Why are functors exact if they preserve all short exact sequences? If a functor $F\colon \mathcal C → \mathcal D$ of abelian categories preserves short exact sequences, why is it exact?
I know the argument is supposed to be that you can split up long exact sequences into short sequences, like it is done here.  But what then?  Say we have an exact sequence
$$M’ \overset α \longrightarrow M \overset β \longrightarrow M’’,$$
which we split up to receive a diagram like shown in the link above.  Next we apply $F$ to end up with a diagram of the same shape with exact diagonals – how does it follow that the vertical arrows yield an exact sequence?
Or can we prove that functors preserving short exact sequences already preserve kernels and cokernels (without the use of additivity)?  Or is this result only true for additive functors?
 A: Proceeding as you say, you have a diagram
$$\begin{matrix}
\ddots &   & \vdots &  &  &  & &  &   \\
 &  \searrow & \downarrow&  &  &  & &  &  \\
 &  & A_{i-1}&  &  &  & 0&  &   \\
 &  & \pi \downarrow & \searrow^d &  &  & \downarrow &  &  \\
0 & \rightarrow & K_i & \overset{\iota}{\longrightarrow} & A_i &  \overset{\pi}{\longrightarrow} & K_{i+1} & \rightarrow & 0 \\
 &  & \downarrow &  &  & \searrow^d & \iota \downarrow & &  \\
 &  & 0 &  &  & & A_{i+1} & &  \\
 &  &  &  &  & & \downarrow & \searrow & \  \\
 &  &  &  &  & & \cdots & & \ddots   \\
\end{matrix}$$
where you know rows and columns are exact and you want to deduce the diagonal is exact. (Sorry for rotating $45^{\circ}$ from your image, but I had to do it to get the diagram with the tools available here.)
Since $\pi$ is epic (think surjective), we know that $$\mathrm{Image}(A_{i-1} \overset{d}{\longrightarrow} A_{i}) = \mathrm{Image}(A_{i-1} \overset{\pi}{\longrightarrow} K_i \overset{\iota}{\longrightarrow} A_i) =\mathrm{Image}(K_i \overset{\iota}{\longrightarrow} A_i) $$
Since $\iota$ is monic (think injective), we know that
$$\mathrm{Ker}(A_{i} \overset{d}{\longrightarrow} A_{i+1}) = \mathrm{Ker}(A_{i} \overset{\pi}{\longrightarrow} K_{i+1} \overset{\iota}{\longrightarrow} A_{i+1}) =\mathrm{Ker}(A_{i} \overset{\pi}{\longrightarrow} K_{i+1}).$$
Since the horizontal row is exact, we know $\mathrm{Image}(K_i \overset{\iota}{\longrightarrow} A_i)=\mathrm{Ker}(A_{i} \overset{\pi}{\longrightarrow} K_{i+1})$. We deduce that $\mathrm{Image}(A_{i-1} \overset{d}{\longrightarrow} A_{i}) = \mathrm{Ker}(A_{i} \overset{d}{\longrightarrow} A_{i+1})$, as desired.
