# 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?

• At the end of this answer to a related question, I post a link to something I wrote up (in Spanish) which contains exactly what you want. If you understand it and write it up in English and post it as an answer, I'd upvote it :) – Bruno Stonek Apr 8 '15 at 19:05
• @lenticcatachresis I neither speak nor read Spanish but I might just be able to guess my way through this with a dictionary. With David’s answer, though, this will be unnecessary. Many thanks eitherway! – k.stm Apr 8 '15 at 19:25

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.

• Thanks! This seems to work for the category of modules, say, where there is a notion of injectivity and kernels/cokernels can be defined as submodules rather than some limits/colimits. Does this approach generalise to arbitrary abelian categories? (Sorry for being so lazy not to think about it in detail.) – k.stm Apr 8 '15 at 19:06
• Modulo some size concerns which aren't relevant here, abelian categories are module categories so if you can prove the statement for module categories then you've proven it for abelian categories. See here or here. – Jim Apr 8 '15 at 19:09
• @Jim I know that, but that is hardly enlightening and not very satisfying unless you prove/understand Mitchell’s embedding theorem first which I haven’t and which seems to me too much of an effort for this problem. – k.stm Apr 8 '15 at 19:11
• Since you work in abelian categories, instead of 'injective / surjective' you can equally well use 'monomorphism / epimorphism' and the used properties hold. – Berci Apr 8 '15 at 19:12
• @Berci Thanks, so generally $\operatorname{ker}~if = \operatorname{ker}~f$ whenever $i$ is a monomorphism and $\operatorname{img}~fp = \operatorname{img}~f$ whenever $p$ is an epimorphism – which can be proven by checking the universal properties, right? – k.stm Apr 8 '15 at 19:16