Everything is taken from Freyd's book Abelian categories.
By a Grothendieck category Freyd means a bicomplete well-powered abelian category such that, for $C,(B_i)_{i \in I}$ being subobject of a fixed object with $(B_i)_{i \in I}$ being a chain we have
$$C\cap\left(\bigcup_{i \in I} B_i\right) = \bigcup_{i \in I} (C\cap B_i).$$
In an abelian category, an extension of an object $A$ is simply a monomorphism $A\to B$. An injective $u\colon A\to B$ is essential if, for any nonzero $v\colon B'\to B$ the intersection of $u$ and $v$, considered as subobjects of $B$, is nonzero.
Let $\mathsf{A}$ be a Grothendieck category with a generator $G$. Then $\mathsf{Hom_A}(G,-)$ is a functor into the category of $R$-modules where $R = \mathsf{End_A}(G)$ with the $R$-module structure on $\mathsf{Hom_A}(G,A)$ being defined by composition: $rf = f\circ r$.
Lemma. If $u\colon A\to E$ is a an essential extension in $\mathsf{A}$, then $\mathsf{Hom_A}(G,u)\colon \mathsf{Hom_A}(G,A)\to\mathsf{Hom_A}(G,E)$ is an essential extension in $R\text{-}\mathsf{Mod}$.
Freyd provides the following proof:
It is not clear to me $P \neq 0$. Indeed, the pullback of $\mathrm{im}(x)$ along $u$ is nonzero as it is their intersection, but here $x$ is not necessary a monomorphism. Moreover, for there to exist $G\to P$ such that $G \to P \to G \xrightarrow{x} E \neq 0$ (as $G$ is a generator), we must have $P \to G \xrightarrow{x} E \neq 0$. Again, for this follow we need either $x$ to be a monomorphism and $P\to G \neq 0$ or to $u$ to be an epimorphism, from which it would follows that it is an isomorphism, which I find unlikely.
Is this proof correct? If yes, what am I missing? If not, can it be salvaged?
Best Answer
Factor $x$ by the image as you suggest:
$$x: G \stackrel{\pi}\twoheadrightarrow \mathrm{Im}\,x \hookrightarrow E. $$
Now one can use the fact that "pullback squares can be pasted together", in other words, the pullback square you have decomposes as pair of pullback squares:
\begin{array} \\ G & \stackrel{\pi}\twoheadrightarrow & \mathrm{Im}\,x & \hookrightarrow & E\\ \uparrow & & \uparrow & &\uparrow \\ P & \stackrel{\pi'}\twoheadrightarrow & P_{im} & \hookrightarrow & A\\ \end{array}
Here $P_{im}$ is nonzero by $A \rightarrow E$ being essential ($\mathrm{Im}\,x$ is nonzero since so is $x$). The map $\pi'$ is an epimorphism since so is $\pi,$ and epimorphisms are preserved by pullbacks in abelian categories. Thus, $P$ admits an epimorphism onto a nonzero object, and it follows that it is itself nonzero.