Equivalent condition to left exactness

Question

I'm trying to prove that an additive functor $F:\mathcal{A}^\text{op}\to \text{AbGrp}$ on an abelian category to abelian groups is left exact if and only if for every epimorphism $p:A\to B$ in $\mathcal{A}$ one has an exact sequence $$0\longrightarrow F(B)\overset{F(p)}\longrightarrow F(A)\overset{F(d_0-d_1)}\longrightarrow F(A\times_BA) $$ where $A\times_B A$ is the fibre product and $d_0,d_1:A\times_BA\to A$ are the usual projection. I managed to prove that if $F$ is left exact then that sequence is exact (simply proving that $B$ is a cokernel for $d_0-d_1$ and then using left exactness), but I am stuck with the reverse implication. In principle it is enough to prove that $F$ preserves kernels, namely that if $f:A\to B$ is a map in $\mathcal{A}$ then $F(\text{coker}f)=\ker F(f^{\text{op}})$. I was trying to use $\text{coker}f$ as $p$, but it doesn't seem really useful since I get a pull-back diagram which does not involve $f$...

Answer 1

The fibre product can be rewritten as

$$\begin{aligned}A\times_B A=\{(a,a')\in A\oplus A\mid a-a'\in\ker p\}&\xrightarrow{\cong} A\oplus \ker p\\(a,a')&\mapsto (a, a-a') \end{aligned}$$

and the short exact sequence $0\to A\times_B A\xrightarrow{d-d'} A\xrightarrow {p} B\to 0$ becomes $0\to A\oplus\ker p\xrightarrow{(0,1)} A\to B\to 0$. $F$ is an additive functor by assumption, so if the sequence

$$\begin{aligned}0\to FB\to FA\xrightarrow{(0,1)} & FA\oplus F\ker p \\&\llap{{}={}}F(A\oplus\ker p)\\&\llap{{}={}}F(A\times_B A)\end{aligned}$$

is exact, then also $0\to FB\to FA\to F\ker p$ is exact.