As we all know, the forgetful functor $\mathsf{Ab} \to \mathsf{CMon}$ from abelian groups to commutative monoids has a left adjoint, the Grothendieck group. I would like to categorify this construction.

For this, abelian groups should be replaced by symmetric monoidal categories in which every object is invertible, let's call them abelian $2$-groups (notice that $2$-groups usually are assumed to be groupoids; if possible I would like to avoid this assumption). And commutative monoids are replaced by symmetric monoidal categories. We have a forgetful functor of $2$-categories $\mathsf{2Ab} \to \mathsf{2CMon}$. I am pretty sure that it has a left adjoint (in the $2$-categorical sense) and how it looks like, but it is quite tedious to verify all the details. Therefore I would like to know if this construction is already known (probably!) and if it has been written down somewhere, so that I may cite it.

Here is a sketch of the construction: Given a symmetric monoidal category $S$, whose tensor product will be denoted simply by $(a,b) \mapsto ab$, we define a symmetric monoidal category $S^{-1} S$ as follows (not to be confused with Quillen's $S^{-1} S$!): Objects are pairs of objects of $S$. One should think of $(a,b)$ as $a^{-1} b$. A morphism $(a,b) \to (c,d)$ is an equivalence class of morphisms $ebc \to ead$ for objects $e$, where two such morphisms $ebc \to ead$, $fbc \to fad$ are called equivalent if there are objects $e',f'$ and an isomorphism $e'e \to f'f$ such that $$\begin{array}{c} e'ebc & \rightarrow & e'ead \\ \downarrow && \downarrow \\ f'fbc & \rightarrow & f'fad \end{array}$$ commutes. The composition of morphisms is a little bit long, but is motivated by the transitivity proof in the construction of the usual Grothendieck group (one also has to check the category axioms ...). The tensor product is $(a,b) (c,d) := (ca,bd)$ on objects (one also has to define it on morphisms ...), the unit is $(1,1)$. Then $(a,b) (b,a) = (ba,ba) \cong (1,1)$ shows that every object is invertible. One has a symmetric monoidal functor $\iota : S \to S^{-1} S$, $a \mapsto (1,a)$. If $F : S \to T$ is a symmetric monoidal functor and every object of $T$ is invertible, then there is a symmetric monoidal functor $\tilde{F} : S^{-1} S \to T$ with $\tilde{F}(a,b) = F(a)^{-1} F(b)$ (one also has to define $\tilde{F}$ on morphisms and check the coherence diagrams ...) and this construction (probably ...) establishes an equivalence of categories $\hom(S,T) \cong \hom(S^{-1} S,T)$. Notice that if $S$ is discrete, $S$ is just a commutative monoid and $S^{-1} S$ is the usual Grothendieck group of $S$.

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    $\begingroup$ I am worried about your universal property. How does your construction avoid the problems with Quillen's $S^{-1}S$-construction? (See Thomason "Beware the phony multiplication on Quillen's $S^{-1}S$") $\endgroup$ Aug 14, 2015 at 9:32
  • $\begingroup$ Quillen only wants multiplications with objects of $S$ to be homotopy equivalences. His $S^{-1} S$ construction is therefore quite different. $\endgroup$ Aug 14, 2015 at 9:36
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    $\begingroup$ I am not saying your construction agrees with Quillen's, but that it is similar. The whole point of Thomason's paper is that despite how it looks at first Quillen's construction in fact does not have the analogous universal property (and so any constructions using it to, say, construct multiplications in algebraic K-theory, are bogus). The problem is that certain morphisms, which look like natural transformations, are not actually natural. Your construction looks similar, and so that raises a red flag. $\endgroup$ Aug 14, 2015 at 9:41
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    $\begingroup$ The universal constructions works with the space of words $a_1^{-1} b_1 \dots a_n^{-1} b_n$. If you work with symmetric monoidal categories then on group completion you get canonical nontrivial isos $b a^{-1} = a^{-1} b$, which allow you to build a path $a_1^{-1} b_1 \dots a_n^{-1} b_n = a^-1 b$, $a = a_1 \dots a_n$, $b = b_1 \dots b_n$. Thus you can get a homotopy equivalence with category of pairs $a^{-1} b$ with some very complex morphisms, and the multiplication is at best homotopy equivalent to the one you would want to write. $\endgroup$ Aug 15, 2015 at 8:20
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    $\begingroup$ @MartinBrandenburg , the problem is multiplication. As stated by you the $a^{-1}b $ braiding is trivial, since $(a, b) \equiv (a, 1)(1, b) \equiv (1, b)(a,1 ) $ on the nose with your definition, thus Thomason's counterexample. Symmetric still means you must keep track of coherence isos. Same problem with your definition of morphisms, you sweep all permutations under the rug which the gods of homotopy forbid. $\endgroup$ Aug 15, 2015 at 10:48

3 Answers 3


Maybe you could be interested in the recent PhD thesis:

Une introduction élémentaire au 2-groupe de Grothendieck by C. Drugmand, 2016, UCL, Louvain-la-Neuve.



There are left adjoints to the forgetful functors from compact categories (what you call 2CMon) to traced symmetric monoidal categories, and from traced symmetric monoidal categories to symmetric monoidal categories. Composing them should give what you want. The former is known as the Int-construction, and is due to Joyal, Street, and Verity. Sources include Traced monoidal categories and Abstract Scalars, Loops, and Free Traced and Strongly Compact Closed Categories (by Abramsky).

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    $\begingroup$ In compact categories objects are assumed to be dualizable; I want them to be invertible. $\endgroup$ Aug 14, 2015 at 9:20
  • $\begingroup$ @MartinBrandenburg So then you are reduced to forcing a compact category into one with all objects invertible. I mean, invertible objects are compact (with your $(a,b)$ being $\hom((1,a),(1,b))$), so after you arrive at a compact category you further have to invert the units $1\to\hom(a,a)$ (and maybe separately the counits $a^*\otimes a\to1$) $\endgroup$ Aug 15, 2015 at 18:20

Here are two comments on generalising Drugmand's results.

From symmetric monoidal groupoids to symmetric monoidal categories

Note that the Grothendieck group functor $$K_0\colon\mathsf{CMon}\longrightarrow\mathsf{Ab}$$ can be constructed via the tensor product of commutative monoids as $$K_0(A)\cong\mathbb{Z}\otimes_{\mathbb{N}}A.$$ This boils down to the fact that a bilinear map $\mathbb{Z}\times A\to G$ is in natural bijection with a morphism of monoids $A\to G$ sending every element of $A$ to an invertible element of $G$. When $G$ is a group, this is just any morphism of monoids, so it follows that $\mathbb{Z}\otimes_{\mathbb{N}}A$ satisfies the universal property of the abelian group completion of $A$.

I believe (but haven't checked) that one could do something similar for symmetric monoidal categories, generalising Drugmand's symmetric $2$-group completion $2$-functor $$ K^{\mathrm{D}}_{0}\colon\mathsf{SymMonGrpd}\longrightarrow\mathsf{2Ab} $$ taking values in symmetric monoidal groupoids to a $2$-functor $$ K_{0}\colon\mathsf{SymMonCats}\longrightarrow\mathsf{2Ab} $$ taking values in symmetric monoidal categories by defining $$ K_0(\mathcal{C})\overset{\mathrm{def}}{=}\mathbb{S}\otimes_{\mathbb{F}}\mathcal{C}, $$ where $\otimes_\mathbb{F}$ is the tensor product of symmetric monoidal categories and $\mathbb{S}$ is the $1$-truncation of the sphere spectrum.

I think $\mathbb{S}\otimes_{\mathbb{F}}-$ would then factor as groupoid completion followed by Drugmand's $2$-functor.

From symmetric monoidal categories to monoidal categories

I think (but again haven't checked) that one could define also a non-abelian Grothendieck $2$-group completion functor $$K_0\colon\mathsf{MonCats}\longrightarrow\mathsf{2Grp}$$ by first delooping a monoidal category $\mathbf{C}$ to a bicategory $\mathbf{B}\mathcal{C}$, taking its classifying space $|\mathrm{N}^{\mathrm{D}}_{\bullet}(\mathbf{B}\mathcal{C})|$, taking the fundamental bigroupoid of that, and then de-delooping back to a $2$-group.

(This is just the categorified version of the non-abelian group completion of a monoid $A$, which is given by $\pi_{1}(|\mathrm{N}_{\bullet}(\mathbf{B}A)|)$, as noted in the comments to another question of Martin here).


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