# Exact sequence and flasque sheaves

What is an easy example of an exact sequence of sheaves (of modules or abelian groups if you prefer) $$0\rightarrow \mathcal F \rightarrow \mathcal G \rightarrow \mathcal H \rightarrow 0$$ such that $$\mathcal G$$ is flasque (also called flabby) but $$\mathcal H$$ is not flasque?

In Harthshorne II.1 exercises if $$\mathcal F$$ and $$\mathcal G$$ are flasque then $$\mathcal H$$ would necessarily be so. And one uses $$\mathcal F$$ is flasque to show that on the level of sections we have exactness as well. I was wondering to what extent this would fail (and this is best explained by an example) if we don't have this.

• If $\mathcal{G}$ is a constant sheaf and $\mathcal{F}$ is the zero section of $\mathcal{G}$, then is it easy example? The definition of flasque is a sheaf $\mathcal{S}$ on a toplogical space $X$ such that every inclusion $V \subset U$ of open sets, the restriction map $\mathcal{S}(U) \to \mathcal{S}(V)$ is surjective. The constant sheaf satisfies this condition. Definition is Hartshorne page 67 exercise 1.15 in section II. Commented Nov 12, 2019 at 9:04
• The constant sheaf is flasque and the sequence $0 \to \mathbb{Z} \to \mathbb{C} \to \mathbb{C}^* \to 0$ is an example. Where we use $\mathbb{C} \to \mathbb{C}^*;z \mapsto exp(2\pi i z)$ and $\mathbb{C}^*$ is the set of real numbers except the zero. Commented Nov 12, 2019 at 9:15
• I think you have to be a bit careful about the constant sheaf in your first comment. The flasqueness would depend on the topology of $X$. I can imagine a case where $X$ is say the one-point compactification of $\mathbb N$ (with discrete topology) and flasqueness does not hold. Commented Nov 12, 2019 at 9:23
• In the problem of Hartshorne, the subproblem (a) says that if the base space $X$ is an irreducible topological space, then any constant sheaf is flasque, but I cannot understand if $X$ is not irreducible, then why generally speaking a constant sheaf is not flasque. Commented Nov 12, 2019 at 9:35
• @JeanBillie let $\Delta$ be a constant sheaf of integers on $[0,1]$. Then $\Gamma\left([0,\frac{1}{10})\cup (\frac{9}{10},1],\Delta\right) = \mathbb{Z}\oplus \mathbb{Z}$ and $\Gamma\left([0,1],\Delta\right) = \mathbb{Z}$. This implies that $\Delta$ is not flabby.
– Slup
Commented Nov 12, 2019 at 10:03

## 3 Answers

Any sheaf $$\mathcal{F}$$ admits an injection $$\mathcal{F\to G}$$ into a flasque sheaf $$\mathcal{G}$$. Take $$\mathcal{H}$$ the cokernel. By the long exact sequence in cohomology, we have isomorphisms $$H^i(X,\mathcal{H})\simeq H^{i+1}(X,\mathcal{F})$$ for any $$i>0$$. Take $$\mathcal{F}$$ a sheaf with a non vanishing $$H^2$$ (for example), then $$\mathcal{H}$$ has a non vanishing $$H^1$$ and thus is not flasque.

• For example, in your notation, if $\mathcal{F}$ is a structure sheaf on a complex manifold $X$, then how to get a such injection $\mathcal{F} \to \mathcal{G}$ (and $0 \to \mathcal{F} \to \mathcal{G} \to cockernel \to 0$). Or the sheaf of holomorphic differential forms on a complex manifold? Commented Nov 12, 2019 at 10:32
• @JeanBillie Pick canonical morphism $\mathcal{F}\rightarrow \prod_{x\in X}\mathrm{sky}_x\left(\mathcal{F}_x\right)$, where $\mathrm{sky}_p(A)$ is the skyscraper corresponding to point $p$ and set $A$.
– Slup
Commented Nov 12, 2019 at 10:36
• This $\mathcal{G}$ as described by Slup is part of the Godement resolution. Essentially, this is the sheaf of discontinuous sections. Commented Nov 12, 2019 at 10:38
• Very nice, Roland: +1 Commented Nov 12, 2019 at 10:38

As a complement, let me give a simple example where we have an exact sequence of sheaves of abelian groups $$0\rightarrow \mathcal F \rightarrow \mathcal G \rightarrow \mathcal H \rightarrow 0$$ with $$\mathcal G , \mathcal H$$ both flasque but with $$\mathcal F$$ not flasque.
Take for $$X=\mathbb A^1(\mathbb C)$$ the complex affine line seen as an algebraic variety endowed with its Zariski topology.
We have an exact sequence of sheaves of abelian groups $$0\rightarrow \mathcal O_X^* \rightarrow \mathcal K^*_X \rightarrow \mathcal Div_X\rightarrow 0$$ where $$\mathcal K^*_X$$ is the sheaf of non-zero rational functions and $$\mathcal Div_X$$ the sheaf of divisors on $$X$$.
Concretely, for an open subset $$U\subset X$$ the group $$\mathcal Div_X(U)=\oplus_{P\in U} \mathbb Z \cdot P$$ is the free group on the points in $$U$$.
The map $$\mathcal K^*_X(U)\rightarrow \mathcal Div_X(U)$$ associates to the rational function $$\phi$$ the difference $$Z(\phi)-P(\phi)$$ of its zero set and its pole set, both counted with multiplicities.
The sheaf $$\mathcal K^*_X$$ is well known to be flasque ( more or less by definition) while $$\mathcal Div_X$$ is trivially flasque because any finite sum $$\Sigma_{P\in U} n_P \cdot P \in \mathcal Div_X(U)$$ can be extended to the sum $$\Sigma_{P\in X} n_P \cdot P \in \mathcal Div_X(X)$$ with $$n_P=0$$ for $$P\in X\setminus U$$.
However the sheaf $$\mathcal O_X^*$$ is not flasque, as witnessed by the section $$z \in \mathcal O_X^*(X\setminus \{O\})$$, not extendable to $$\mathcal O_X^*(X)=\mathbb C^*$$.

• Flasque is also the name given in French, yes? Modern German ("welk" used mostly. I have also seen "wacklig" in newer lit.) and English ("flabby" used also) don't seem to agree on one single word.
– Jose
Commented Nov 13, 2019 at 6:42
• +1 The example is very clear and with good detail! It seems that this exact sequence appears very often in algebraic and arithmetic geometry. Commented Nov 13, 2019 at 6:43
• This is a little off-topic, but: Do you know a simple reason why the sheaf $K_X^*$ in the complex category, even though not flasque anymore, is still acyclic (am I right about that?) for a Riemann surface $X$? That is to say, replace nonzero rational functions by nonzero meromorphic functions. I need not be able to extend meromorphic functions off smaller open sets to larger open sets because of essential singularities. Commented Dec 7, 2019 at 14:59
• I suppose I can argue by applying the sheaf long exact sequence on the exact sequence you described to get $0 \to K^*_X(X) \to Div_X(X) \to H^1(X, O^*_X) \to H^1(X, K^*_X) \to H^1(X, Div_X)$ and then observe that since $H^1(X, O^*_X) = \text{Pic} X = Div_X(X)/K^*(X)$, the penultimate map is zero, and further $Div_X$ is clearly flasque, so $H^1(X, K^*_X) = 0$. But this is already so much harder than the algebraic category. Maybe there's some GAGA argument (treading on murky waters here, I know nothing of this stuff) to show $K^*_X$ is still acyclic? Commented Dec 7, 2019 at 15:20
• @Balarka Sen: You certainly cannot use GAGA, since the affine line is not projective. The sheaf $\mathcal K^*_X$ is constant, hence flasque, hence acyclic. But this is irrelevant for my post, since acyclic does not imply flasque. Commented Dec 7, 2019 at 20:50

$$\newcommand\N{\mathbb{N}} \newcommand\Z{\mathbb{Z}}$$This is NOT an answer to my question. It is just to clarify with Jean Billie that you cannot assume the constant sheaf to be flasque if your topology is arbitrary. I would suggest Jean Billie to post this as a question in another thread. Meanwhile I write it here because it is too long for a comment:

Here is a case where you don't have flasque: Let $$\beta\N$$ be the (Alexandroff) one-point compactification of $$\N$$ (with discrete topology). Take your group to be $$(\Z,+)$$. $$\N$$ is an open and discrete subset of $$\beta\N$$. Since $$\N$$ is discrete, from the sheafification of the constant presheaf $$\Z$$ over $$\beta\N$$ you can define a section in $$s\in \underline{\Z}(\N)$$ by $$s:\N\rightarrow \bigcup_{x\in \N} {\Z_x} \qquad s(i) = i$$ where the $$\Z_x$$ is just a copy of $$\Z$$ (it is the stalk at $$x\in \N$$). There is no global section that restricts to $$s$$ because otherwise this section restricted to a small enough neighbourhood of $$\infty$$ (that extra point that compactifies $$\N$$) coincides with a section of the constant presheaf. But we know that all open sets containing $$\infty$$ must be cofinite. So, such a global section must be constant over all but finite number of elements in $$\N$$ but $$s(i)$$ is different for every $$i$$. So you have a contradiction.

For an irreducible topology you cannot do this. So you need some kind of reducibility or discreteness to make such an example.

• You are right. That is too complex. Please ignore. The easiest discrete toplogy would do indeed. By the way, would you like to post your comment as an answer? Commented Nov 12, 2019 at 10:26
• @Roland I don't agree that a constant sheaf on $\{0,1\}$ (or $\mathbb{N}$ with discrete topology) is not flabby. I even think that every sheaf on a discrete space is flabby.
– Slup
Commented Nov 12, 2019 at 10:34
• @Slup Yes you are right, I don't know what I was thinking, I remove my comment. Commented Nov 12, 2019 at 10:36
• I won't remove my comment. But I think Slup is right. In fact, I also believe if it is discrete then the constant presheaf is also trivially the constant sheaf. Commented Nov 12, 2019 at 11:06
• The constant sheaf $\underline {\mathbb Z}$ on the circle $S^1$ is not acyclic since $H^1(S^1,\underline {\mathbb Z})=\mathbb Z$. So I think that your example is a bit of an overkill... Commented Dec 7, 2019 at 20:56