Braiding (up to equivalence?)

Question

I am slightly confused by the definition of a braiding in a monoidal category. It says that it is a natural isomorphism $c_{X,Y}\colon X\otimes Y\to Y\otimes X$. If I understand this correctly, this means that taking any $f\colon X\to X'$, $g\colon Y\to Y'$, we must have $$c_{X',Y'}(f\otimes g)=(f\otimes g)c_{X,Y}.$$

My question is: What if I find a class of isomorphisms $c_{X,Y}$ such that the above equation holds up to a monoidal autoequivalence. That is, suppose we have a monoidal equivalence $F\colon C\to C$ and $$c_{X',Y'}(f\otimes g)=(T(g)\otimes T(f))c_{T(X),T(Y)}.$$

(For simplicity, we may also assume that $T$ acts on objects as identity.)

So, this is not a braiding? Is there a canonical way, how to construct an actual braiding out of this or it need not exist at all?

Answer 1

Too long for a comment.

A braiding is not just a natural isomorphism between the functors $(-\otimes=)$ and $(=\otimes-)$. It is a natural isomorphism satisfying important coherence relations which one should always rigorously check before we call something a braiding (I have a vendetta against abuse of the isomorphism symbol in category theory literature).

The equation $c_{X',Y'}(f\otimes g)=(f\otimes g)c_{X,Y}$ is not correct in general. Let's denote by $\otimes'$ the twisted functor $(a,b)\mapsto b\otimes a$. We want $c_{X,Y}:\otimes(X,Y)\to\otimes'(X,Y)$ to be a natural transformation, so that we want $c_{X',Y'}\circ(\otimes(f,g))=(\otimes'(f,g))c_{X,Y}$ i.e. $c_{X',Y'}(f\otimes g)=\color{red}{(g\otimes f)}c_{X,Y}$.

Your autoequivalence statement does not make sense unless $T$ really is the identity on objects, since the RHS runs $T(X)\otimes T(Y)\to T(Y')\otimes T(X')$ but the LHS runs $X\otimes Y\to Y'\otimes X'$. To remove the asymmetry with $T$ on one side but not on the other, I think the "correct question to ask" should involve applying $T$ to make the domains and codomains align (requiring equality of objects is unnatural in category theory):

Can we get a natural isomorphism $(-\otimes=)\cong(=\otimes-)$ if there exist arrows (isomorphisms) $c_{A,B}:A\otimes B\to B\otimes A$ satisfying: $$T(c_{X',Y'}(f\otimes g))\mu_{X,Y}=\mu_{Y',X'}(T(g)\otimes T(f))c_{T(X),T(Y)}:T(X)\otimes T(Y)\to T(Y'\otimes X')$$Where $\mu:T(-)\otimes T(=)\implies T(-\otimes =)$ is the monoidal multiplication associated to the monoidal functor $T$?

My hunch is that the answer is no. For a little while I hit this equation with the antiequivalence and the natural isomorphism maps but really only succeeded in converting the equation above for $T$ into an analogous one with its antiequivalence. I suspect a positive answer could be possible under extra conditions, if you found a useful coherence relation between $c_{TX,TY}$ and $T(c_{X,Y})$.