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A braided monoidal category is a monoidal category with a braiding (a commutativity constraint $\gamma_{A,B}: A \otimes B \to B \otimes A$) which satisfies the hexagon identities: $$ \gamma_{A, B \otimes C} = (1 \otimes \gamma_{A,C})(\gamma_{A,B} \otimes 1) \quad (1) \\ \gamma_{A \otimes B, C} = (\gamma_{A,C} \otimes 1) (1 \otimes \gamma_{B,C}). \quad (2) $$ How to interpret the braiding $\gamma_{A,B}: A \otimes B \to B \otimes A$ as a natural transformation?

By definition, a natural transformation between two functors $F,G: \mathcal{C} \to \mathcal{D}$ is a map $\gamma: F \to G$ such that for every morphism $f: X \to Y$ in $\mathcal{C}$, $\gamma_Y \circ F(f) = G(f) \circ \gamma_X$.

In a monoidal category, we have two functors $F,G: \mathcal{C} \times \mathcal{C} \to \mathcal{C}$, $F(A, B) = A \otimes B$, $G(A,B) = B \otimes A$. Can $\gamma_{A,B}: A \otimes B \to B \otimes A$ be interpreted as a natural transformation: $\gamma: F \to G$? It seems that the condition $\gamma_Y \circ F(f) = G(f) \circ \gamma_X$ does not translate to (1), (2)? Thank you very much.

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Yes, exactly $\gamma:F\to G$ is meant here, and the hexagon identities are required above the naturality condition, they indeed don't follow from that.

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  • $\begingroup$ thank you very much. So (1) and (2) do not follow from naturality? $\endgroup$ – LJR May 29 '17 at 22:00
  • $\begingroup$ Yes, they do not follow. $\endgroup$ – Berci May 29 '17 at 23:32

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