In Mitchell's book "Theory of Categories", Corollary I.16.8 (page 24) states that the following holds in any exact category: 



> 
> 
> Let 
> $$
> 0 \to A \to B \to C \to 0 
> $$
> $$
> 0 \to B^' \to B \to B^{''} \to 0 
> $$
> be short exact sequences. Then $B^' \to B \to C$ is epi iff $A \to B \to B^{''}$ is epi. 
> 


It seems to me that Mitchell's proof requires the existence of pushouts and pullbacks. Therefore I wonder if the corollary actually is true for any exact category. Can someone acknowledge this corollary ? 


The reason why I think Mitchell's proof requires pushouts and pullbacks is as follows: In a first step the two short exact sequences are embedded crosswise into a commutative diagram with three short exact columns and three short exact rows. But according to Proposition I.16.5 such a diagram only exists if some of its squares are a pushout or a pullback. 



On a corollary in Mitchell's book 
I don't remember what an exact category is in this context, but in any case this seems to be true in any pointed category. Name the arrows $i:A\to B$, $p:B\to C$, $j:B'\to B$, and $q:B\to B''$, and suppose that $pj$ is epi. If
$fqi=0$ then by the universal property of $p:B\to C$ we have 
$fq=gp$ for a unique $g$. Now $gpj=fqj=f0=0$ and $pj$ is epi so $g=0$. Thus $fq=gp=0$,
but $q$ is epi so $f=0$. This proves that $qi$ is epi.


**Edit:** the comments below point out that this only shows that if $fqi=0$ then $f=0$. So this argument, as is, would need the category to be **Ab**-enriched. If instead it is exact, in the sense specified in the comments, then to finish I should prove:


**Lemma:** Let $h$ be an arrow with the property that if $fh=0$ then $f=0$. Then $h$ is epi. 


**Proof:** The assumptions tell us that the cokernel of $h$ is $0$. Factorize $h$ as $me$,
with $e$ epi and $m$ mono. Since $e$ is epi, we have $cok(m)=cok(me)=0$. But $m$ is a mono, so is the kernel of its cokernel, in this case 0, so $m$ must be invertible. This now proves that $h$ is epi.


