$K$-theory is often billed as the "universal way to split exact sequences". But it seems we're too anxious to group-complete things to actually take the slogan at face value.

Consider the following $\infty$-categories:

  • $\mathcal{W}$ - Waldhausen categories (or Waldhausen $\infty$-categories, if you prefer)

  • $\mathcal{C}_1$ - symmetric monoidal $\infty$-categories

  • $\mathcal{C}_2$ - $E_\infty$-spaces

  • $\mathcal{C}_3$ - infinite loop spaces

Say that a functor $F: \mathcal{W} \to \mathcal{C}_i$ is additive if $F\mathcal{E} W\to F W \times F W$ is an equivalence for all $W \in \mathcal{W}$, where $\mathcal{E}W$ the Waldhausen category of exact sequences $w' \to w \to w''$ in $W$ and the map projects onto $(w',w'')$. Consider the following functors:

  • $K_\oplus^1: \mathcal{W} \to \mathcal{C}_1$ - sending $W$ to its simplicial localization, with $E_\infty$ structure given by coproduct
  • $K_\oplus^2: \mathcal{W} \to \mathcal{C}_2$: - sending $W$ to the nerve of its category of weak equivalences (the core of $K_\oplus^1 W$)
  • $K_\oplus^3 : \mathcal{W} \to \mathcal{C}_3$: - sending $W$ to the group completion of $K_\oplus^2 W$

The universal property of $K$-theory is that it is a functor $K^3: \mathcal{W} \to \mathcal{C}_3$ which constitutes a reflection of $K^3_\oplus$ into the category of additive functors $\mathcal{W} \to \mathcal{C}_3$ (I think this is Clark Barwick's formuation). Analogously, I ask:


  1. Does $K^1_\oplus$ admit a reflection $K^1$ into the category of additive functors $\mathcal{W} \to \mathcal{C}_1$? If so, is it modeled by a variant of the $S_\bullet$ construction?

  2. Does $K^2_\oplus$ admit a reflection $K^2$ into the category of additive functors $\mathcal{W} \to \mathcal{C}_2$? If so, is it modeled by the $S_\bullet$ construction itself?

  • 1
    $\begingroup$ I believe that the answer to 2 is yes and that this is in Waldhausen's foundational paper (usually phrased as saying that one iterate of the S. construction suffices). However the S. construction implicitly deloops once. If you want to recover K-theory to have to take loops on S., making it difficult to get something that's not group-complete. $\endgroup$ Jan 3, 2017 at 16:45
  • $\begingroup$ @TylerLawson When I say "$K$ is a reflection of $K_\oplus$", I mean there is a map $K_\oplus \to K$ with the obvious universal property. I see that Waldhausen does prove that $S_\bullet$ is additive, but I don't see him proving a universal property. I see your point about $E_\infty$-valued "loops" -- I suppose one way to do it would be to fibrantly replace $wS_\bullet \mathcal{C}$ with a 1-object Segal space (regarding the "$\bullet$" direction as the "categorical" direction) and then take the endomorphism object, although getting the full $E_\infty$ structure will still take more work. $\endgroup$
    – Tim Campion
    Jan 3, 2017 at 18:30
  • $\begingroup$ Hi Tim. I don't quite understand the functor in bullet point 1. Let's say we consider the Waldausen category $Perf_k$. Maybe we regard this as a discrete Waldhausen category (though you can also do $\infty$-Waldhausen). What do you send this to? Something like invert all the weak equivalences or something? $\endgroup$ Jun 13, 2017 at 3:27
  • $\begingroup$ Hi Elden! Yeah, I just mean "localize at the weak equivalences" (in the $\infty$-categorical sense). So this is the most vanilla of the above functors, the only twist is remembering the symmetric monoidal structure. If you model $\infty$-categories by simplicially-enriched categories, it's Dwyer-Kan localization; if you model $\infty$-categories by marked simplicial sets, it's fibrant replacement of (nerve of category, weak equivalences), etc. In particular, it forgets the cofibration part of the Waldhausen structure (though cofibrations still figure in the definition of $\mathcal{E} W$). $\endgroup$
    – Tim Campion
    Jun 13, 2017 at 4:20

1 Answer 1


I'm not sure about Waldhausen categories in general, but if you restrict attention to stable $\infty$-categories (with trivial Waldhausen structure in which all maps are cofibrations) then group completion is essentially forced on you by the splitting of exact sequences. In principle, this happens because for every object $X \in {\cal C}$, splitting the exact sequence $$ X \to 0 \to \Sigma X $$ means that $\Sigma X$ becomes the inverse of $X$. In particular, taking the maximal subgroupoid of a stable $\infty$-category ${\cal C}$ (considered as an $\mathbb{E}_\infty$-monoid in spaces with respect to direct sum), and then forcing the relations given by splitting of exact sequences, yields an $\mathbb{E}_\infty$-monoid which is already group-like.

This idea can also be phrased from the perspective of additivity. Observe first that the $\infty$-category ${\rm Cat}^{\rm ex}_\infty$ of small stable $\infty$-categories is semi-additive in the sense that it has a zero-object (an initial object which is also final) and biproducts (coproducts which are also products). Since the $\infty$-category of $\mathbb{E}_\infty$-monoids is also semi-additive it is somewhat natural to restrict attention to functors $F:{\rm Cat}^{\rm ex}_\infty \to {\rm Mon_{\mathbb{E}_\infty}}$ which are themselves semi-additive, in the sense that they preserve zero-objects and biproducts. Note that so far we have not enforced any group-likeness condition. Note also that such a semi-additive $F$ will be additive in the sense you describe if and only if it sends the map $p:{\cal EC} \to {\cal C}\times{\cal C}$ given by $[x' \to x \to x''] \mapsto (x',x'')$ to an equivalence of $\mathbb{E}_\infty$-monoids. We now claim the following: if a semi-additive functor $F:{\rm Cat}^{\rm ex}_\infty \to {\rm Mon_{\mathbb{E}_\infty}}$ is additive then it automatically takes values in group-like $\mathbb{E}_\infty$-spaces. To see this observe that when ${\cal C}$ is stable the $\infty$-category ${\cal EC}$ of fiber sequences has an automorphism $T: {\cal EC} \to {\cal EC}$ sending $[x'\to x \to x'']$ to $[x \to x'' \to {\rm cof}(x\to x'')]$. Furthermore, the functor $p:{\cal EC} \to {\cal C}\times{\cal C}$ above has a one-sided inverse $i: {\cal C} \times {\cal C} \to {\cal EC}$ sending $(x,y)$ to $[x \to x\oplus y \to y]$. In particular, if $F(p)$ is an equivalence then $F(i)$ is an equivalence as well. It then follows that if $F$ is additive then it sends the composed functor $$ {\cal C} \times {\cal C} \stackrel{i}{\to} {\cal EC} \stackrel{T}{\to} {\cal EC} \stackrel{p}{\to} {\cal C} \times {\cal C} \stackrel{\sigma}{\to} {\cal C} \times {\cal C} $$ to an equivalence, where $\sigma$ is the equivalence $\sigma(x,y) = (\Omega y,x)$. But this composed functor is just the shear map $(x,y) \mapsto (x,x\oplus y)$ and since $F$ is semi-additive it follows that the shear map of the $\mathbb{E}_\infty$-monoid $F({\cal C})$ is an equivalence, i.e., $F({\cal C})$ is group-like.

  • $\begingroup$ Thanks - this is beautiful! So you're observing that $K^2_\oplus = K^3_\oplus$ when we restrict the domain to stable $\infty$-categories, which basically answers Question 2. $\endgroup$
    – Tim Campion
    Jun 24, 2018 at 19:13
  • $\begingroup$ The same argument also shows that $K^1_\oplus(\mathcal C)$ is grouplike. I once gave an argument that if $Id \Rightarrow F$ is a strong monoidal transformation of monoidal endofunctors of symmetric monoidal categories, and if $C$ is a monoid in semiadditive categories, and if $F(C)$ is grouplike, then $F(C)$ is trivial -- in particular, this should apply to $F = K^1_\oplus$, partially answering Question 1. Do you know of weaker conditions implying that $K^1_\oplus(C)$ is trivial? $\endgroup$
    – Tim Campion
    Jun 24, 2018 at 19:13
  • $\begingroup$ Yonatan's observation is actually reminiscent of John Berman's discussion of what happens when you have a covariantly functorial negation operation on a symmetric monoidal category. I think the key here is that a stable $\infty$-category has a covariant endofunctor (namely $\Sigma$) which gets identitifed with the negation operation by the $K$-theory functor. $\endgroup$
    – Tim Campion
    Jun 24, 2018 at 19:50
  • $\begingroup$ Sorry, where I said above that $K^1_\oplus(\mathcal C)$ is grouplike, I meant to say that the additive reflection of $K^1_\oplus$ applied to $\mathcal C$ is grouplike, and hence by the linked argument trivial. $\endgroup$
    – Tim Campion
    Jun 25, 2018 at 3:01
  • $\begingroup$ Yes, I agree. Indeed, first a similar argument shows that any semi-additive additive functor from stable $\infty$-categories to symmetric monoidal $\infty$-categories takes values in group like symmetric monoidal $\infty$-categories. Then one can show that $F(\Sigma): F(C) \to F(C)$ provides a covariant inverse functor which means that $F(C)$ is in fact an $\infty$-groupoid. $\endgroup$ Jun 25, 2018 at 7:40

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