How to prove that a faithful functor reflects exact sequences? I am trying to solve Popescu, Abelian Categories with Applications to Rings and Modules, exercise 4, page 68. Assume that we have abelian categories $A,B$ and an additive faithful functor $T:A\to B$. We need to show that if 
$$
T(a) \stackrel{T(f)}{\longrightarrow} T(b)\stackrel{T(g)}{\longrightarrow} T(c)
$$
is exact, then $a \stackrel{f}{\longrightarrow} b \stackrel{g}{\longrightarrow} c$ is exact as well. Thanks to the answer to this question, I know how to check that faithful functors reflect kernels and cokernels, but this doesn't seem to be enough to claim that they reflect exact sequences in general. In particular, I have some problems in checking that $\ker(T(g))\cong T\left(\ker(g)\right)$.
Then I found Eilenberg, Moore, Foundations of Relative Homological Algebra, Proposition 1.2, in which they explicitly ask for $T$ being kernel preserving in order to prove the foregoing fact (even if in a less general context).
Question: Do I really need $T$ being kernel preserving? Is there a way to deduce it from the hypotheses?
 A: Here is a proof.  Suppose $$T(A) \stackrel{T(f)}{\longrightarrow} T(B)\stackrel{T(g)}{\longrightarrow} T(C)$$ is exact.  Then $T(gf)=0$ so $gf=0$.
Now let $k: K\to B$ be a kernel of $g$, let $j:J\to T(B)$ be a kernel of $T(g)$, let $d:B\to D$ be a cokernel of $f$, and let $e:T(B)\to E$ be a cokernel of $T(f)$.  Since $T(A)\to T(B)\to T(C)$ is exact, $j$ is also the inclusion of the image of $T(f)$ and thus $e$ is also a cokernel of $j$; in particular, $ej=0$.  Since $T(g)T(k)=0$, there is $\alpha:T(K)\to J$ such that $j\alpha=T(k)$.  Similarly, there is $\beta:E\to T(D)$ such that $\beta e=T(d)$.  Thus $T(dk)=\beta e j\alpha=0$.  Since $T$ is faithful, $dk=0$.  But that says exactly that the image of $f$ is contains the kernel of $g$, and so $A\to B\to C$ is exact.
Note that these hypotheses do not imply that $T$ preserves kernels.  For instance, $T:Ab\to Ab$ could be the functor $T(A)=A\oplus A/2A$, which is faithful because of the first summand but does not preserve kernels because of the second.  More generally, $T$ could have the form $T_0\oplus T_1$ where $T_0$ is faithful and $T_1$ is an arbitrarily ill-behaved additive functor.
