## Cofinal inclusions of Waldhausen categories

### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points)

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 $K(\mathcal{B})\rightarrow K(\mathcal{C})$?

-

Probably any exact functor is a subcategory up to K-equivalence, so it is probably not possible to say anything. – Ben Wieland May 12 2010 at 19:21

## 2 Answers

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).

-

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 2010 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 2010 at 18:12

### You can accept an answer to one of your own questions by clicking the check mark next to it. This awards 15 reputation points to the person who answered and 2 reputation points to you.

This might not answer your question exactly, but hopefully it might nevertheless shed some light on the situation. This paper,

Sagave, Steffen On the algebraic K-theory of model categories. J. Pure Appl. Algebra 190 (2004), no. 1-3, 329--340

explores the relation between Quillen model categories and Waldhausen categories. Obviously the two notions are very close. Basically, in a model category, the full subcategory of finite or homotopically finite objects forms a Waldhausen category, and Quillen equivalences induce K-theory equivalences. One of the ideas of the paper is that when a Waldhausen category comes from a model category then you get a bit of extra structure on it (coming from the fact that you can do cofibrant replacements), and this extra structure allows you to prove under various mild assumptions that a functor of Waldhausen categories is a K-theory equivalences.

In particular, look at the Approximation Theorem (Thm 2.8), which gives a nice criterion for an exact functor to be a K-theory equivalence.

-
{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 39, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9237678647041321, "perplexity_flag": "head"}