Let $\mathcal{A}$ and $\mathcal{B}$ be Waldhausen or exact categories, so that we can take the $K$-theory spectrum of $\mathcal{A}$ and $\mathcal{B}$. An exact functor $F: \mathcal{A} \to \mathcal{B}$ induces a morphism of $K$-theory spectra $K(\mathcal{A}) \to K(\mathcal{B})$.

Under what conditions does a non-exact functor $\mathcal{A} \to \mathcal{B}$ induce a morphism of $K$-theory spaces $\Omega^\infty K(\mathcal{A}) \to \Omega^\infty K(\mathcal{B})$?

To give an example, suppose $\mathcal{A}$ is the category of finite sets and $\mathcal{B}$ is the category of finite $C_p$-sets. Then there is a functor $\mathcal{A} \to \mathcal{B}$ given by multiplicative induction (it sends a set $S$ to $\mathrm{Map}(C_p, S)$) and that should induce a morphism on $K$-theory spaces. This gives a map $\Omega^\infty S^0 \to \Omega^\infty S^0_{C_p}$, which should arise from the Hill-Hopkins-Ravenel "multiplicative norm" construction. (This sort of construction is used in Segal's paper on the Kahn-Priddy theorem in the 1974 "New developments in topology.")

Has this been written down somewhere?

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    $\begingroup$ If you use the construction of K-theory of Cisinski and Blumberg-Mandell, which doesn't use cofibrations, you see that it's enough that your functor preserves weak equivalences. $\endgroup$ – Fernando Muro Jun 19 '16 at 11:43
  • $\begingroup$ Sorry, the functor should also preserve homotopy push-out squares and the zero object (up to homotopy). $\endgroup$ – Fernando Muro Jun 23 '16 at 19:21
  • $\begingroup$ @FernandoMuro: I think that's exactly the problem. The non-exact functors I have in mind need not preserve homotopy pushouts. $\endgroup$ – Akhil Mathew Jun 26 '16 at 0:59

I think the idea should be that "polynomial maps" from A to B induce maps on the K-theory spaces. Here's one possible suggestion of a way to implement this, based on Segal's proof of the Kahn-Priddy theorem.

First I should pick a context for algebraic K-theory. For the purpose of getting the idea across let me take the simplest: $A$ will be an $E_\infty$-space, and K-theory will be group-completion $A \mapsto A^{gp}$. Though the outline should work in any context. However, I should stress that I haven't checked the details of any of this; I'm more just brainstorming.

There is a tensor product on E_\infty-spaces satisfying the obvious universal property. In particular we can construct the "cofree coalgebra" on $A$, namely

$$Power(A) = \prod_{n\geq 0} (A^{\otimes n})^{S_n}.$$

E.g., when $A$ is finite sets under disjoint union, $Power(A)$ is functors from finite sets to finite sets, under pointwise disjoint union.

Consider furthermore the obvious diagonal map (of spaces)

$$exp: A\rightarrow Power(A).$$

Now, we can define a map $A\rightarrow B$ to be pre-polynomial if it arises as the composition of $exp$ with an $E_\infty$ map $Power(A)\rightarrow B$, and define it to be polynomial if the composition $A\rightarrow B\rightarrow B^{gp}$ is pre-polynomial. I believe this definition covers the examples that at least I have in mind.

Then the claim is that a polynomial map induces a map of spaces $A^{gp}\rightarrow B^{gp}$ in a natural way. But there is a subtlety here: More properly one needs a polynomial structure on a map, meaning a factorization through exp. There can be more than one such structure on any a given map $A\rightarrow B$: for an example, which I have a sneaking suspicion you're familiar with, see the bottom of this post. It is then a crucial question whether or not the induced map on group completions depends on this extra structure -- hopefully not.

To prove the claim, it clearly suffices to handle the universal case of $exp: A\rightarrow Power(A)$. For this, we can note that $Power(A)$ is a commutative algebra object of $E_\infty$-spaces under tensor product, and $exp$ turns the addition in $A$ to multiplication in $Power(A)$. It follows that the group-completion of $Power(A)$ is an $E_\infty$-ring, and there is a map of spaces $$exp: A\rightarrow Power(A)^{gp}$$ which is $E_\infty$ for the multiplicative structure on $Free(A)^{gp}$.

Now, the crucial claim is that this map in fact lands in the units of $Power(A)^{gp}$, so that it automatically extends to the group completion. This is is motivated by the fact that $e^T$ is invertible as a formal power series with inverse $e^{-T}$, but it needs to be checked rigorously.

And then of course more work needs to be done to express and prove functoriality, and one should think about the business of polynomial vs. polynomial structure, etc. But I hope this might provide some sort of guideline, even if it's not fully expressed.

An example of many different polynomial structures on the identity map $fSets\rightarrow fSets$: for any $n>0$, project $Power(fSets)$ to its $n^{th}$ component to get a finite set with $S_n$-action, then take $S_n$-fixed points.

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