$\mathcal{F}$ left-exact functor of abelian categories, then $\mathcal{F}(\ker (f))\cong\ker (\mathcal{F}(f))$

Question

Let $\mathcal{C},\mathcal{D}$ be two abelian categories. Fix a morphism $f: M\rightarrow N$ in $\mathcal{C}$ and write $(\ker(f),\iota)$ for its kernel. Let $\mathcal{F}:\mathcal{C}\longrightarrow \mathcal{D}$ be a left-exact functor. I want to show that $\mathcal{F}(\ker (f))\cong\ker (\mathcal{F}(f))$. My work so far is the following. I know that $$0\rightarrow \ker(f)\xrightarrow{\iota} M \xrightarrow{f} N$$ is a left-exact sequence. Thus, left-exactness of $\mathcal{F}$ gives me that $$0\rightarrow \mathcal{F}(\ker(f))\xrightarrow{\mathcal{F}(\iota)} \mathcal{F}(M) \xrightarrow{\mathcal{F}(f)} \mathcal{F}(N)$$ is also left-exact. In particular, this means that $\ker(\mathcal{F}(f))=\text{im}(\mathcal{F}(\iota))$, i.e. $\mathcal{F}(\iota)(\mathcal{F}(\ker(f)))=\ker(\mathcal{F}(f))$. If I manage to prove that $\mathcal{F}(\iota)$ is an isomorphism then I am done. I know that in an abelian category, epi $+$ mono $\Leftrightarrow$ isomorphism. $\mathcal{F}$ is left-exact so it preserves monomorphisms. So if I proved that $\mathcal{F}(\iota)$ is epi, I would have finished. However, I do not know how to prove this (or if it can be proved). Can someone help me solve this problem?

Answer 1

Generally, $\mathcal{F}(\iota)$ will not be an epimorphism. You want to prove that $\mathcal{F}(\mathrm{ker}(f))\cong \mathrm{im}(\mathcal{F}(\iota))$ via the map $\mathcal{F}(\iota)$, and to show this you only need that $\mathcal{F}(\iota)$ is a monomorphism. Indeed, for a general monomorphism $g\colon A\to B$ in an abelian category $\mathcal{C}$, we have that $\mathrm{im}(g)\cong \mathrm{coker}(\mathrm{ker}(g)\to A)\cong \mathrm{coker}(0\to A)\cong A$ and you can see that it is the map $g\colon A\to \mathrm{im}(g)$ that witnesses this.

Put differently, you know by exactness of $0\to\mathcal{F}(\mathrm{ker}(f))\xrightarrow{\mathcal{F}(\iota)}\mathcal{F}(M)\to\mathcal{F}(N)$ that $\mathrm{im}(\mathcal{F}(\iota))\cong\mathcal{F}(\mathrm{ker}(f))/\mathrm{ker}(\mathcal{F}(\iota))\cong\mathcal{F}(\mathrm{ker}(f))/0\cong \mathcal{F}(\mathrm{ker}(f))$, as witnessed by $\mathcal{F}(\iota)$.