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I'm wondering whether the category whose objects are short exact sequences of abelian groups, and whose morphisms are commutative diagrams of such short exact sequences, is cocomplete. Working naively, it seems you can get coproducts by taking them componentwise. However, for coequalizers, I think we are not so lucky. Consider the two short exact sequences $0 \rightarrow 0 \rightarrow \mathbb{Z} \rightarrow \mathbb{Z} \rightarrow 0$ and $0 \rightarrow \mathbb{Z} \rightarrow \mathbb{Z} \rightarrow 0 \rightarrow 0$ with maps between integers being the identity. Consider the morphism from the first sequence to the second which is the identity on the middle map between the integers and obviously zero elsewhere. Then taking cokernels componentwise would give $0 \rightarrow \mathbb{Z} \rightarrow 0 \rightarrow 0 \rightarrow 0$, which clearly cannot be exact.

So the obvious candidate for cokernels is not the correct one, but perhaps there is another not-so-obvious choice. I'm wondering how I might go about showing whether or not there are coequalizers.

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At least one can prove that the cokernel exists in your example. It is given by the zero sequence. (exercise) – Martin Brandenburg Oct 24 '12 at 17:55
Have you tried the short exact sequence 0->coker/ker->coker->coker->0 from the snake lemma? That seems to fit in with Martin's comment.... – Jason Polak Oct 24 '12 at 18:03
up vote 6 down vote accepted

This is an expansion of Jason's comment.

The category of short exact sequences is obviously additive and has arbitrary coproducts. Thus, cocompleteness is equivalent to the existence of cokernels. So let us given a morphism $f_* : A_* \to B_*$ of short exact sequences:

$\begin{array}{ccccccccc} 0 & \rightarrow & A_1 & \rightarrow & A_2 & \rightarrow & A_3 & \rightarrow & 0 \\ & & ~ ~ \downarrow f_1 & &~~ \downarrow f_2 & & ~~ \downarrow f_3 & & \\ 0 & \rightarrow & B_1 & \rightarrow & B_2 & \rightarrow & B_3 & \rightarrow & 0 \end{array}$

The snake lemma gives us an exact sequence

$0 \to \mathrm{ker}(f_1) \to \mathrm{ker}(f_2) \to \mathrm{ker}(f_3) \to \mathrm{coker}(f_1) \to \mathrm{coker}(f_2) \to \mathrm{coker}(f_3) \to 0.$

Let $K$ be the kernel of $\mathrm{coker}(f_1) \to \mathrm{coker}(f_2)$. Equivalently, $K \cong \mathrm{ker}(f_3) / (\mathrm{ker}(f_2) / \mathrm{ker}(f_1))$. Then we have a short exact sequence $C_*$ defined by

$0 \to \mathrm{coker}(f_1) /K \to \mathrm{coker}(f_2) \to \mathrm{coker}(f_3) \to 0$

together with a morphism $p_* : B_* \to C_*$. I claim that this is the cokernel of $f_*$. It is an epimorphism since it components are epimorphisms. It factors through the non-exact cokernel of $f_*$, therefore we have $p_* f_* = 0$. Now let $D_*$ be another exact sequence and $g_* : B_* \to D_*$ be a morphism satisfying $g_* f_*$. Then $g_*$ factors through the non-exact cokernel of $f_*$.

We are left to prove that $g_1 : B_1 \to D_1$ vanishes on $K$, so that it even factors through $C_1= \mathrm{coker}(f_1) /K$. But this is a diagram chase: An element in $K$ is represented by an element in $B_1$ whose image in $B_2$ comes from an element in $A_2$. Thus the image in $D_2$ vanishes. Since $0 \to D_1 \to D_2$ is exact, this means that the image in $D_1$ already vanishes, qed.

Of course this reasoning can also be done with arrows. Therefore, if $\mathcal{A}$ is an arbitrary abelian category, then the category $S(\mathcal{A})$ of short exact sequences in $\mathcal{A}$ has cokernels and (by duality also) kernels. If $\mathcal{A}$ has coproducts and satisfies has AB4, then $S(\mathcal{A})$ has arbitrary coproducts and is therefore cocomplete.

The category of short exact sequences is never abelian, in fact not balanced: From the above description of cokernels we see that $f_*$ is an epimorphism in that category iff $f_2,f_3$ are epimorphisms. Intuitively it is ok that $f_1$ doesn't appear here since $f_1$ is uniquely determined by $f_2,f_3$ due to the exactness! Similarily, $f_*$ is a monomorphism iff $f_1,f_2$ are monomorphisms. Thus, $f_*$ is a mono- and an epimorphism iff $f_2$ is an isomorphism, $f_1$ is a monomorphism and $f_3$ is an epimorphism. However, $f_*$ is an isomorphism iff $f_1,f_2,f_3$ are isomorphisms.


We can simplify these arguments a lot: Consider the category $E(\mathcal{A}) \subseteq \mathrm{Mor}(\mathcal{A})$ of epimorphisms in $\mathcal{A}$. The morphisms are commutative diagrams. Obviously, coproducts and cokernels exist in this category, since the category is closed under these operations in $\mathrm{Mor}(\mathcal{A})$. There is a forgetful functor $S(\mathcal{A}) \to E(\mathcal{A})$, which turns out to be an equivalence of categories! The quasi-inverse chooses for every epimorphism $A_2 \to A_3 \to 0$ a kernel $0 \to A_1 \to A_2$. For every morphism $(f_2,f_3) : A_* \to B_*$ in $E(\mathcal{A})$ the universal property of the kernel yields a unique $f_1 : A_1 \to B_1$ such that $(f_1,f_2,f_3)$ becomes a morphism in $S(\mathcal{A})$. Since $E(\mathcal{A})$ is cocomplete, the same is true for $S(\mathcal{A})$. If one unwinds the definitions, one gets the same cokernels as described above explicitly.

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$S(\mathcal{A})$ is never abelian if $\mathcal{A} \neq 0$. It is easy to show that a morphism $(a,b,c)$ of short exact sequences is monic if $a,b$ are monic and, dually, it is epic if $b,c$ are epic (that's because: if the middle map and one of the outer maps is zero then so is the third). Taking the morphism $(0,1,0)$ between $(0 \to A \to A)$ and $(A \to A \to 0)$ with nonzero $A$ you find a monic and epic morphism that isn't an isomorphism. However, $S(A)$ is exact in the sense of Quillen with respect to the short exact sequences given by $3 \times 3$-diagrams with exact rows and columns. – commenter Oct 24 '12 at 19:23

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