# Adding inverses to a symmetric monoidal category (Reference?)

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 $S$.

• 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 $S^{-1}S$") – Chris Schommer-Pries Aug 14 '15 at 9:32
• Quillen only wants multiplications with objects of $S$ to be homotopy equivalences. His $S^{-1} S$ construction is therefore quite different. – Martin Brandenburg Aug 14 '15 at 9:36
• 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. – Chris Schommer-Pries Aug 14 '15 at 9:41
• 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. – Anton Fetisov Aug 15 '15 at 8:20
• @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. – Anton Fetisov Aug 15 '15 at 10:48

• @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 $a^*\otimes a\to1$) – მამუკა ჯიბლაძე Aug 15 '15 at 18:20