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Let $\mathcal A$, $\mathcal A\prime$ be two abelian categories and $F:\mathcal A\to\mathcal A\prime$ an additive functor. My question is:

suppose F is faithful, can we deduce that $F$ reflects short exact sequences?

What I can assure is that this is the case when $F$ is either left exact or right exact.

Censi LI
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    Using the Mitchel Embedding theorem, you can reduce to looking at additive faithful functors between categories of modules over rings. – D_S Mar 09 '17 at 13:26

1 Answers1

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It suffices to show that $F$ reflects kernels, since then dually it will also reflect cokernels and $0\to A\to B\to C\to 0$ is exact iff $A\to B$ is the kernel of $B\to C$ and $B\to C$ is the cokernel of $A\to B$.

So suppose we have $$A\stackrel{f}{\to}B\stackrel{g}{\to} C$$ in $\mathcal{A}$ such that $F(f)$ is a kernel of $F(g)$. Let $h:K\to B$ be a kernel of $g$. Since $F(gf)=0$ and $F$ is faithful, $gf=0$, so there is a unique $i:A\to K$ such that $f=hi$. We wish to show $i$ is an isomorphism.

Since $F(f)$ is a kernel of $F(g)$, there is a unique $j:F(K)\to F(A)$ such that $F(h)=F(f)j$. Note that $$F(f)jF(i)=F(h)F(i)=F(f),$$ so $jF(i)=1_{F(A)}$ since $F(f)$ is monic. It follows that $F(i)$ is monic, and hence $i$ is monic as well since $F$ is faithful.

Now let $p:K\to K/A$ be a cokernel of $i$ and $q:B\to B/A$ be a cokernel of $f$. The map $h:K\to B$ gives a unique map $h':K/A\to B/A$ such that $h'p=qh$, and $h'$ is monic because $h$ is monic. Now note that $$F(h'p)=F(qh)=F(q)F(f)j=0.$$ Thus $h'p=0$, so $p=0$ since $h'$ is monic. But $p$ was a cokernel of $i$, so this means $i$ is epic.

Thus $i$ is epic and monic, and so is an isomorphism. It follows that $f$ is a kernel of $g$, and $F$ reflects kernels.

Eric Wofsey
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  • Thank you! Very inspiring! In fact I figured that your proof can be modified to give that $F$ preserves all exact sequences. – Censi LI Mar 10 '17 at 09:33