# Exact sequence of sheaves with non exact sequence of global sections

Let $X$ be some topological space. By $\mathcal{F}_i$ we denote some sheaves of abelian groups on $X$. The sequence of sheaves and morphisms $$\mathcal{F}_1\longrightarrow \mathcal{F}_2\longrightarrow \mathcal{F}_3\longrightarrow...$$ is said to be exact if for each $x\in X$ the corresponding sequence of stalks $$(\mathcal{F}_1)_x\longrightarrow (\mathcal{F}_2)_x\longrightarrow (\mathcal{F}_3)_x\longrightarrow...$$ is exact. However if the sequence of sheaves is exact than the sequence of global sections is not necessarily exact! (The most famous example is the sequence of sheaves $$0\longrightarrow\mathbb{Z}\hookrightarrow \mathcal{O}\stackrel{\exp}{\longrightarrow} \mathcal{O}^*\longrightarrow0$$ considered as sheaves on $\mathbb{C}-\{0\}$, where $\mathcal{O}$ is a sheaf of holomorphic functions, $\mathcal{O}^*$ is a sheaf of holomorphic functions with no zeros).

So, could you give me some easy examples of such phenomenon ?

• What's wrong with the exponential sequence as an example? Nov 22 '13 at 22:01
• Take a non-zero sheaf $F$ with $F(X)=0$, take a closed point $x\in X$ and consider $G$ the skyscraper sheaf $F_x$ supported at $\{ x\}$. Then we have a canonical surjective map $F\to G$ which is not surjective when passing to sections on $X$. Nov 22 '13 at 22:02
• @Cantlog: Why not making this an answer? Nov 23 '13 at 0:19

If $p$ is a point on a compact Riemann surface of genus one, we have the exact sequence of sheaves $0 \to \mathcal O \to \mathcal O(p)\to \mathbb C_p \to 0$ (the last non zero sheaf being a sky-scraper sheaf) .
The sequence of global sections is $$0 \to \mathbb C = \mathbb C \stackrel {0} {\to}\mathbb C \to 0$$ and is thus not exact.

• Dear Georges: your sequence does not appear to be exact at the last $\mathbb{C}$. Worse, you have suddenly turned blue! Is this by choice?
– user64687
Nov 23 '13 at 0:31
• Dear @Asal: no, it is not my choice at all ! I suppressed an invalid e-mail address in the about me window and some #*~§☢☹ software turned me as blue as a smurf . Do you know if I can revert to my former avatar ? Nov 23 '13 at 0:44
• Dear @Asal: as for the (less important ?!) mathematics, the fact that my displayed sequence is not exact at the last $\mathbb c$ is precisely the point of my answer. Do we have a misunderstanding? Nov 23 '13 at 0:56
• Dear Georges: unfortunately I have no idea, but surely an email to a moderator (or a flag on this answer?) will lead to the answer. It's funny, but in my mind the little red thingy is so strongly linked to your excellent answers and courteous comments, that seeing this blue one in its place is a severe psychic jolt...
– user64687
Nov 23 '13 at 0:56
• About your sequence: sorry, I didn't actually read what you wrote; of course you are right. By now I am so conditioned to expect to see the long exact sequence of cohomology that no new information can override it. Let my excuse for sloppiness be the shock caused by your strange new appearance!
– user64687
Nov 23 '13 at 0:59

Let $M$ be a smooth manifold and consider the de Rham sheaf sequence on $M$:

$$0\to\mathbf{R} \to \mathcal{O}_M \to \Omega^1_M\to \Omega^2_M\to \dots$$

where the first map is inclusion and the other maps are exterior differentiation. It is exact as a sequence of sheaves by the Poincaré lemma, but on global sections it is the usual de Rham complex, whose cohomology is the de Rham cohomology. Of course unless $M$ is 1-dimensional, is not a short exact sequence, but nevertheless, the phenomenon is the same.

• You can always split long exact seq into many short ones. Nov 23 '13 at 8:10
• @Martin Indeed, that is what I meant :) Nov 23 '13 at 12:32

Let $X=\mathbb{A}^1_k$ where $k$ is an infinite field. Take $Y=\{p,q\}$ where $p,q$ are two distinct closed points of $X$ and let $U=X-Y.$ Take $\mathbb{Z}_Y=j_{\star}(\mathbb{Z}|_Y)$ where $j: Y \hookrightarrow X$ is the inclusion and $\mathbb{Z}_U=i_{!}(\mathbb{Z}|U)$ for the inclusion $i:U \hookrightarrow Z.$

Show that the following is an exact sequence

$$0 \longrightarrow \mathbb{Z}_U\longrightarrow \mathbb{Z}\longrightarrow \mathbb{Z}_Y\longrightarrow 0$$

and $H^1(X,\mathbb{Z}_U) \neq 0.$

I'm sure you can find tones of more non-trivial examples in literature.

• Dear @Ehsan, I don't think your second sentence is correct. Nov 23 '13 at 0:06
• I agree with Georges. It's the other way round: If $H^1(X,F_1)=0$, then $H^0(X,F_2) \to H^0(X,F_3)$ is surjective. More precisely, $H^0(X,F_2) \to H^0(X,F_3)$ is surjective if and only if $H^1(X,F_1) \to H^1(X,F_2)$ is the trivial map. I assume here that $0 \to F_1 \to F_2 \to F_3 \to 0$ is exact. Nov 23 '13 at 0:17
• Dear Georges and Martin, you're absolutely right, my bad! edited. Nov 23 '13 at 0:49