$(\operatorname{Im}\alpha)(\operatorname{Spec}A)=\operatorname{Im}(\alpha(\operatorname{Spec}A))$?

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 $\alpha: \widetilde L \to \widetilde M$, do we have $(\operatorname{Im}\alpha)(\operatorname{Spec}A)=\operatorname{Im}(\alpha(\operatorname{Spec}A))$?

Answer 1

This follows from the fact that the global section functor $F := \Gamma(\operatorname{Spec} A,-): \textbf{QCoh}(\operatorname{Spec} A) \to \textbf{Mod}(A)$ is exact, in particular it commutes with images, i.e.

$$\Gamma(\operatorname{Spec}A,\operatorname{Im}(\alpha))=F(\operatorname{Im}(\alpha)) = \operatorname{Im}(F(\alpha))=\operatorname{Im}(\Gamma(\operatorname{Spec} A,\alpha))$$ , where $\Gamma(\operatorname{Spec} A,\alpha)$ is the map on global sections, i.e. the map $\Gamma(\operatorname{Spec} A,\widetilde L) \to \Gamma(\operatorname{Spec} A,\widetilde M)$.