Let G
 be a groupoid. I'm wondering how to construct the free [2-group](http://ncatlab.org/nlab/show/2-group) on G.


By the free 2-group I mean a 2-group $\mathcal{F}\left(G\right)$
 equipped with a functor $i:G\longrightarrow\mathcal{F}\left(G\right)$
 such that for any 2-group $\mathcal{G}$
 and any functor $F:G\longrightarrow\mathcal{G}$
 there is a monoidal functor $F':\mathcal{F}\left(G\right)\longrightarrow\mathcal{G}$
 and an isomorphism $\alpha:F'\circ i\Longrightarrow F$
 such that the functor $F':\mathcal{F}\left(G\right)\longrightarrow\mathcal{G}$
 is unique up to coherent isomorphism. Here is my attempt at a description of $\mathcal{F}\left(G\right)$
 :


The objects are defined inductively by requiring that $\mathcal{F}\left(G\right)$
 contain an object 1
 , the objects of G
 and for each object $g\in G$
 an object $\overline{g}\in G$
 such that for any two objects $x,y\in\mathcal{F}\left(G\right)$
 there is an object $x\otimes y$
 in $\mathcal{F}\left(G\right)$
 . 


The 'prearrows' in $\mathcal{F}\left(G\right)$
 are built as follows. We require that every arrow from G
 be a prearrow in $\mathcal{F}\left(G\right)$
 . For each pair of prearrows $a:x\longrightarrow z$
 and $b:y\longrightarrow w$
 in $\mathcal{F}\left(G\right)$
 there is a prearrow $a\otimes b:x\otimes y\longrightarrow z\otimes w$
 . In addition we adjoin a prearrow $e\_{g}:g\otimes\overline{g}\longrightarrow1$
 for each $g\in G$
 and we adjoin prearrows $\alpha\_{x,y,z}:\left(x\otimes y\right)\otimes z\longrightarrow x\otimes\left(y\otimes z\right)$
 , $\lambda\_{x}:1\otimes x\longrightarrow x$
 , and $\rho\_{x}:x\otimes1\longrightarrow x$
 for each $x,y,z\in\mathcal{F}\left(G\right)$
 . The arrows are then given by equivalence classes of prearrows generated by the requirement that $\alpha,\rho,\lambda$
 and e
 be isomorphisms, the naturality conditions on $\alpha,\rho$
 and $\lambda$
 , the condition that $\otimes$
 is a functor, and the axioms of a monoidal category.


My question is, does this construction work? If not, can anybody give an indication of a construction that does? Also, to define the free symmetric 2-group on a groupoid, does it suffice to add prearrows $\gamma\_{x,y}:x\otimes y\longrightarrow y\otimes x$
 for each $x,y\in\mathcal{F}\left(G\right)$
 and generate the equivalence relation from the axioms of a symmetric monoidal category?



How to construct a free 2-group on a groupoid?
let me offer you an alternative. The free 2-group on a groupoid is well defined up to equivalence (not isomorphism), hence I'll offer you a strict model (which is known to exist by abstract reasons, since any 2-group can be strictified).


A strict 2-group is essentially the same thing as a crossed module $\partial\colon\mathcal F(G)\_1\rightarrow\mathcal F(G)\_0$. It is defined as follows. Let $BG$ be the classifying space (simplicial set) of G. Let $FBG$ be its Milnor construction (obtained by taking free group dimension-wise and killing the base point), which is a simplicial group. This is a simplicial group. Consider its Moore complex $N\_\*FB(G)$, which is a non-negatice non-abelian chain complex with certain features. The crossed module $\partial\colon\mathcal F(G)\_1\rightarrow\mathcal F(G)\_0$ is the following truncation of the Moore complex


$$\bar d\_1\colon N\_1FB(G)/d\_2(N\_2FB(G))\longrightarrow N\_0FB(G).$$


In particular $\mathcal F(G)\_0$ is the free group on the set of objects of $G$. This crossed module can also be defined by generators and relations. This is in my paper with Baues entitled 'Secondary homotopy groups', $\S6$.


The previous construction is a strict left adjoint of the forgetful functor from crossed modules to groupoids. You may then wonder why it is also a 2-adjoint. This follows from the following facts:


* this adjoint pair is a Quillen pair for the usual model structures in these categories.
* the 2-categorical structures can be recovered from the model structures via mapping spaces.
* All groupoids are cofibrant (also fibrant, but this is not relevant here) and all crossed modules are fibrant (but not cofibrant, but it doesn't matter).


The symmetrization has an intermediate step, which is to make it braided. Let me tell you how this works for any crossed module $\partial\colon C\_1\rightarrow C\_0$. The result is a reduced quadratic module, in the sense of Baues's 'Combinatorial homotopy and 4-dimensional complexes'. Such an object is a diagram of groups of the form
$$D\_0^{ab}\otimes D\_0^{ab}\stackrel{\langle-,-\rangle}\longrightarrow D\_1\stackrel{\delta}\longrightarrow D\_0$$
satisfying some equations, $\delta$ is a (special kind of) crossed module, and $D\_0^{ab}$ denotes the abelianization of $D\_0$. The left adjoint of the forgetful functor from reduced quadratic modules to crossed modules send $\partial\colon C\_1\rightarrow C\_0$ to the reduced quadratic module where $D\_0$ is the quotient of $C\_0$ by triple commutators and $D\_1$ is a quotient of $C\_1\times (C\_0^{ab}\otimes C\_0^{ab})$ by some relations. See my previous paper with Baues. The symmetrization of a reduced quadratic module consists of forcing the bracket to be antisymmetric $\langle a,b\rangle=-\langle b,a\rangle$ by taking quotients.


For the relation between these strict categories of (symmetric, braided) 2-groups and the usual non-strict version see the survey in $\S 3$ of my joint paper with Tonks and Witte, 'On determinant functors and K-theory'. Everything contained in that section is kind of folklore, but it's either scattered through old literature or we do not know of any other published proof.


