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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).

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It is similar to the question mathoverflow.net/questions/11674/… Are you sure that Positselski's answer does not answer your question negatively as well? –  Bugs Bunny Jul 12 '10 at 10:26
    
I do not think so. I am not looking for a strictification of a braided monoidal category. On the contrary, I want to "destrictify": I want to make the set of objects "smaller", by identifying all isomorphic objects, at the expense of introducing a nontrivial associator. –  Anton Kapustin Jul 12 '10 at 15:39
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3 Answers

Yes. This follows from two simple facts:

  1. 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.
  2. 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.

  1. 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.
  2. 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.
  3. 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.

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Just a note to say that there is a more general context in which this fact can be placed. Just as an algebra structure for a monad can be transferred across any isomorphism, a pseudoalgebra structure for a 2-monad can be transferred across any (adjoint) equivalence. An "unbiased" structure of braided monoidal category is literally a pseudoalgebra structure for some 2-monad (the strict 2-monad whose strict algebras are braided strict-monoidal categories), so it transfers easily.... –  Mike Shulman Jul 10 '11 at 7:05
    
... A "biased" structure of braided monoidal category is not literally a pseudoalgebra structure for some 2-monad, but it is a strict algebra structure for some strict 2-monad. And since that strict 2-monad is "flexible" (=cofibrant), any pseudoalgebra structure for it is equivalent to a strict one. Thus, we can transfer such a structure in two steps: first transfer it as a pseudoalgebra structure, then strictify it using flexibility. –  Mike Shulman Jul 10 '11 at 7:06
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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.

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I think $\beta_{A,B}$ in general does not satisfy the hexagon axiom. Here is an example. Consider the category of Z-graded complex vector spaces, with the strict monoidal structure given by the convolution: $V_x\otimes V_y=V_{x+y}$. Let us define a nontrivial braiding $\beta(V_x,V_y)=\exp(a xy)$ for some complex number $a$. Now let $k$ be an integer and let's modify the category by introducing morphisms between vector spaces in degree $x$ and vector spaces in degree $x+k$, so that nonisomorphic objects live in the range $0\leq x\leq k-1$. For obvious choices of $p$ and $i$, the hexagon fails. –  Anton Kapustin Jul 12 '10 at 15:50
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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.

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