# Examples of surjective sheaf morphisms which are not surjective on sections

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?

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.

• 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).

• 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
• I know de Rham cohomology, but no nontrivial complex analysis, and this helped me :) – Soham Chowdhury Aug 25 '16 at 11:17

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).

• I was meaning to ask a question about $j_!$, and this answers it! Good example too. – Zhen Lin Aug 18 '11 at 14:41
• Dear @Akhil, you can't partition the irreducible space $X$ into two proper open subsets ! – Georges Elencwajg Nov 11 '16 at 9:49

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.

• 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

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.)

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.)

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.