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Suppose $I$ is a finite $\infty$-category and $F:I\rightarrow\text{fCW}$ is a functor that takes values in finite CW complexes. For each $X\in I$, let $[F(X)]$ denote the class of $F(X)$ in $K_0(\text{fCW})\cong\mathbb{Z}$.

Since evaluation at $X$ is a right exact functor, $[F(-)]$ is a homomorphism $K_0(\text{fCW}^I)\rightarrow\mathbb{Z}$. As we let $X$ vary over equivalence classes of objects in $I$, these assemble into a homomorphism $$K_0(\text{fCW}^I)\xrightarrow{ev}\bigoplus_{X\in I/\cong}\mathbb{Z}.$$ Question: Is this an isomorphism?

Remark: There is a splitting like this one when fCW is replaced by abelian/stable categories. See for example https://arxiv.org/abs/0908.3417

Edit: Tom Goodwillie points out this is not the isomorphism one would expect, but I am still curious whether $K_0(\text{fCW}^I)$ splits according to the objects of $I$ (and maybe their automorphism groups, as in the paper linked above), or whether it can be computed at all in general.

K-theory can mean so many things, I want to be clear: When I write $K_0(\mathcal{C})$, I mean the abelian group supporting a universal function $[-]:\mathcal{C}\rightarrow K_0(\mathcal{C})$ such that:

  • if $X\cong Y$, then $[X]=[Y]$,
  • $[X\amalg Y]=[X]+[Y]$,
  • if $A\rightarrow B\rightarrow C$ is a cofiber sequence, $[B]=[A]+[C]$.
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  • $\begingroup$ I'm not sure I understand your definitions, but it seems that if $I$ is an ordinary category then the left hand side is invariant under equivalence of categories while the right hand side depends on the object set. $\endgroup$
    – Tom Goodwillie
    Commented Oct 30, 2018 at 16:08
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    $\begingroup$ You're right! I meant to sum over each equivalence class of $X\in I$. I will correct that. $\endgroup$
    – John Berman
    Commented Oct 30, 2018 at 16:29
  • $\begingroup$ If you take $I$ to be the standard simplicial circle then I think the functor category has infinitely many isomorphism classes. However the indexing simplicial set is not a quasi-category, so this is maybe not what you want. $\endgroup$
    – K.J. Moi
    Commented Oct 31, 2018 at 1:15

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If $I = \mathbb Z/2\mathbb Z$ counts as a finite $\infty$ category (Edit: it doesn't), then the answer is no. The space $* \sqcup *$ admits both a free and a trivial $I$ action. We have an additive map given by taking homology and pairing with the alternating representation $$X \mapsto \chi(H^*(X, \mathbb C), {\rm alt}) = \sum_{i} (-1)^i \dim (H^i(X, \mathbb C) \otimes{ \rm alt})^{\mathbb Z/2}$$ which distinguishes between the two actions on $* \sqcup *$.

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    $\begingroup$ It's a finite category, but not a finite $\infty$-category, because the nerve $\mathbb{R}P^\infty=B\mathbb{Z}/2\mathbb{Z}$ is not a finite cell complex. If you want to think about ordinary categories, I am interested in categories like finite posets, or groups like $\mathbb{Z}$, $\mathbb{Z}\times\mathbb{Z}$, or $\mathbb{Z}\ast\mathbb{Z}$. (Finiteness is rather different for categories and $\infty$-categories.) $\endgroup$
    – John Berman
    Commented Oct 30, 2018 at 20:29
  • $\begingroup$ OK thanks, I thought finite might mean postnikov finite instead of CW finite. $\endgroup$
    – Phil Tosteson
    Commented Oct 30, 2018 at 21:14
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    $\begingroup$ Even for $I=\mathbb Z$ the answer is no. A diagram is a finite complex $X$ with automorphism $a:X\to X$, right? And there are at least two independent maps from the $K$-group to $\mathbb Z$, the Euler characterisitc of $X$ and the Lefschetz number of $a$. $\endgroup$
    – Tom Goodwillie
    Commented Oct 30, 2018 at 23:41
  • $\begingroup$ Great, I think this right. I also think answer would change if we took the K theory of finitely generated (i.e. compact) $I$ spaces. $\endgroup$
    – Phil Tosteson
    Commented Oct 31, 2018 at 0:06
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    $\begingroup$ @MarcHoyois $B\Bbb Z$ turns out to be finite. There's a presentation of $B\Bbb Z$ with one object, three morphisms $l, f, r$, and two homotopies $lf = id$, $fr = id$. (If you demand that the left inverse and right inverse are equal then you get the wrong answer and have to add higher cells; this is related to $B\Bbb Z$ not having a finite $C_2$-equivariant description) $\endgroup$
    – Tyler Lawson
    Commented Nov 2, 2018 at 9:45

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