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

Question

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?

Answer 1

$\forall f\in A, D(f)\to \operatorname{Im}(\alpha(D(f)))=(\operatorname{Im}(\alpha(\operatorname{Spec}A)))_f$, 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.

Answer 2

@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 $\mathscr{O}_{\mathbb{C}}^{\ast}(\mathbb{C})$.