# Is the presheaf defined by $U\to \operatorname{Im}(\alpha (U))$ indeed a sheaf?

Let $A$ be a commutative ring with unity and $L,M$ be $A$-modules. Given a morphism of $\mathcal O_{\operatorname{Spec}A}$-modules $\alpha: \widetilde L \to \widetilde M$, for any open subset $U\subset \operatorname{Spec}A$, is the presheaf of modules defined by $U\to \operatorname{Im}(\alpha (U))$ indeed a sheaf of modules?

• I think you want $A$ to be noetherian? – MooS Nov 30 '17 at 16:23
• I think this is a great question, because the fact that the image presheaf is not a sheaf seems to be related to the section functor failing to be right exact. Hence there is hope in a situation, where the section functor is exact. – MooS Nov 30 '17 at 16:33

@MooS is correct, the sheaf $\text{im}(\alpha)$ is not the same as the presheaf of images. To see this, consider the following exact sequence, where $\mathscr{O}_{\mathbb{C}}$ is the sheaf of holomorphic functions on $\mathbb{C}$:

$$0 \rightarrow \mathbb{Z} \rightarrow \mathscr{O}_{\mathbb{C}} \rightarrow \mathscr{O}_{\mathbb{C}}^{\ast} \rightarrow 1$$

where the first map is multiplying by $2 \pi i$ and the second map is $\text{exp}$. Locally, one can take logarithms, making this sequence exact, but local existence of logarithms does not entail global existence. Now because this is exact, we know that the sheaf $\text{im}(\text{exp}) \simeq \mathscr{O}_{\mathbb{C}}^{\ast}$, so $\text{im}(\text{exp})(\mathbb{C})\simeq \mathscr{O}_{\mathbb{C}}^{\ast}(\mathbb{C})$, but $\text{exp} \left( \mathscr{O}_{\mathbb{C}}(\mathbb{C}) \right)$ is not isomorphic to $\mathscr{O}_{\mathbb{C}}^{\ast}(\mathbb{C})$.

$\forall f\in A, D(f)\to \operatorname{Im}(\alpha(D(f)))=(\operatorname{Im}(\alpha(\operatorname{Spec}A)))_f$, so the presheaf is isomorphic to $\widetilde{\operatorname{Im}(\alpha(\operatorname{Spec}A))}$, and it is indeed a sheaf. Therefore, neither $A$ nor $\operatorname{Spec}A$ need be noetherian.

• It seems wrong. – Born to be proud Nov 30 '17 at 19:45
• This only shows that this is a sheaf on basic open sets. But this does not yield the result. – MooS Nov 30 '17 at 21:27
• @MooS If $\operatorname{Spec} A$ is noetherian, how to show the presheaf is a sheaf? – Born to be proud Nov 30 '17 at 21:57
• @MooS Is the sheaf $\operatorname{Im}\alpha$ associated to the presheaf isomorphic to $\widetilde{\operatorname{Im}(\alpha(\operatorname{Spec}A))}$? – Born to be proud Dec 1 '17 at 0:43
• I do not think that it is true, even in the noetherian case. The problem is that the global section functor is known to be exact in this case, but if you take a non-affine open, the corresponding section functor might be non-exact. – MooS Dec 1 '17 at 9:29