Inverting objects in a symmetric monoidal category In Voevodsky’s ICM address:
https://www.uio.no/studier/emner/matnat/math/MAT9580/v18/documents/voevodsky-a1-homotopy-theory-icm-1998.pdf
In theorem 4.3 it is claimed that given a symmetric monoidal category $(C, \wedge, 1)$ and an object $ X \in C$ in order for $C[X^{-1}]$ be be symmetric monoidal it is enough for $ (123) : X^3 \rightarrow X^3$ to be the identity in $C[X^{-1}]$.
Question: Is there a way to see why this condition on $(123)$ is enough to give us the monoidal structure in an explicit example? Or at least an example of why it is necessary?
Thanks!
 A: To be clear, this claim refers to a very specific construction of $\mathcal{C}[X^{-1}]$, where you copy the construction of the localization of the ring and defines it as the colimits of:
$$ \mathcal{C} \overset{\_ \otimes X}{\to} \mathcal{C} \overset{\_ \otimes X}{\to} \mathcal{C} \overset{\_ \otimes X}{\to} \mathcal{C} \dots $$
If you are just looking at constructing a $\mathcal{C}[X^{-1}]$ which is universal for adding an inverse to the object $X \in \mathcal{C}$ then this exists without no assumption by the (higher categorical version of the) special adjoint functor theorem.
The condition mentioned there is only important for the localization being computed in the expected way...
For maybe a concrete explanation of why this (123) conditions appears, it is because if $X$ is an invertible object in a symetric monoidal category, then you can show that it satifies the (123) conditions, so when we construct $\mathcal{C}[X^{-1}]$, we are going to have at some point to "kill" the action of (123) on $X^{\otimes 3}$ and the naive iteration presented above doesn't do that at all, so we at least need to assume that this (123) action is already trivial (or more precisely that it is trivial on $X^{\otimes 3} \otimes X^{\otimes n}$ for $n$ large enough).
To give an example let's look at the free symetric monoidal category on on object, i.e. the category $\mathcal{S}$ of finite set and bijection being them with the tensor product being the disjoint union. In this situation Voevodsky's condition isn't satisfied.
If I take the colimit:
$$ \mathcal{S} \overset{ \coprod \{1\}}{\longrightarrow} \mathcal{S} \overset{ \coprod \{1\}}{\longrightarrow} \mathcal{S} \overset{ \coprod \{1\}}{\longrightarrow} \mathcal{S} \overset{ \coprod \{1\}}{\longrightarrow} \mathcal{S} \dots $$
I get a category whose (isomorphisms class of ) objects are indexed by $\mathbb{Z}$, with no non-invertible maps and where each object has as endomorphisms monoids $\Sigma_\infty = \text{colim } \Sigma_n$.
This can't be a symetric monoidal category: if it were the unit would have a commutative monoid of endomorphisms and here each object has this non-commutative monoid $\Sigma_\infty$.
Now $\mathcal{S}[\{1\}^{-1}]$ exists : it is the free symetric monoidal category generated by an invertible object, so it is the free symetric monoidal groupoid on object, so you get the 1-truncation of the sphere spectrum, i.e. a category that has object indexed by $\mathbb{Z}$ (with monoidal structure the addition) and where each object has a $\pi_{n+1}(S^n) \simeq \mathbb{Z}/2\mathbb{Z}$ of automorphisms, which is involved in the braiding of symetric structure.
Note that, there is a map from the naive iterative construction to the correct localization, that sends each $\Sigma_\infty$ to $\mathbb{Z}/2\mathbb{Z}$ through the signature of a permutation, and it corresponds to the fact that the localization can be obtained from the naive iteration by impossing a few more relations in the colimit which among other things kills the action of (123) (it is also related to Quillen's $+$ construction).
I have a work in progress with Mathieu Anel where we generalized this condition and which I think make it a bit more clear how it appears and how we can still construct the localization iteratively when it is not satisfied, but it won't be out before a few month.. so I guess I'll come back when it is available.
