Exact sequence and flasque sheaves

Question

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.

Answer 1

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.

Answer 2

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 $\mathcal O_X^*(X)=\mathbb C^*$.

Answer 3

$\newcommand\N{\mathbb{N}} \newcommand\Z{\mathbb{Z}}$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.