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?



Quasi-isomorphisms in exact categories
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.


