) @>>> \varinjlim_{V \ni P}H^{i}(V,\mathscr{F}\rvert_V) @>>> \varinjlim_{V \ni P}H^{i}(V^P,\mathscr{F}\rvert_{V^P}) @>>> \cdots\\ @. @VVV @VVV @VVV @VVV @VVV\\ \cdots @>>> H^{i-1}(X_P,\mathscr{F}_P) @>>> H^{i-1}(X_P^P,\mathscr{F}_P\rvert_{X_P^P}) @>>> H^{i}_P(X_P,\mathscr{F}_P) @>>> H^{i}(X_P,\mathscr{F}_P) @>>> H^{i}(X_P^P,\mathscr{F}_P\rvert_{X_P^P}) @>>> \cdots \end{CD} $$ Now, we see that if the vertical arrows above are isomorphisms except for the middle arrow, then by the five lemma [Weibel, Exc. 1.33], the middle one would be as well. Thus, it suffices to show that $$\tag{*}\label{eq:III.2.5} \varinjlim_{V \ni P} H^i(V,\mathscr{F}\rvert_V) \cong H^i(X_P,\mathscr{F}_P), \quad \varinjlim_{V \ni P} H^i(V^P,\mathscr{F}\rvert_{V^P}) \cong H^i(X_P^P,\mathscr{F}_P\rvert_{X_P^P}). $$ We first show the isomorphisms for $i = 0$ (using the Lemma below): $$ \varinjlim_{V \ni P} \Gamma(V,\mathscr{F}\rvert_V) = \varinjlim_{V \ni P} \Gamma(V,\mathscr{F}) = \varinjlim_{V \supset X_P} \Gamma(V,\mathscr{F}) = \Gamma(X_P,\mathscr{F}_P)\\ \varinjlim_{V \ni P} \Gamma(V^P,\mathscr{F}\rvert_{V^P}) = \varinjlim_{V \ni P} \Gamma(V^P,\mathscr{F}) = \varinjlim_{V \supset X_P} \Gamma(V^P,\mathscr{F}) = \varinjlim_{V^P \supset X_P^P} \Gamma(V^P,\mathscr{F}) = \Gamma(X_P^P,\mathscr{F}_P\rvert_{X_P^P}) $$ and so it suffices to show them for $i \ge 1$.

As in the proof of Prop. III.2.9, since the isomorphisms hold for $i = 0$, to show the functors agree for $i \ge 1$, it suffices to show they are both effaceable as functors $\mathfrak{Ab}(X) \to \mathfrak{Ab}$, since in that case they are both universal, hence must be isomorphic by Thm. III.1.3A. So let $\mathscr{G}$ be the sheaf of discontinuous functions of $\mathscr{F}$ from Exercise II.1.16(e). Then, $\mathscr{G}$ is flasque and there is a natural inclusion $\mathscr{F} \hookrightarrow \mathscr{G}$. Now the sheaves $\mathscr{G}\rvert_V$, $\mathscr{G}\rvert_{V^P}$, $\mathscr{G}\rvert_{X_P}$, and $\mathscr{G}\rvert_{X_P^P}$ are flasque by the Lemmas below. Thus, every functor in \eqref{eq:III.2.5} is zero when applied to $\mathscr{G}$ by Prop. III.2.5, hence the functors are effaceable so we are done. $\blacksquare$

EDIT. Andreas pointed out a gap in the proof, and provided the following Lemma to fix it:

Lemma. If $\mathscr{F}$ is a sheaf on a Zariski space $X$, and $Y \subset X$ is a subset closed under generization, then $$\varinjlim_{\tilde{U} \supset U} \Gamma(\tilde{U},\mathscr{F}) \cong \Gamma(U,\mathscr{F}\rvert_Y)$$ where $\tilde{U}$ are open in $X$, and so there is no need to sheafify in the definition of restriction.

Proof. Because open subsets $U$ of $Y$ are closed under generization in $X$, it suffices to show the claim for $U = Y$. Consider the natural map $$\varinjlim_{Y\subseteq U}\Gamma(U,\mathscr{F}) \to \Gamma(Y,\mathscr{F}\rvert_Y)$$ obtained from the definition of sheafification.

Injectivity. This follows from the fact that a separated presheaf injects into its sheafification [Stacks, Tag 0082].

Surjectivity. We want to show every global section of $\mathscr{F}\rvert_Y$ extends to some open neighborhood of $Y$. So let $s \in \Gamma(Y,\mathscr{F}\rvert_Y)$. Since $Y$ is noetherian, there is a maximal open subset $U \subset Y$ such that $s\rvert_U$ lifts to a section $t\in\Gamma(\tilde{U},\mathscr{F})$ with $\tilde{U} \cap Y=U$. Note it is nonempty since choosing an arbitrary point $z \in Y$, the germ of $s_z$ lives in $\mathscr{F}_z = (\mathscr{F}\rvert_V)_z$, and $s_z$ extends to some open subset of $X$.

Suppose $U \ne Y$, and let $y \in Y \setminus U$. Then, by the argument above using stalks, there exists an open neighorhood $V$ of $y$ such that $s\rvert_V$ lifts to a section $t'\in\Gamma(\tilde{V},\mathscr{F})$ with $\tilde{V} \cap Y=V$.

Now because $t,t'$ are both local lifts of $s$, there exists an open subset $W$ in $X$ such that $U \cap V \subset W \subset \tilde{U} \cap \tilde{V}$ and $t\rvert_W = t'\rvert_W$ (since the support of $t-t'$ is closed in $\tilde{U} \cap \tilde{V}$). Now let $A = \tilde{U} \cap \tilde{V} \setminus W$, and let $A=\bigcup_{i=1}^k A_i$ be an irreducible decomposition. We want to show $\overline{A} \cap Y = \emptyset$, where $\overline{A}$ denotes the closure of $A$ in $X$. For suppose $\overline{A} \cap Y$ is non-empty; then, since $Y$ is closed under generization, we can assume the generic point $x_i$ of some $A_i$ also lies in $\overline{A} \cap Y$. But this is a contradiction, since then, $x_i \in \tilde{U} \cap \tilde{V} \cap Y = U \cap V \subset W$ implies that $x_i\notin A_i$.

Now the two sections $t\rvert_{\tilde{U} \setminus \overline{A}}$ and $t'\rvert_{\tilde{V}\setminus\overline{A}}$ match on the overlap of their domains, which contains $U \cap V$, and hence patch together to give a lift of $s\rvert_{U\cup V}$, contradicting the maximality of $U$. $\blacksquare$

This is used in the following way:

Lemma. If $\mathscr{F}$ is a flasque sheaf on a Zariski space $X$, and $Y \subset X$ is a subset closed under generization, then $\mathscr{F}\rvert_Y$ is flasque.

Proof. This follows since direct limits are exact, and using the previous Lemma. $\blacksquare$

  • $\begingroup$ I realized after a few days that you can salvage the other answer by showing the functors $H^i_P(X_P,\cdot_P)$ and $\varinjlim_{V \ni P} H^i_P(V,\cdot\rvert_V)$ are effaceable as I did, by using the sheaf of discontinuous sections, and using that flasque sheaves are acyclic for cohomology with supports by Exercise III.2.3(c). $\endgroup$ – Takumi Murayama Jan 3 '16 at 2:04
  • $\begingroup$ Thanks for your answer. I had totally given up on this. What I am still missing is why $\mathcal G|_{X_P}$ is flasque if $\mathcal G$ a sheaf of discontinuous functions. This is NOT similar to showing that $\mathcal G|_V$ is flasque for an open subset $V$, because that is just trivial. $\endgroup$ – Andreas Mar 5 '16 at 14:55
  • $\begingroup$ @Andreas This was my argument: if you think of the sheaf $\mathscr{G}$ of discontinuous functions on $\mathscr{F}$ as a sheaf in the espace étalé sense, then if you restrict to $X_p$ (which, as you point out, need not be open), the sheaf $\mathscr{G}\rvert_{X_P}$ is the same sheaf as if you took $\mathscr{F}\rvert_{X_P}$ and took its sheaf of discontinuous functions. So $\mathscr{G}\rvert_{X_P}$ is flasque. I think what's confusing is that I don't know how to make this argument unless I use the espace étalé description for sheaves. $\endgroup$ – Takumi Murayama Mar 5 '16 at 15:03
  • $\begingroup$ But if I take $X=\mathbb A^1_{\mathbb C}$, $P=0$, and $\mathcal F=\mathbb Z$, then the restriction of $\mathcal F$ to $X_0$ is $\mathbb Z$ again, and hence its sheaf of discontinuous functions has stalk $\mathbb Z^2$ at $0$, whereas the stalk at $0$ of the sheaf of discontinuous functions of $\mathcal F$ is much larger. $\endgroup$ – Andreas Mar 5 '16 at 15:28
  • $\begingroup$ I was a bit confused by your example since I didn't know which topology you were using (it's the scheme $\operatorname{Spec} k[x]$, right?). In any case, your objection seems to show that $\mathscr{G}\rvert_{X_P}$ is much bigger than the sheaf of discontinuous functions of $\mathscr{F}\rvert_{X_P}$, which I agree with. But this sheaf $\mathscr{G}\rvert_{X_P}$ is still flasque: a section $s\colon U \to \mathscr{G}\rvert_{X_P}$ can be extended to a section $\tilde{s}\colon \tilde{U} \to \mathscr{G}$ where $\tilde{U} \cap X_P = U$ by having $\tilde{s}(x) = 0$ for all $x \in \tilde{U} \setminus U$ $\endgroup$ – Takumi Murayama Mar 5 '16 at 16:09

Here is how I would prove this:

Let $X$ be a Zariski space. Let $ p \in X$ be a closed point and let $X_{p} \subset X$ be the set of all points $ q \in X$ such that $p \in (q)^{-}$. We observe that $X_{p}$ has the induced topology. Now let $j: X_{p}\hookrightarrow X$ be the inclusion map. For any sheaf $\mathcal F$, let $\mathcal F_{p}=j^{*}\mathcal F$. We then state our claim as: $\Gamma_{p}(X, \mathcal F)=\Gamma_{p}(X, \mathcal F_{p})$. Any open set $U$ containing $p$ also contains $X_{p}$ so gluing sheaves will not affect $\Gamma(X_{p}, \mathcal F_{p})= \lim\underset {\rightarrow p \in U} \Gamma (U, \mathcal F)$. Taking $\pi \in \Gamma_{p} (X, \mathcal F)$, we get a section $\pi^{\prime} \in \Gamma_{p}(X_{p}, \mathcal F_{p})$. We may represent $\pi^{\prime}$ by $\pi \in \Gamma(U, \mathcal F)$. Shrink $U$ such that we may assume $\mathrm{Supp}(\pi)=p$. Glue $\pi$ and $0\in \Gamma (X/p, \mathcal F)$ to obtain a global section. Then there exists a bijection $\Gamma_{p}(X, \mathcal F_{p})\leftrightarrow \Gamma_{p}(X_{p}, \mathcal F_{p})$.

Finally, if $0 \rightarrow \mathcal F \rightarrow I_{0} \rightarrow I_{1} \rightarrow \cdots$ is a flasque resolution of $\mathcal F$, then $0 \rightarrow \mathcal F_{p} \rightarrow I_{0,p} \rightarrow I_{1,p} \rightarrow \cdots$ is an injective resolution of $\mathcal F_{p}$. By repeating this argument we obtain $\Gamma_{p}(X, I_{i}) \cong \Gamma_{p}(X_{p}, I_{i,p})$ and we have that $H^i_p(X, \mathcal F)=H^{i}_{p}(X_{p}, \mathcal F_{p})$.

I hope this helps!

  • 2
    $\begingroup$ Thanks for your answer. This is basically how I tried to do it as well. The part I do not understand is why $I_i$ being flasque implies that $(I_i)_P$ is injective. $\endgroup$ – Andreas Jan 18 '14 at 10:22

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