equivalence between subcategories in abelian categories keeps exact sequence Let $A$ be an abelian category and $B$ a subcategory, not necessary abelian. Let $C^\bullet$ be a exact complex in $A$ with $C^i\in B$. Suppose there is another abelian category $A'$ and $B'$ a subcategory, and $f: B\cong B'$ an equivalence, and  $D^\bullet=f(C^\bullet)$. Then must $D$ be exact?
The point is that we can not form kernel in $B$.  
 A: No.  A counter-example is the Yoneda embedding:  take $A=B=Ab$ to be the category of abelian groups, and let $A'=Fun(A^{op}, Ab)$ be the category of additive contravariant functors from $A$ to $Ab$.  Then $A'$ is an abelian category, and the Yoneda functor
$$ h:A\to Fun(A^{op},Ab): X\mapsto Hom(-,X) $$
is fully faithful.  Let $B'$ be the image of $h$.  Then $h$ restricts to an equivalence of categories from $B$ to $B'$.
Now, consider the exact sequence
$$ 0\to \mathbb{Z} \stackrel{2}{\to} \mathbb{Z} \stackrel{\pi}{\to} \mathbb{Z}/2\mathbb{Z} \to 0.$$
Applying $h$, this becomes the complex
$$ 0\to Hom(-,\mathbb{Z}) \stackrel{2^*}{\to} Hom(-,\mathbb{Z}) \stackrel{\pi^*}{\to} Hom(-,\mathbb{Z}/2\mathbb{Z}) \to 0.$$
This complex is not exact anymore, because $\pi^*$ is not an epimorphism; this can be seen by replacing the "$-$" by $\mathbb{Z}/2\mathbb{Z}$, to get
$$ 0\to 0 \stackrel{0}{\to} 0 \stackrel{\pi^*}{\to} Hom(\mathbb{Z}/2\mathbb{Z},\mathbb{Z}/2\mathbb{Z}) \to 0.$$
Thus $\pi^*$ is not surjective.
Therefore, $h$ does not take exact complexes to exact complexes.
