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John Berman
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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, $[B]=[A]+[C]$.

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

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, $[B]=[A]+[C]$.

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, $[B]=[A]+[C]$.
added 49 characters in body
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John Berman
  • 889
  • 4
  • 11

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

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, $[B]=[A]+[C]$.

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?

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, $[B]=[A]+[C]$.

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

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, $[B]=[A]+[C]$.
added 49 characters in body
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John Berman
  • 889
  • 4
  • 11

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 $$K_0(\text{fCW}^I)\xrightarrow{ev}\bigoplus_{X\in I}\mathbb{Z}.$$$$K_0(\text{fCW}^I)\xrightarrow{ev}\bigoplus_{X\in I/\cong}\mathbb{Z}.$$ Question: Is this an isomorphism?

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, $[B]=[A]+[C]$.

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, these assemble into a homomorphism $$K_0(\text{fCW}^I)\xrightarrow{ev}\bigoplus_{X\in I}\mathbb{Z}.$$ Question: Is this an isomorphism?

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, $[B]=[A]+[C]$.

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?

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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John Berman
  • 889
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  • 11
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