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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.

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2 Answers 2

up vote 6 down vote accepted

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.

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Steve, thanks for the argument, like it. In fact, it shows: If the category is Ab (Hom's are abelian groups with bilinear composition), $p$ is the cokernel of $i$ and $q$ the cokernel of $j$ then the statement is true. However, if there is no additive structure on the Hom's your argument doesn't seem to work. For, then we start with $fqi = f^'qi$ and the universal property of $p$ couldn't be used in the same way as above. –  Ralph Mar 10 '11 at 6:59
For the sake of completeness: A category is exact (in the sense of Mitchell) if (i) kernels and cokernels exist, and (ii) each monomorphism is a kernel and dually each epimorphism is a cokernel, and (iii) each morphism can be written as a composition of an epimorphism followed by a monomorphism. –  Ralph Mar 10 '11 at 7:10
Ralph: yes, of course, you're quite right, this argument needs the category to be enriched over abelian groups, not just over pointed sets. Is this part of the Mitchell definition? –  Steve Lack Mar 10 '11 at 8:09
No, this isn't part of the definition. Informally, Mitchell's exact category is just a plain framework to work reasonable with exact sequences, i.e. such that $A \to B \to 0$ is exact iff $A \to B$ is epi, $0 \to A \to B$ is exact iff $A \to B$ is mono, etc. For instance, an abelian category is an exact additive category. –  Ralph Mar 10 '11 at 8:45
OK, thanks Ralph, I've now given a fix. –  Steve Lack Mar 10 '11 at 9:38

In the Mitchell book I.16.8 (page 24) is a Corollary of the $3\times3$ (or nine) Lemma, and this Lemma is true in a exat category (see the Book: H. Shubert "CAtegories" p. 136), then I see that Mitchell dont use addittivity (for demostrate the corillary from the Lemma) then the corollary follow also for exat categories.

Is you dont have the H. Schubert Book I can post a proof here.

PS. for "exat category" Shubert mean: pointed, with $(regular-Epi, regular-Mono)$ factorization with finite limits and colimits.

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Thanks for your hint, Sergio. Mitchell also offers a nine lemma (that's Proposition 16.1 on page 20). But the proof of the above 16.8 says: "Form the diagram (2) according to 16.5." Thereafter the proof relies on commutativity and exactness of (2). This diagram (2) is $3 \times 3$ (3 columns, 3 rows) and 16.5 says: "The diagram is commutative with exact rows and columns if and only if II [the left upper square] is a pullback, IV [the right lower square] is a pushout ...". –  Ralph Mar 10 '11 at 23:58
(continue) So it's not obvious (to me at least) that such a commutative diagram with exact rows and columns can be formed in general (while it's obvious if the category has pullbacks and pushouts). Probably I'm missing something or in the situation of 16.8 the pushout and pullback needed for (2) can be constructed directly. Apart from that I more like Steve's direct proof rather than Mitchell's nested one (16.8 uses 16.5 that uses 13.2). –  Ralph Mar 10 '11 at 23:59
Ralph, you can still construct the 3x3. Factorize $pj:B'\to C$ as an epi $p':B'\to C'$ followed by a mono $k:C'\to C$. Form the cokernel $r:C\to C''$ of $k$, and the kernel $i':A'\to B'$ of $p'$. Then $ji':A'\to B$ factorizes through $i$ as some mono $h:A'\to A$, and $rp:B\to C''$ factorizes through $q$ as some epi $p'':B''\to C''$. Form the cokernel $s:A\to A''$ of $h$. Then $qi:A\to B''$ factorizes through $s$ as some $i'':A''\to B''$. Now (i) check that $i''$ is mono, and (ii) check that $p''$ is the cokernel of $p''$. –  Steve Lack Mar 11 '11 at 10:50
Steve, thanks for the construction. In fact that's a nice exercise in using kernels, cokernels and the universal properties of monos and epis. –  Ralph Mar 11 '11 at 14:28

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