# Cofinal inclusions of Waldhausen categories

Let $\mathcal{C}$ be a Waldhausen category. Suppose that $\mathcal{B}$ is a subcategory of $\mathcal{C}$, and that $\mathcal{B}$ is closed under extensions. If $\mathcal{B}$ is strictly cofinal in $\mathcal{C}$ (in the sense that given any $C\in \mathcal{C}$ there exists a $B\in \mathcal{B}$ such that $C\amalg B\in \mathcal{B}$), can we say anything about $K(\mathcal{B}) \rightarrow K(\mathcal{C})$?

In Waldhausen's paper "Algebraic K-theory of spaces" Waldhausen claims that the inclusion $\mathcal{B}\rightarrow \mathcal{C}$ induces a weak equivalence $wS_\bullet \mathcal{B}\rightarrow wS_\bullet\mathcal{C}$ (and thus an equivalence on K-theories), but I'm not sure that this is right, as $\mathcal{B}$ does not need to be a full subcategory. In particular, if there are objects $C,C'$ which are in $\mathcal{B}$ but are not isomorphic in $\mathcal{B}$ they may well be isomorphic (or at least weakly equivalent) in $\mathcal{C}$.

Consider the following example. Let $\mathcal{C}$ be the category of pairs of pointed finite sets, whose morphisms $(A,B)\rightarrow (A',B')$ are pointed maps $A\vee B\rightarrow A'\vee B'$, and let $\mathcal{B}$ be the category of pairs of pointed finite sets whose morphisms $(A,B)\rightarrow (A',B')$ are pairs of pointed maps $A\rightarrow B$ and $A'\rightarrow B'$. We make $\mathcal{C}$ a Waldhausen category by defining the weak equivalences to be the isomorphisms, and the cofibrations to be the injective maps. $\mathcal{B}$ is clearly cofinal in $\mathcal{C}$, but $K_0(\mathcal{B}) = \mathbf{Z}\times \mathbf{Z}$, while $K_0(\mathcal{C}) = \mathbf{Z}$. Going even further, the Barratt-Priddy-Quillen theorem should tell us that $K(\mathcal{B}) = QS^0\times QS^0$, while $K(\mathcal{C}) = QS^0$.

If we add the condition that $\mathcal{B}$ needs to be a full subcategory of $\mathcal{C}$, then I believe that Waldhausen's paper is correct. But even without that, it is possible to say anything about the map $K(\mathcal{B})\rightarrow K(\mathcal{C})$?

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Probably any exact functor is a subcategory up to K-equivalence, so it is probably not possible to say anything. –  Ben Wieland May 12 '10 at 19:21

If I recall correctly, Waldhausen states this theorem in the context of a "subcategory with weak equivalences and cofibrations" of C. This is a somewhat stronger condition than having an exact inclusion functor from B to C; it stipulates that a map in B is a cofibration if it is a cofibration in C with cofiber in B, and a weak equivalence in B if it is a weak equivalence in C. In particular, the condition excludes the kind of example you're thinking about.

In general, given an exact functor $B \to C$ you can study the cofiber $K(B) \to K(C)$ as the $K$-theory of an explicit simplicial Waldhausen category (consisting of sequences $C_0 \to C_1 \to \ldots C_k$ where the cofibers $C_{i+1} / C_i$ are in B).

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I'm pretty sure that in my example B is a subcategory with weak equivalences and cofibrations. A map in B is a cofibration if it is a pair of injective maps... which when you take the unions of source and target are still injective, so the image of this under inclusion is still a cofibration. Analogously for a weak equivalence. And the cofiber of a map is just the coimage, which will be in B because you can take it componentwise. (The example was constructed to have exactly this property, so if I'm making an obvious mistake please point it out.) –  Inna May 5 '10 at 3:58
Yes, you're right; I'm sorry, I was mistaken in my assertion above. Presuming then that your counterexample stands, this raises two interesting questions: 1) Can you construct an example like this where the categories (or at least C) have factorization? 2) What does Waldhausen's cofiber theorem say in this setting? –  Andrew May 5 '10 at 18:12