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I am trying to understand quasi-isomorphisms in an exact category as defined via the mapping cylinder. I would like to know whether these form a category of weak equivalences in the sense of Waldhausen.

Some background.

Recall that in an exact category in the sense of Quillen we have a good notion of a long exact sequence that is simply a chain complex whose differentials factor through short exact sequences whose kernels and cokernels are also objects of the exact category. Call such sequences acyclic.

By the Gabriel-Quillen embedding theorem we can consider every exact category $\mathcal{E}$ as a subcategory of some abelian category $\mathcal{A}$, with short exact sequences inherited from $\mathcal{A}$. In general it is not the case that the acyclic sequences of $\mathcal{E}$ are exactly the long exact sequences of $\mathcal{A}$ whose objects are in $\mathcal{E}$ (consider the exact category of free $R$-modules sitting inside the abelian category of all $R$-modules, for example).

In an abelian category a morphism of chain complexes is a quasi-isomorphism iff its mapping cone is exact. Since we don't have homology in exact categories it is sensible to define quasi-isomorphisms via the mapping cone. But now there are two possible classes of quasi isomorphism I can choose, either $q_{1}$, the morphisms between chain complexes of exact categories whose mapping cones are acyclic in $\mathcal{E}$, or $q_{2}$, the morphisms whose mapping cone is a long exact sequence when viewed in the ambient abelian category.

Clearly $q_{1} \subseteq q_{2}$ but I see no reason for them to be equal. So, after all this set-up, my question is: is there a standard choice here? Do either form a category of weak equivalences in the sense of Waldhausen? I'd like to work with $q_{1}$ as it avoids extrinsic information about the exact category but I can pass to $q_{2}$ if necessary.

Any insights would be appreciated.

A bonus question, in case both form weak equivalences, do the corresponding $K$-theories differ? How?

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As an aside, I wanted to tag this as 'exact categories' but no such tag exists and I have insufficient reputation to create it. If someone thinks that they warrant a category then please feel free to tag it as such. –  Tom Harris Jun 26 '13 at 17:27
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Tom, there you go. –  Mariano Suárez-Alvarez Jun 26 '13 at 18:50
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1 Answer

up vote 2 down vote accepted

If $\mathcal E$ is karoubian, i.e. idempotent complete, then $q_1=q_2$ on bounded complexes. This was noticed in 1.11.8 of Thomason-Trobaugh's paper, see also 1.11.5. On unbounded complexes, $q_1\subsetneq q_2$ even in the karoubian case, e.g. in the category $\mathcal E$ of free $\mathbb Z/4$-modules, the complex

$$\cdots\rightarrow \mathbb Z/4\stackrel{2}\rightarrow \mathbb Z/4\stackrel{2}\rightarrow \mathbb Z/4\rightarrow \mathbb \cdots$$

is $\sim 0$ in $q_2$ but not in $q_1$. If $\mathcal E$ is the category of projective objects in an abelian category where any object has projective dimension $\leq d$, then you can check as an exercise that $q_1=q_2$.

If you're interested in $K$-theory then you're probably interested just in bounded complexes, but also in non-karoubian categories. You can combine the fact that $q_1=q_2$ on karoubian categories with the cofinality theorem to deduce that both weak equivalences yield the same $K$-theory in all cases.

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Thanks this is great. Just for clarification, is it clear that $q_{1}$ forms a category of weak equivalences? I'm unsure whether it is closed under compositions. –  Tom Harris Jun 26 '13 at 23:23
    
Thanks! The class $q_1$ is closed under compositions since the full subcategory of acyclic complexes is a triangulated subcategory of the homotopy category of complexes in $\mathcal E$, see Neeman's 'The derived category of an exact category'. You can also produce an elementary proof, but it would be a little bit involved. –  Fernando Muro Jun 27 '13 at 7:07
    
Sorry to return to this after the point, but how does this imply that $q_{1}$ is closed under compositions on the category of all chain complexes in $\mathcal{E}$? I don't have much experience with triangulated categories, I can only see that it is closed under compositions on those chain complexes that are already acyclic. –  Tom Harris Aug 13 '13 at 14:41
    
The mapping cone of a composition fits into an exact triangle with the mapping cones of the factors –  Fernando Muro Aug 13 '13 at 18:38
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