Does faithful additive functor between abelian categories reflect short exact sequence? 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.
 A: 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.
