Let $X$ a sheaf and $ \{\mathcal{F}_{\lambda} \}_{\lambda}$ a arbitrary family of $ \mathcal{O}_X$-Modules. Because thats an abelian category it contains products and coproducts (=sums).

We considering the product presheaf $\prod \mathcal{F}_{\lambda}$ by defining $U \to \prod \mathcal{F}_{\lambda}(U)$ for euch open $U$ (resp. the sum presheaf $\bigoplus \mathcal{F}_{\lambda}$ by defining $U \to \bigoplus \mathcal{F}_{\lambda}(U)$).

My question is why the product presheaf as defined above is automatically a sheaf (therefore the sheaf axiom holds) where the sum is only a presheaf (therefore have to be sheafificated)?

  • $\begingroup$ Maybe it's interesting to say that even though the direct sum presheaf isn't a sheaf in general, it is if $X$ is a Noetherian topological space. More generally, for $X$ Noetherian, a colimit taken in presheaves of a diagram of sheaves is a sheaf [ref]. $\endgroup$ Oct 21, 2023 at 16:56

2 Answers 2


Garbificated ? Did you try to prove that the product is a sheaf ?

Here is a counter-example for the direct sum : let $X=\mathbb{R}$ and $U=\mathbb{R}\setminus\mathbb{Z}$. Then $U$ is the disjoint union of the intervals $(n,n+1)$.

Let $\mathcal{F}_n=\mathbb{Z}$ be the constant sheaf and $s_n$ be the section of the presheaf $\bigoplus\mathcal{F}_k$ on $(n,n+1)$ such that every component of $s_n$ is zero except the $n$-th component which is $1$. In other words, $s_n\in(\bigoplus\mathcal{F}_k)((n,n+1))=\bigoplus\mathbb{Z}$ is the sequence $(...,0,0,1,0,0,...)$ where $1$ is at the $n$-th position.

I claim that these sections can't be glued to a section on $U$ (though they vacuously agreed on overlaps). If you want to glue these sections, you will need a section which have an infinite number of non zero component. And this won't be an element of $\bigoplus \mathcal{F}_n$.

  • $\begingroup$ Yes, for arbitrary product (and therefore indeed for finite sum too because they coinside in finite case) if there is given a cover $\{U_i\}$ of $X$ and local sections $s_i = (s_{i, \lambda})_\lambda \in \prod \mathcal{F}_{\lambda}(U_i)$ such that they coinside in comon sections $\prod \mathcal{F}_{\lambda}(U_i \cap U_j)$ then that means that the $s_i$ are componentwise compatible. $\endgroup$
    – user267839
    Oct 31, 2017 at 16:30
  • $\begingroup$ Thaterefore for every $\lambda$ the $s_{i, \lambda}$ are compatible and because $\mathcal{F}_{\lambda}$ is a sheaf, they glue together to a $s_{\lambda}$. It's easy to proof that $s:=(s_{\lambda})_{\lambda}$ is the glued global section of the $s_i$ s. So for product presheaf the sheaf axiom holds. $\endgroup$
    – user267839
    Oct 31, 2017 at 16:30
  • $\begingroup$ @KarlPeter Yes, that is correct. You should show your attempts in your questions, or in that case, what you proved. $\endgroup$
    – Roland
    Oct 31, 2017 at 17:19

Let $X$ be an infinite set with the discrete topology. Let $\mathbb C_x$ be the skyscraper sheaf on $x\in X$, and let $\mathcal F:=\bigoplus_{x\in X}\mathbb C_x$. Then $\mathcal F$ is the presheaf of functions on $X$ with finite support: for $U\subset X$, $$\mathcal F(U)=\{f\colon U\to \mathbb C:\mathrm{supp}(f)\text{ is finite}\}.$$ But $\mathcal F$ is not a sheaf. Indeed, for any $x\in X$, consider $s_x\in\mathcal F(\{x\})$, defined by $s_x(x)=1$.

Now the sheaf axiom predicts a function $s\in\mathcal F(X)$ such that $s|_{\{x\}}=s_x$, i.e., $s(x)=1$ for all $x\in X$. That is, $s$ is just the constant function $1$. But such a function does not exist as a function of $\mathcal F(X)$.

In fact, the sheafification $\mathcal F^\#$ of $\mathcal F$ is just the constant sheaf $\underline{\mathbb Z}_X$: for any $U\subset X$, $$\mathcal F^\#(U)=\{f\colon U\to \mathbb C\}\simeq\mathbb Z^U.$$


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