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Let $X$ be an object in a monoidal category $({\cal C}, \otimes)$, and $\gamma:X \otimes X \to X \otimes X$ a braiding (that is to say a morphism in ${\cal C}$ from $X \otimes X$ to itself that satisfies the braid relation $$ (\gamma \otimes \text{id}) \circ (\text{id} \otimes \gamma) \circ (\gamma \otimes \text{id}) = (\text{id} \otimes \gamma) \circ (\gamma \otimes \text{id}) \circ (\text{id} \otimes \gamma). $$

For any other object $Y$ in ${\cal C}$ that is isomorphic to $X$, via an isomorphism $f:X \to Y$, it is clear that $$ (f^{-1} \otimes f^{-1}) \circ \gamma \circ (f \otimes f) $$ defines a braiding for $Y$. Moreover, for any non-equivalent isomorphism $g:X \to Y$, we get a different braiding on $Y$.

Thus, a single braiding for an object in ${\cal C}$ can induce a number of different braidings on any other object in its isomorphism class (given the existence of a non-equivalent isomorphisms that is). Is there a canonical way to extend a braiding from an object to its isomorphism class?

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When you say "satisfies the hexagonal identities", how do you define $\gamma_{X,X\otimes X}$ ? Defining a braiding needs some kind of naturality, I think (at least on "one leg", and you can take the hexagon for the other leg as an actual definition, like in the construction of the Drinfeld center). –  Adrien Sep 5 '12 at 19:28
    
You're quite right, what I asked didn't really make sense. Now that I reflect I see that what I meant to say was braid relations. I've changed the question accordingly. I hope not too much. –  Mihail Matrix Sep 6 '12 at 0:00
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Actually, the phrase I grew up with, instead of 'braiding' here, is Yang-Baxter operator $R: X \otimes X \to X \otimes X$. That's what Joyal and Street call it, anyway. –  Todd Trimble Sep 6 '12 at 0:14
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Anyway, I can't imagine the answer is any but 'no'. –  Todd Trimble Sep 6 '12 at 0:21
    
E.g. $\mathbb{Z}$ has two braidings in the category of abelian groups: $\pm1$. –  Fernando Muro Sep 6 '12 at 5:43
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