# Category of sheaves on a topological space is not $\mathbf{AB4^{*}}$

I want to show that the category of sheaves on a topological space ($$\mathbf {Sh}(X)$$) is not $$\mathbf{AB4^{*}}$$. Recall that $$\mathbf{AB4^{*}}$$ means that the product of epimorphisms is an epimorphism.

What I have tried:

Let $$X = \mathbb{C}$$ and consider the sheaf of holomorphic functions: that is, for every open set $$U$$, we have the abelian group $$\mathcal{O}(U) := \left\{f\colon U \subseteq \mathbb{C} \longrightarrow \mathbb{C} : f \text{ is holomorphic in } U\right\}$$ Now consider the sheaf morphism that, given in components, is the complex differentiation map: $$\begin{array}{rcl} d(U) \colon \mathcal{O}(U)& \longrightarrow &\mathcal{O}(U)\\ f & \longmapsto & d(U)(f) := \dfrac{df}{dz} \end{array}$$ I've seen that $$d$$ is an epimorphism of sheaves (not of presheaves: the function $$f(z) = 1/z$$ in $$\mathbb{C}\setminus{\left\{0\right\}}$$ is a classical counterexample), but I suspect that the product morphism is not epic. Any ideas? I accept other counterexamples, my goal is to show that $$\mathbf{AB4^{*}}$$ doesn't hold. $$\mathbf{EDIT:}$$

Given $$\left\{f_{i} \colon A_{i} \longrightarrow B_{i}\right\}_{i \in \mathcal{I}}$$, the product morphism is the unique morphism $$\varphi \colon \prod_{i \in \mathcal{I}} A_{i} \longrightarrow \prod_{i \in \mathcal{I}} B_{i}$$ so in my example, by product morphism I mean $$d = f_{i}$$ for $$i \in \mathcal{I}$$, where $$\mathcal{I}$$ is an infinite set.

• What do you mean "the product morphism" ? Commented Jun 23, 2019 at 18:08
• Post edited!!!! Commented Jun 23, 2019 at 18:22
• I was asking specifically about your example (although that wasn't clear): do you think it doesn't hold for any product ? Because in an abelian category, a finite product of epimorphisms is just as well a finite coproduct, and in any category a coproduct of epimorphisms is epic. So you're thinking of an infinite product of $d$'s ? Commented Jun 23, 2019 at 18:28
• Sorry, of course I was thinking of an arbitrary product. Commented Jun 23, 2019 at 18:32

I think that your example does not work. Indeed, we have the following lemma.

Lemma (Uniform surjectivity implies product surjectivity)

Let $$\varphi_i : \mathcal{F}_i \to \mathcal{G}_i$$ be a collection of sheaf morphisms, with base space $$X$$. Suppose that there is a basis $$\mathcal{B}$$ of $$X$$ such that $$\varphi_i(U):\mathcal{F}_i(U) \to \mathcal{G}_i(U)$$ is surjective for all $$U \in \mathcal{B}$$. Then $$\prod_i \mathcal{F}_i \to \prod_i \mathcal{G}_i$$ is sheaf surjective.

Proof. Sheaf surjectivity is equivalent to surjectivity on stalks. Take $$[(g_i,U)]$$ be an element of the stalk $$(\prod \mathcal{G}_i)_x$$. Take $$x \in V \subset U$$ such that $$V \in \mathcal{B}$$. By hypothesis, there exist $$[(f_i, V)]$$ that are sent to $$[(g_i |V, V)]$$, and we are done.

Corollary. Your example does not work. Indeed, if I well remember from complex analysis, your sheaf function is surjective on opens that are disks, and these constitute a basis.

Having in mind that we must force some disuniformity in $$i$$ with respect to the opens where we have a lifting, we build the following example.

1. For $$U \subset X$$, define $$\mathbb{Z}_U$$ as the sheafification of the presheaf $$Z_U$$ such that $$Z_U(V) = \mathbb{Z}$$ if $$V \subset U$$, and 0 otherwise. Restriction functions are idenitites when possible and zeros otherwise. To be honest, it is possible that $$Z_U$$ is already a sheaf, but we don't care since we are only interested in stalks and these are preserved by sheafification.

2. We have that $$(\mathbb{Z}_U)_x = \mathbb{Z}$$ if $$x \in U$$, and 0 otherwise, as a direct computation yields.

3. If $$\mathbb{Z}_{U}(V) \neq 0$$ and $$V$$ is connected, then $$V \subset U$$. Indeed, a section $$s \in \mathbb{Z}_{U}(V)$$ is given by a cover $$\mathcal{V}_{i \in I}$$ of $$V$$ and coherent sections $$s_i \in \mathcal{V}_i$$. Suppose $$V \not \subset U$$. Define $$\Sigma$$ as the poset of subsets $$Q \subset I$$ such that $$s_i=0$$ for all $$i \in Q$$. Then it is non empty, because there exist i s.t $$V_i \not \subseteq U$$, and evidently every ascending chain has an upper bound. We can henceforth apply Zorn lemma a find a maximal $$J$$. Suppose by contradiction that $$J \neq I$$. There exist $$i \in I \setminus J$$ and $$j\in J$$ such that $$V_i \cap V_j \neq \emptyset$$, otherwise they would disconnect $$V$$. $$V_i$$ must be contained in $$U$$, otherwise $$s_i$$ would be zero. But then $$s_i = s_i | V_i \cap V_j = s_j | V_i \cap V_j = 0$$.

4. Take $$X=\mathbb{R}^k$$, and $$U_n =$$ disk of radius $$1/n$$ around 0. Take the skyscraper sheaf $$\mathcal{G}$$ centered at zero with stalk $$\mathbb{Z}$$, i.e. $$\mathcal{G}(U) = \mathbb{Z}$$ if $$0 \in U$$ and 0 otherwise, with restriction given by identities when possible and zero otherwise. Define $$\varphi_n: \mathbb{Z}_{U_n} \to \mathcal{G}$$ in the following way. It is enough to define $$\phi_n: Z_{U_n} \to \mathcal{G}$$, and then $$\varphi_n$$ follwos by universal property of sheafification. If $$V \subset U_n$$, define $$\phi_n(V) = 1_{\mathbb{Z}}$$ and zero otherwise. It commutes with restrictions. Indeed, let $$V \subset V'$$ be subsets. If $$V' \not \subset U_n$$, the diagram starts with a zero. If $$0 \not \in V$$, the diagram ends with a zero. If $$0 \in V \subset V' \subset U_n$$, then the commuting diagram is made of all identities.

1. Consider the sheaf morphism

$$\prod \varphi_n: \prod_{n \in \mathbb{N}} \mathbb{Z}_{U_n} \to \prod_{n \in \mathbb{N}} \mathcal{G}$$

Take the element $$[(\{1\}_{n \in \mathbb{N}}, U_1)]$$ in $$( \prod_n\mathcal{G} )_0$$, and suppose that there exist $$[(\{f_n\}_{n \in \mathbb{N}},V)]$$ that is sent to $$[(\{1\}_{n \in \mathbb{N}}, U_1)]$$. This means that, after further restricting to $$V'$$ if necessary, $$f_n | V' = 1$$ for all $$n \in \mathbb{N}$$. Take $$V''$$ to be the connected component of $$V'$$ containing $$0$$. Recall that $$V''$$ is open. This imples in particular $$\mathbb{Z}_{U_n}(V'') \not 0$$ for all $$n \in \mathbb{N}$$, i.e. $$V'' \subset \bigcap_n U_n = \{0\}$$; which is a contradiction.

On the other hand, the functions $$\varphi_n$$ are surjective, because they are surjective on stalks, and we are done.

• Your example should work for any locally connected first countable space, I think. Do you agree? Even more generally we could say a locally connected space which contains at least one point which has a countable basis Commented Dec 28, 2023 at 14:40

Another example: let $$\mathbb{N}^+$$ be the set of positive integers. Take $$X=\{\infty\}\cup \mathbb{N}^+$$, and give it the topology generated by opens $$U_k=\{\infty,k,k+1,\dots\}$$ and $$V_k=U_k\setminus\{\infty\}$$ for $$k=1,2,\dots$$. Then $$\{\infty\}=X\setminus V_1$$ is closed in $$X$$. Every pair of nonempty open subsets of $$X$$ has a nonempty intersection, so $$X$$ is irreducible.

Let $$i:\infty\to X$$ be the closed embedding. Let $$G=i_*\mathbb{Z}$$. For every integer $$n\ge 1$$, let $$F_n$$ be the following presheaf on $$X$$: $$V\mapsto \begin{cases}0, & V\not\subset U_n,\\ \mathbb{Z}, & V\subset U_n, \end{cases}$$ where $$V$$ runs through nonempty open subsets of $$X$$. By this link, $$F_n$$ is a sheaf on $$X$$. The restriction of $$F_n$$ to $$\infty$$ is $$\mathbb{Z}$$. By adjunction, there is a canonical surjection $$f_n:F_n\to G$$ of sheaves.

We prove that $$h=\prod_{j\ge 1}f_j:\prod_{j\ge 1}F_j\to \prod_{j\ge 1}G$$ is not surjective. Suppose to the contrary that it is surjective. Let $$t=(1,1,\dots)\in\prod_{j\ge 1}\mathbb{Z}=\Gamma(X,\prod_{j\ge 1}G)$$. Since the morphism $$h_{\infty}$$ between the stalks at $$\infty$$ is surjective, there is an open neighbourhood $$U$$ of $$\infty\in X$$, and $$s\in \Gamma(U,\prod_{j\ge 1}F_j)$$ with $$h_{\infty}(s_{\infty})=t_{\infty}$$. There is an integer $$N>0$$ with $$U_N\subset U$$. One may assume that $$U=U_N$$. Then for every $$j>N$$, one has $$U_N\not\subset U_j$$, so $$\Gamma(U,F_j)=0$$. Then $$s_{\infty}^{(j)}=0$$ and $$1=t_{\infty}^{(j)}=f_j(s_{\infty}^{(j)})=f_j(0)=0$$, which is a contradiction.

• But the constant presheaf is famously not a sheaf. I also don’t think your $f_j$ are sheaf morphisms; consider for $k<j$ the allegedly commutative diagram of restrictions associated to $U_k\supseteq U_j$. $$\require{AMScd}\begin{CD}\Bbb Z@>1>>\Bbb Z\\@V0VV@VV1V\\\Bbb Z@>>1>\Bbb Z\end{CD}$$This clearly does not commute! Commented Dec 28, 2023 at 14:30
• Dear @FShrike, thanks a lot for pointing out my mistake. Is the new example satisfying?
– Doug
Commented Jan 1 at 9:19
• I don’t actually know what the definition of $G$ is. Is it $G(U)=\begin{cases}0&\infty\notin U\\\Bbb Z&\infty\in U\end{cases}$? Commented Jan 2 at 15:29
• Dear @FShrike, Yes.
– Doug
Commented Jan 2 at 15:42
• ok! I agree with this now. +1 Commented Jan 2 at 16:02