Showing that a functor is Exact Let $R$ and $S$ be rings, and let $F : Mod(R) \to Mod(S)$ be a functor which sends zero to zero.  Given that $F$ is exact on short exact sequences (that is  with zeros on two ends, and 3 nonzero terms in middle) how to show that 

for all $A \to B \to C$ exact sequence, $F(A) \to F(B) \to F(C)$ is exact.

In the thread
Equivalent definition of exactness of functor?
I think that the author 'SL2' who answers the question is implicitly assuming that $F$ commutes with kernels and images when he applies $F$ to $C_i$. Can someone help me in pointing  out why is this so?
Please help. Thanks in advance.
 A: Take an exact sequence $A\overset{f}\to B \overset{g}\to C$.
Let $A' = A/\ker{f}$, $C'=g(B)$.
Then $0\to A' \to B \to C' \to 0$ is exact.
I think that this is what you had in mind.  As you say, if $F$ commutes with kernels and images, then the above will give us exactly what we want.
But it does!  Consider the exact sequence $0\to\ker{f} \to A \to f(A) \to 0$...
A: In SL2's answer that you reference, the following diagram is constructed using kernels and cokernels in the first category $\operatorname{Mod}(R)$:



*

*The long sequence is exact

*The diagonal sequences are exact

*The triangles commute


So after applying $F$, we obtain a corresponding diagram with $A_i$ replaced by $F(A_i)$ etc.  and note


*

*The diagonals still start and end with $0$

*The diagonals are still exact

*The triangles still commute

*The long sequence is a complex (because $F(A_{i-1})\to F(A_{i+1})$ factors over via $F(C_i)\to F(C_{i+1})$


Do the rest (for exactnes at $F(A_2)$, say) by diagram chasing (because we can in $\operatorname{Mod}(S)$): If $x\in F(A_2)$  is mapped to $0\in F(A_{3})$, then it is also mapped to $0\in F(C_3)$, hence comes from some $y\in F(C_2)$, which comes form some $z\in F(A_1)$. So $z\mapsto x$ as required.
Note that we can work completely inside the given diagram and do not need that $F(C_i)$ are really kernels/cokernels (as in: For every morphism  $X\to F(A_1)$ such that $X\to F(A_1)\to F(A_2)$ is zero there exists a unique $X\to F(C_1)$ ...)!
