Let $A\to B\to C\to 0$ be exact, and $F$ a right-exact functor. Why must $F(A)\to F(B)\to F(C)\to 0$ be exact?

Question

Fix two Abelian categories $\mathscr{A}$ and $\mathscr{B}$, and let $F:\mathscr{A}\to\mathscr{B}$ be a right-exact functor. Suppose $$A\xrightarrow{f}B\xrightarrow{g}C\to 0$$ is an exact sequence in $\mathscr{A}$. I know that if $f$ is injective, then $$F(A)\xrightarrow{F(f)}F(B)\xrightarrow{F(g)}F(C)\to 0$$ must be exact, but what about for arbitrary $f$? My current approach is to replace $f:A\to B$ with $i:\ker g\hookrightarrow B$ so that we have the short exact sequence $$0\to\ker g\xrightarrow{i}B\xrightarrow{g}C\to 0$$ which induces an exact sequence $$F(\ker g)\xrightarrow{F(i)}F(B)\xrightarrow{F(g)}F(C)\to 0$$ but I dont see why we should have that $\mathrm{im}\,F(f)\subseteq\mathrm{im}\,F(i)$, which is what we require to show exactness.

Answer 1

Consider the image factorization of $f$: $$A \xrightarrow{p} \operatorname{Im} f \xrightarrow{i} B$$ and consider the sequence $\operatorname{Im} f \xrightarrow{i} B \xrightarrow{g} C \to 0$. We know that $i$ is mono. Moreover $\operatorname{Im} f = \operatorname{Im} i$ so the sequence is exact. Hence we have a right exact sequence $$F(\operatorname{Im} f) \xrightarrow{F(i)} F(B) \xrightarrow{F(g)} F(C) \to 0.$$

We need to check that $\ker F(g) = \operatorname{Im} F(f)$. To prove it we shall show $\operatorname{Im} F(f) = \operatorname{Im} F(i)$. Since the sequence $$\ker f \rightarrow A \xrightarrow{p} \operatorname{Im}f \to 0$$ is exact, we can conclude that $F(p)$ is epi. (Why?) Moreover $F(f) = F(i)\circ F(p)$ so they have same image. Thus we have a right-exact sequence $$F(A) \xrightarrow{F(f)} F(B) \xrightarrow{F(g)} F(C) \to 0.$$