Q:

Property of finitely presented objects

Let $\mathcal{C}$ be a Grothendieck category. Take a short exact sequence in this category:
$$ 0 \to X \to Y \to Z \to 0 $$
where the object $Y$ is finitely presented. Is true that $Z$ is finitely presented if and only if $X$ is finitely generated?
The implication $(\Rightarrow)$ is rather trivial. I can't see a way to prove also the other.

A:

Let $Y,Z$ be finitely presented. I claim that $X$ is finitely generated, i.e. if $(A_{\alpha})$ is a directed system of monomorphisms, then
$$\mathrm{colim}_\alpha \hom(X,A_\alpha) \to \hom(X,\mathrm{colim}_\alpha A_\alpha)$$
is bijective, resp. an isomorphism of abelian groups. We cannot conclude that that $X$ is finitely presented: Take any ring $R$ which is not coherent, it has a finitely generated ideal $I$ which is not finitely presented, and then $0 \to I \to R \to R/I \to 0$ provides a counterexample in $\mathsf{Mod}(R)$.
Injectivity. Take some element in the kernel, say $X \to A_{\alpha}$. We obtain $Y \to A_{\alpha} \sqcup_X Y$ and the composition $Y \to \mathrm{colim}_\alpha A_{\alpha} \sqcup_X Y$ vanishes on $X$. Hence, it lifts to a morphism on $Z$. Since $Z$ is finitely presented, it factors as $Z \to A_{\beta} \sqcup_X Y \to \mathrm{colim}_\alpha A_\alpha \sqcup_X Y$ for some $\beta \geq \alpha$. Consider the composition $Y \to Z \to A_{\beta} \sqcup_X Y$. A priori this isn't the canonical morphism, but it becomes so in the colimit. Since $Y$ is finitely presented, we may choose $\gamma \geq \beta$ such that $Y \to Z \to A_{\gamma} \sqcup_X Y$ is the canonical morphism. It follows that the canonical morphism $Y \to A_\gamma \sqcup_X Y$ vanishes on $X$, i.e. that $X \to A_\gamma \to A_\gamma \sqcup_X Y$ vanishes. But since we are in an abelian category, pushouts of monomorphisms are monomorphisms, so that in particular $A_\gamma \to A_\gamma \sqcup_X Y$ is a monomorphism. It follows that $X \to A_\gamma$ vanishes, as desired.
Surjectivity. Let $f : X \to \mathrm{colim}_\alpha A_\alpha$ be a morphism. Since we are in a Grothendieck category, each $A_\alpha$ is a subobject of the colimit, and if $X_\alpha \subseteq X$ denotes the preimage or pullback, then $X = \mathrm{colim}_\alpha X_\alpha$. Then $f$ is the colimit of the maps $f_\alpha : X_\alpha \to A_\alpha$. Consider the isomorphism $Z \cong Y/X = \mathrm{colim}_\alpha Y/X_\alpha$. Since $Z$ is finitely presented, this factors as $Z \to Y/X_\alpha \to Y/X$ for some $\alpha$. By an argument similar to the one above, we may assume that the composition $Y \to Z \to Y/X_\alpha$ is the canonical projection (after increasing $\alpha$). Hence, the canonical projection $Y \to Y/X_\alpha$ vanishes on $X$, i.e. $X=X_\alpha$, so that $f$ factors through $A_\alpha$.
Hope it's correct.

