Skeleton of a braided monoidal category Does every (lax) braided monoidal category  have a braided monoidal skeleton? That is, I want to define  a (lax) braided monoidal structure on a skeleton so that it is braided monoidal equivalent to the original category. If this is always possible, how does one accomplish it?
An extreme case of this question (which I most interested in) is this: suppose the original braided monoidal category was strict, in the sense that the associators were trivial. I would like to replace this category with its skeleton, perhaps at the expense of making it nonstrict (generating nontrivial associators).
 A: I think the answer will be yes. Assume $D,C$ are categories and $i:D\to C, p:C\to D$ are
functors, with $p \circ i=id_D$, and that $C$ is braided monoidal. For $A,B \in Ob(D)$, 
set $A \otimes B := p(iA \otimes iB)$. Define the braiding $\beta_{A,B}:= p(\beta_{iA,iB})$, 
$a_{A,B,C}:= p(a_{iA,iB,iC})$. Then I think that $D$ satisfies the braided monoidal axioms. 
A: In the framework of my answer, I think the hexagon holds since (in the strict case)
$(\tilde\beta_{A,C}\otimes id_B)(id_A\otimes\tilde\beta_{B,C})=
p(i(\tilde\beta_{A,C})\otimes id_{iB})p(...)=
p(\beta_{iA,iC}\otimes id_{iB})p(id_{iA}\otimes\beta_{iB,iC})$ $=p((\beta_{iA,iC}\otimes id_{iB})(id_{iA}\otimes\beta_{iB,iC}))=p(\beta_{iA\otimes iB,iC})=p(\beta_{ip(iA\otimes iB),iC})=p(\beta_{i(A\otimes B),iC})= \tilde\beta_{A\otimes B,C}$. 
The problem might be in the axioms of a skeleton. Actually the framework is the following. $C$ is a (small) category and $D$ is a full subcategory such that 
$Ob(D)\to Ob(C)\to Ob(C)/iso$ is a bijection (this induces a map $\pi:Ob C\to Ob D$). In that case we don't have $p,i$
as above. What one can do is introduce $\tilde C$ whose objects are pairs $(X,f)$,
where $X\in Ob(C)$ and $f\in Iso_C(X,\pi X)$ and $Hom_{\tilde C}((X,f),(Y,g))=
Hom_{C}(X,Y)$. We then have functors $i:D\to \tilde C$ (taking $X$ to $(X,id_X)$)
and $p:\tilde C\to D$ (taking $X$ to $\pi X$ and $\phi\in Hom_{\tilde C}((X,f),(Y,g))=
Hom_{C}(X,Y)$ to $g\phi f^{-1}$). Then $p\circ i=id_D$, we also have a forgetful
functor $\tilde C\to C$. But I don't see how to lift the braided monoidal strcture of $C$ to $\tilde C$ which makes it
difficult to apply my proposal. 
A: Yes. This follows from two simple facts:


*

*If $F:C \simeq D$ is an equivalence of categories, and $D$ has a braided monoidal structure, then there exists a braided monoidal structure on $C$ and an enhancement of $F$ to a braided monoidal functor such that $F$ is an equivalence of braided monoidal categories. 

*Every category is equivalent to a skeletal category. 


Number 2 is standard, so I won't dwell on it. If you don't believe number 1, try proving it. It is a great exercise. Here is a sketch of how to go about proving it. 


*

*Replace $F$ by an adjoint equivalence, i.e. pick an adjoint inverse equivalence to $F$. You'll need this to transfer the structure from D to C. 

*Now transfer the structure from D to C. There is not much to say here. Just follow your nose. For example the tensor product in C is defined by
$$ x \otimes_C y := G(F(x) \otimes_D F(y))$$
There is an associator because we have chosen F and G to be adjoint equivalences to each other. The braiding is similar. It is given by,
$$G(\beta_{F(x), F(Y)}): G(F(x) \otimes_DF(y)) \to G(F(y) \otimes_DF(x)). $$
It is tedious to verify, but this does actually satisfy the hexagon identities. 

*Thus we've constructed a braided monoidal structure on C. To show that this new braided monoidal category C is equivalent as a braided monoidal category to D, we need to augment F and G to braided monoidal functors. To do this we just keep playing the same game. For example we need an equivalence $$F(x) \otimes_D F(y) \to F(x \otimes_C y) = FG(F(x) \otimes_D F(y)).$$
This is given by the inverse of the unit/counit of the adjoint equivalence between F and G. Constructing the other structure is no different. 


This is morally the strategy advocated by Benjamin Enriquez. He tried to do this with less then an equivalence and ran into trouble. For the question as stated, you really only need the case that C and D are equivalent. Notice also that this still works when the monoidal structures are just lax. It still produces a strong monoidal equivalence between C and D.  
