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$\def\O{\mathcal{O}} \def\M{\mathcal{M}} \def\N{\mathcal{N}} \def\P{\mathcal{P}} $Given a ringed space $(X,\O{_X})$, an $\O_X$-module $\P$ and $\mathcal{O}_X$-submodules $\M,\N\subset\P$ we define the module sum presheaf $\M+_p\N$ as $$ U\subset X\mapsto\M(U)+\N(U). $$ It is easy to verify that $\M+_p\N$ is a subpresheaf of $\mathcal{O}_X$-modules of $\P$. My question is: is $\M+_p\N$ a sheaf? Since it is a subpresheaf of a separated presheaf (namely, $\P$), it is separated, but when I try to check the gluing axiom, I fail. I suspect that $\M+_p\N$ is not a sheaf, but I cannot find any counterexample. I've thought of trying to recycle the counterexamples given here and here, but I think they don't work for this case.

Do you have some counterexample at hand?

Elías Guisado Villalgordo
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  • It’s easier to see this thinking of the sum as the product. Sheaves are closed under arbitrary limits. But sheaves aren’t closed under arbitrary direct sums of modules; they are closed under finite direct sums because direct sums are products. – Mark Saving Nov 26 '22 at 19:46

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I hesitate to answer this question because it is a duplicate but I think there is a much simpler example which doesn't necessitate the use of cohomology.

Take $X = \{a, b, c\}$ with the topology given by $\mathcal{T} = \{X, \{a, b\}, \{b, c\}, \{b\}, \emptyset\}$, and fix the following subsheaves of the constant sheaf of $\mathbb{Z}$, which we take to be $\mathcal{O}_X$. Take $\mathscr{F} = i_a\mathbb{Z}$ and $\mathscr{G} = i_c\mathbb{Z}$ the skyscraper sheaves of $\mathbb{Z}$ at $a$ and $c$. Then, $\mathscr{F} + \mathscr{G}$ does not satisfy gluing.

Indeed $(\mathscr{F} + \mathscr{G})(\{b\}) = 0$ and the global sections are $\mathbb{Z}$. As such, $1 \in (\mathscr{F} + \mathscr{G})(\{a, b\})$ and $0 \in (\mathscr{F} + \mathscr{G})(\{b, c\})$ should glue together if it were a sheaf, but they can't since the restriction maps of this presheaf from $X$ to $\{a, b\}, \{b, c\}$ are the identity.

Daniel
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