# Extending braidings to tensor powers

Given a braiding $\Psi: X \otimes Y \to Y \otimes X$ for two objects $X,Y$ in a monoidal category, it seems reasonable to assume that $\Psi$ extends uniquely to a braiding $X^k \otimes Y^l \to Y^l \otimes X^k$, for all $k,l \in {\mathbb N}$. The proof would surely be based upon the Yang--Baxter property of $\Psi$ and the fact that one can express any permutation as a product of transpositions. However, I can't seem to write it down exactly.

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$\newcommand{\id}{\mathrm{id}} \newcommand{\ot}{\otimes}$ Your assumption is correct. $\Psi$ does extend uniquely to the map that you want. By drawing string diagrams and playing with them, you can intuitively see what to do, namely: every time you see an $X$ strand to the left of a $Y$ strand, use $\Psi$ to braid $X$ over $Y$. However, there may be many ways to do this. For example if $k=l=2$, you can do $$(\id_Y \ot \Psi \ot \id_X) \circ (\id_Y \ot \id_X \ot \Psi) \circ (\Psi \ot \id_X \ot \id_Y) \circ (\id_X \ot \Psi \ot \id_Y),$$ or you can do $$(\id_Y \ot \Psi \ot \id_X) \circ (\Psi \ot \id_Y \ot \id_X) \circ (\id_X \ot \id_Y \ot \Psi) \circ (\id_X \ot \Psi \ot \id_Y),$$ and the braid relations show that those are the same map. Obviously this is easier to see if you draw a picture.
The challenge is how to efficiently prove that, given any $k$ and $l$ and any possible choice of way you build your braiding, that you get the same map. This is the sort of thing that is known as a "coherence theorem." These are often stated in the fashion: Every diagram in (some category) commutes, where the category is a sort of "free category" whose morphisms are the structural ones that you are talking about.