# Mac Lane‘s proof of coherence theorem for symmetric monoidal categories.

In [CWM, Ch. XI, §1], Mac Lane prove the coherence theorem for symmetric monoidal categories by assuming the strictness. Thus we have a $n-$ary tensor functor $T$, and the theorem states that any two chains of braiding isomorphisms connecting two permutations of $T$ must be equal.

Mac Lane claims that any closed chain corresponds to a relation of the generators of the $n$th symmetric group. But how?

It seems that the corresponding is given by sent a braiding isomorphism to a transposition. For example, let $12\cdots n$ denote the functor $X_1,X_2,\cdots,X_n\mapsto X_1\otimes X_2\otimes\cdots\otimes X_n$ and others likewise. Then the braiding isomorphism $12\cdots n\to21\cdots n$ should be corresponded to $(12)$. But, for example, $23145\cdots n\to32145\cdots n$ should also be corresponded to $(12)$. Note that every functor can be obtained by applying a permutation on $12\cdots n$, thus we can corresponding them to the permutation. So a braiding isomorphism should correspond to the quotient of the permutations corresponded to its domain and codomain.

But then every closed chain would produce the trivial relation, which is not as desired.

So, what's wrong?

• I might be wrong but I think you are assuming that for a braiding $\gamma$, we have $\gamma_{a,b} = \gamma_{b,a}^{-1}$, this is true for symmetric monoidal categories when you consider the braiding to be the symmetry isomorphism, but not for general braided monoidal categories. – John C Oct 3 '14 at 12:59

Secondly, instead of taking products in the $n$th symmetric group $S_n$, one should taking the products formally, i.e. in the free group $F$ generated by the adjacent transpositions. In this way, any chain of braiding isomorphisms will be labeled by a formal product of adjacent transpositions.
Then a chain is closed if and only if its label belongs to the kernel of the canonical map $F\to S_n$. That shuld be what Mac Lane means.