$\DeclareMathOperator{\id}{id}$Let $C$ be an abelian category and suppose we have the following diagram in $C$ $\require{AMScd}$
\begin{CD} @. 0 @. 0 @. 0 \\ @. @VVV @VVV @VVV \\ 0 @>>> X_0'@>c'_0>>X_0@>c_0>>X_0'' \\ @. @Va'VV @VaVV @Va''VV \\ 0 @>>> X_1'@>c'_1>>X_1@>c_1>>X_1'' \\ @. @Vb'VV @VbVV @Vb''VV \\ 0 @>>> X_2'@>c'_2>>X_2@>c_2>>X_2'' \\ \end{CD}
$\DeclareMathOperator{\Img}{Im}$Furthermore, assume the three rows are exact and the second and third columns are exact. My goal is to show that the first column must then be exact as well. I think I have a proof but at some point I use the Freyd-Mitchell theorem in order to be able to work with elements of modules. For most of the proof I think I manage just fine without it. Let's see:
- $\ker a' = 0$: Let $\iota:\ker a'\to X_0'$ be the canonical morphism. We have $c_1'a'\iota = 0$ (because by definition $a'\iota = 0$) but this implies $ac'_0\iota=0$. By exactness, $a$ is a monomorphism hence $c'_0\iota = 0$ and again by exactness we get $\iota=0$ whence $\ker a'=0$. (Here I manage without the embedding theorem)
- $\ker b' = \Img a'$ : By exactness $ba=0$ hence $bac_0'=0$. This implies $c_2'b'a'=0$ but by exactness we have $b'a'=0$ hence $a'$ factorises through $\ker b'$. So we have the following diagram :
$\require{AMScd}$ \begin{CD} X_0' @>e>> \Img a' @>m>> X'_1\\ @| @VuVV @|\\ X_0' @>>> \ker b' @>j>> X'_1 \end{CD}
where both $m,j$ are monic. The morphism $u$ exists by universal property of the image.
Now to finish the proof I would need another morphism $v:\ker b' \to \Img a'$ and argue that $uv = \id$ and $vu=\id$. But this is where my categorical argument stops, I can show that in $R$-$Mod$ $\ker b' \subset \Img a'$ but that's it. To be entirely fair, even the previous part, i.e. finding $u:\Img a'\to \ker b'$ I had originally used elements and concluded that $\Img a' \subset \ker b'$ but I realised as I was writing this post that I could adapt my argument.
So finally here's how I can show what I want using the embedding theorem :
Take $x\in \ker b'$, i.e. $b'(x)=0$, then $c'_2b'(x)=0=bc'_1(x)$, by exactness $\ker b = \Img a$ whence $c'_1(x)=a(y)$ for $y\in X_1$. By commutativity of the diagram $a''c_0(y) = c_1a(y)=c_1c_1'(x)=0$ where the last equality follows again by exactness. but $a''$ is monic hence $c_0(y)=0$ and by exactness this means $y=c'_0(z)$. Finally $a(y)=ac_0(z)=c_1'a'(z)=c_1'(x)$ and since $c_1'$ is monic $a'(z)=x$ and we're done.
So first, are each individual part of the argument correct ? Second, how would one carry out the last part without using elements, with a categorical argument ? Any help is appreciated.