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Let $X$ be a topological space, and let $\mathscr{F}, \mathscr{G}$ be sheaves of sets on $X$. It is well-known that a morphism $\varphi : \mathscr{F} \to \mathscr{G}$ is epic (in the category of sheaves on $X$) if and only if the induced map of stalks $\varphi_P : \mathscr{F}_P \to \mathscr{G}_P$ is surjective for every point $P$ in $X$, but the section maps $\varphi_U : \mathscr{F}(U) \to \mathscr{G}(U)$ need not be surjective. I know of a couple of examples from complex analysis:

  1. Let $X$ be the punctured complex plane, $\mathscr{F}$ the sheaf of meromorphic functions, $\mathscr{G}$ the sheaf of differential $1$-forms, and $\varphi$ the differential map; then $\varphi$ is epic and indeed the sequence $0 \to \mathscr{F} \to \mathscr{G} \to 0$ is even exact, but there are global sections of $\mathscr{G}$ which are not the image of a global section of $\mathscr{F}$, e.g. $z \mapsto \frac{1}{z} \, \mathrm{d}z$.

  2. Let $X$ be the punctured complex plane again, $\mathscr{F}$ the sheaf of meromorphic functions, $\mathscr{G}$ the sheaf of nowhere-zero meromorphic functions, and let $\varphi : \mathscr{F} \to \mathscr{G}$ be composition with $\exp : \mathbb{C} \to \mathbb{C}$; then $\varphi$ is epic but again fails to be surjective on (global) sections: after all, there is no holomorphic function $f : X \to \mathbb{C}$ such that $\exp f(z) = z$ for all non-zero $z$.

Question. Are there simpler examples which do not require much background knowledge beyond knowing the definition of sheaves and stalks?

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up vote 12 down vote accepted

Take $X=\mathbb R$ for your topological space, the constant sheaf $\underline {\mathbb Z} $ for $\mathcal F$ and for $\mathcal G$ the direct sum of two skyscraper sheaves with fibers $\mathbb Z$ at two distinct points $P,Q\in \mathbb R$, that is $\mathcal G=\mathbb Z^P \oplus \mathbb Z^Q$.
The natural restriction $\mathcal F=\underline {\mathbb Z} \to \mathcal G=\mathbb Z^P \oplus \mathbb Z^Q$ is a surjective sheaf morphism but the associated group morphism on global sections $\mathcal F(X)=\mathbb Z \to \mathcal G (X)=\mathbb Z \oplus \mathbb Z$ is not surjective [its image is the diagonal of $\mathbb Z \oplus \mathbb Z$, consisting of pairs $(z,w)$ with $z=w$].

Edit This example can easily be adapted to a three point space space: thanks to Pierre-Yves who, in a comment to Alex's answer, suggested that.

Take $X=\{P,Q, \eta\}$ with closed sets $X,\emptyset, \{P\}, \{Q\}, \{P,Q\}$ (this is the same space as Alex's).The rest is exactly the same as above. Namely $\mathcal F=\underline {\mathbb Z} $, $\mathcal G=\mathbb Z^P \oplus \mathbb Z^Q$, $\mathcal F=\underline {\mathbb Z} \to \mathcal G=\mathbb Z^P \oplus \mathbb Z^Q$ the restriction, which is again a surjective sheaf morphism, and $\mathcal F(X)=\mathbb Z \to \mathcal G (X)=\mathbb Z \oplus \mathbb Z \:$ not surjective (the image being again the diagonal of $\mathbb Z \oplus \mathbb Z$).
The main point is that the stalks of $\mathbb Z^P$ are:
$(\mathbb Z^P)_P=\mathbb Z,(\mathbb Z^P)_Q=0, (\mathbb Z^P)\eta=0$, because $P$ is a closed point. Ditto for $\mathbb Z^Q$.

Tangential remark It might be of some interest to notice that there is a scheme structure on $X$ which makes it the smallest possible non affine scheme. This is explained in the book The Geometry of Schemes by Eisenbud and Harris, on page 22.

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Hi Georges! That's me again. Would it be right to say that a sheaf of sets (say) over your space is just given by three sets and two maps like this: $A\to B\leftarrow C$? And that a section is given by a couple $(a,c)\in A\times C$ satisfying $a\to b\leftarrow c$? [That is $a\mapsto b$ and $c\mapsto b$. (I can write $\to$, $\mapsto$, and $\leftarrow$, but I haven't been able to write an arrow of the fourth type.)] – Pierre-Yves Gaillard Aug 18 '11 at 19:26
Cher @Pierre-Yves: yes, that would fall under the heading "sheaves over a basis of open sets". The basis here would consist of the three open sets different from $X$ and $\emptyset$. – Georges Elencwajg Aug 18 '11 at 21:12

Let $M$ be a smooth manifold, and consider the sheaf of closed 1-forms. There is a surjection from the sheaf of smooth functions to the sheaf of closed 1-forms (namely, the exterior derivative, $f \mapsto df$), which is surjective in the category of sheaves (by the Poincare lemma), but which is in general not surjective (if $M$ has nontrivial first de Rham cohomology).

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A nice generalisation of example 1 of my post, but I'm quite sure this counts as requiring more background knowledge! I'm looking for simple examples. – Zhen Lin Aug 18 '11 at 14:23

The statement that section maps are not always surjective for surjective map of sheaves is equivalent to non-exactness of the functor of global sections — or equivalently, to non-triviality of sheaf cohomology.

Now it's easy to construct any number of explicit examples. Say, take $X=S^1$, $\mathcal F$ to be the sheaf of $\mathbb R$-valied functions and $\mathcal G$ to be the sheaf of $\mathbb R/\mathbb Z\cong S^1$-valued functions. Locally the map is surjective, but $\operatorname{Coker}(\Gamma(\mathcal F)\to\Gamma(\mathcal G))$ is, of course, $H^1(S^1;\mathbb Z)=\mathbb Z$.

(Well, arguably, it's just an instance of the example Akhil gives: $\mathcal G$ can be identified with $\Omega^1(S^1)$ and the map with the de Rham differential. On the other hand, take any finite model of $S^1$ to get completely finite example.)

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Ok, here is another answer. Let for $j$ an open immersion, $j_!$ be the "lower shriek" or "extension by zero" functor. Note that $j_!$ is left-adjoint to the functor $j^*$ of restriction from sheaves on $X$ to sheaves on $U$, and that there is a natural transformation $j_!j^* \to \mathrm{Id}$; note also that the stalks of $j_!$ of a sheaf are the same as the sheaf on $U$, and zero outside.

Then, if $\mathcal{F}$ is any locally constant sheaf on the irreducible space $X$, and $X = U_1 \cup U_2$ is a partition of $X$ into two proper open subsets with inclusions $j_1, j_2$, then the map $j_{1!}(j_1^*\mathcal{F}) \oplus j_{2!} (j_2^*\mathcal{F}) \to \mathcal{F}$ is a surjection (as one checks stalkwise). However, I claim that $\Gamma(X, j_{1!} (j_1^* \mathcal{F})) = 0$ and similarly for the other factor. Here's the justification: given a nonzero section of $j_! j_1^* \mathcal{F}$ is the same as giving an open cover $\{V_\alpha\}$ of the space, and sections of $\mathcal{F}$ over $V_\alpha$ for each $V_\alpha \subset U_1$ and zero for other $V_\alpha$'s (by definition of extension by zero); however, at least one of these open sets (say, $V_\beta$) must not be contained in $U_1$, and this will intersect all the other $V_\alpha$'s (even those contained in $U_1$). So since the section must be zero on $V_\beta$, it must be zero on all the other $V_\alpha$'s (by local constancy).

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I was meaning to ask a question about $j_!$, and this answers it! Good example too. – Zhen Lin Aug 18 '11 at 14:41

Let me try as elementary as is humanly possible:

$X=\{p,q_1,q_2\}$, consisting of only three elements! The open sets are $U_0=\{p\}$, $U_1=\{p,q_1\}$, $U_2=\{p,q_2\}$, and of course the empty set and $X$. Define $\mathscr{F}(U)=\mathscr{G}(U)=\mathbb{Z}$ on all non-empty open sets $U$. Now, the trick is going to be in the restriction maps and the morphism. Define all the restriction maps on $\mathscr{G}$ to be the identity, but the restrictions $\mathscr{F}(X)\rightarrow \mathscr{F}(U_i)$, $i=1,2$ are multiplication by 2. The restrictions $\mathscr{F}(U_i)\rightarrow \mathscr{F}(U_0)$ are again the identity maps (this forces the restriction from $X$ to $U_0$ to also be multiplication by 2).

Define $\phi:\mathscr{F}\rightarrow \mathscr{G}$ to be the identity on all open sets except for $X$, where it is multiplication by 2. Then $\phi(X)$ is not surjective, but it is surjective on all the stalks (check!). I hope I haven't made a mistake.

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Seems correct to me. I was hoping for a finitary example but I thought it might be too much to ask for; this fits the bill perfectly! (I wonder if someone will think of something even simpler.) – Zhen Lin Aug 18 '11 at 14:51
@Zhen: Couldn't Georges's example be "finitarized"? (By replacing $\mathbb R$ with a three point space.) – Pierre-Yves Gaillard Aug 18 '11 at 14:55
Ah, wait a second, I just realised I was careless. $\mathscr{F}$ as defined isn't a sheaf, because it doesn't have the collation property. For example, what is the global section restricting to $1$ on $U_1$ and on $U_2$? But I think this could be fixed by adding a fourth point which is not in any open set. – Zhen Lin Aug 18 '11 at 15:08
@Zhen You are right. Adding a fourth point that is not in any open (apart from $X$) won't do because the stalk at that point would be the global section, and $\phi$ isn't surjective there. That's also the reason why two points won't be enough. But I am sure that something like the above should work. Let me think about this tomorrow. – Alex B. Aug 18 '11 at 16:15
Dear @Pierre-Yves: you are right (I hadn't noticed this possibility!).Thanks a lot for your very pertinent comment: I have added the details you suggest in an edit to my answer. – Georges Elencwajg Aug 18 '11 at 16:15

Alex B. gave an example above that was supposed to be surjective on all stalks but not surjective. According to Hartshorn exercise II.1.2(b) a morphism is surjective $\Leftrightarrow$ it is surjective on all stalks. So that pursuit is in vain.

But the question was to find a morphism that is surjective but not surjective on sections. Sections are on open sets, not stalks so this doesn't contradict the statement above. It would be cool to have a very simple example for this.

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Here is a simple 'hands-on' example. Let $X=\{A,B,C\}$ be the three-point space with open sets $\{A,B,C\}$, $\{A,B\}$, $\{B,C\}$, $\{B\}$, and $\emptyset$, and let $\mathscr F$ be the sheaf of abelian groups on $X$ 'generated' by the $\mathbf Z$-valued functions $f\in\mathscr F(\{A,B\})$ and $g\in\mathscr F(\{B,C\})$, defined by

$f : (A,B)\mapsto(1,2)$

(which is shorthand for the function defined on points by $f(A)=1, f(B)=2$), and

$g : (B,C)\mapsto(0,1)$.

By 'generated,' I mean $\mathscr F$ is the sheaf associated to the presheaf obtained by all restrictions of $f$ and $g$ (in this case, the only nontrivial restriction is to $\{B\}$). In particular, using the notation $\langle f\rangle$ for the cyclic group generated by $f$, with $0$ element the $0$ function, we have that

$\mathscr F(\{A,B,C\})=\langle (A,B,C)\mapsto(0,0,1)\rangle$.

Now consider the presheaf induced by the map on functions induced by $\mathbf Z\mapsto\mathbf Z/2\mathbf Z$. That is, $f\mapsto\overline f$, where $\overline f$ is the function $\overline f : (A,B)\mapsto(1,0)$ valued in $\mathbf Z/2\mathbf Z$, and $\overline g : (B,C)\mapsto(0,1)$.

Then the presheaf image $\mathscr G$ has $\mathscr G(\{A,B,C\})=\langle(A,B,C)\mapsto(0,0,1)\rangle$, but the associated sheaf $\mathscr G^+$ has, for example, the function $(A,B,C)\mapsto(1,0,1)$ in $\mathscr G^+(\{A,B,C\})$, so considered as a morphism of sheaves $\mathscr F\rightarrow\mathscr G^+$, we are (tautologically) surjective (and, of course, surjective on stalks) but not surjective on sections.

($X$ is the same space as in Alex and Georges' answers, so perhaps this example is the same, but maybe not exactly.)

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