Setup. Let $k$ be a field, $\mathcal{C}$ be a $k$-linear abelian category, $\mathcal{D}$ an arbitrary $k$-linear category. Let $\mathcal{C}'$ be a full subcategory of $\mathcal{C}$ with the following property: For every $M \in \mathcal{C}$ there is some $A \in \mathcal{C}'$ and an epimorphism $A \twoheadrightarrow M$. I have proven the following result:
Theorem. If $F : \mathcal{C}' \to \mathcal{D}$ is a $k$-linear functor with the pushout property $(\star)$ below, then $F$ extends to a right exact $k$-linear functor $\overline{F} : \mathcal{C} \to \mathcal{D}$.
$(\star)$ Given a pushout square in $\mathcal{C}$ with objects in $\mathcal{C}'$
$\begin{array}{c} A' & \rightarrow & A \\ \downarrow && \downarrow \\ B' & \rightarrow & B \end{array}$
in which $A' \to B'$ (and hence $A \to B$) is an epimorphism in $\mathcal{C}$, then also
$\begin{array}{c} F(A') & \rightarrow & F(A) \\ \downarrow && \downarrow \\ F(B') & \rightarrow & F(B) \end{array}$
is a pushout in $\mathcal{D}$ and $F(A') \to F(B')$ (and hence $F(A) \to F(B)$) is an epimorphism.
Question. Is this Theorem already known? Does it appear in the literature?
Comments. It is clear that $(\star)$ is necessary for the extension of $F$, but the sufficiency requires a proof. Of course one idea is to choose exact sequences $A' \to A \to M \to 0$ and define $\overline{F}(M)$ to be the cokernel of $F(A') \to F(A)$. But then one has to show that 1) there is an action on morphisms, 2) $\overline{F}$ is, indeed, a functor, 3) it is $k$-linear, 4) it is right exact.
Surprisingly, each step does not seem to be trivial, and my proof takes several pages. The main technical input is the notion of a weak pullback square, by which I mean a commutative square as above with the property that $A' \to A \times_B B'$ is an epimorphism. By assumption, all morphims $B' \to B \leftarrow A$ can be completed to a weak pullback square. If $A \to B$ is an epimorphism, it is actually a pushout square - this is where $(\star)$ comes into play. For 1), given a morphism $M \to N$ and short exact sequences $A' \to A \to M \to 0$ and $B' \to B \to N \to 0$ with $A',A,B',B \in \mathcal{C}'$, the methods sketched above produce a commutative diagram
$\begin{array}{c} A' & \rightarrow & A & \rightarrow & M & \rightarrow & 0 \\ \uparrow && \uparrow \\ C' & \rightarrow & C & & \downarrow \\ \downarrow && \downarrow \\ B' & \rightarrow & B & \rightarrow & N & \rightarrow & 0 \end{array}$
such that the diagram with $A',A,C',C$ is a pushout square and $C' \to A'$ is an epimorphism, hence the cokernel of $F(C') \to F(C)$ is isomorphic to the cokernel $\overline{F}(M)$ of $F(A') \to F(A)$. Hence there is a unique morphism $\overline{F}(M) \to \overline{F}(N)$ such that the composition with the epimorphism from $F(C)$ equals $F(C) \to F(B) \to \overline{F}(N)$. This suggests that $\mathcal{C}$ is a localization of a suitably defined catergory of morphisms in $\mathcal{C}'$.
