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.

  • $\begingroup$ What do you mean "the product morphism" ? $\endgroup$ Commented Jun 23, 2019 at 18:08
  • $\begingroup$ Post edited!!!! $\endgroup$
    – user_12345
    Commented Jun 23, 2019 at 18:22
  • $\begingroup$ 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 ? $\endgroup$ Commented Jun 23, 2019 at 18:28
  • $\begingroup$ Sorry, of course I was thinking of an arbitrary product. $\endgroup$
    – user_12345
    Commented Jun 23, 2019 at 18:32

2 Answers 2


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.

  • $\begingroup$ 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 $\endgroup$
    – FShrike
    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.

  • $\begingroup$ 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! $\endgroup$
    – FShrike
    Commented Dec 28, 2023 at 14:30
  • 1
    $\begingroup$ Dear @FShrike, thanks a lot for pointing out my mistake. Is the new example satisfying? $\endgroup$
    – Doug
    Commented Jan 1 at 9:19
  • $\begingroup$ I don’t actually know what the definition of $G$ is. Is it $G(U)=\begin{cases}0&\infty\notin U\\\Bbb Z&\infty\in U\end{cases}$? $\endgroup$
    – FShrike
    Commented Jan 2 at 15:29
  • $\begingroup$ Dear @FShrike, Yes. $\endgroup$
    – Doug
    Commented Jan 2 at 15:42
  • $\begingroup$ ok! I agree with this now. +1 $\endgroup$
    – FShrike
    Commented Jan 2 at 16:02

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