In MacLane's *Categories for the working mathematician*, the author shows that the evaluation at 1 gives an equivalence of categories $\mathrm{hom}\_{\mathrm{BMC}}(B,M)\simeq M\_0$ where $B$ is the braid category, $M$ is a braided monoidal category and $M\_0$ is the underlying (ordinary) category to $M$. As a result of that, he states a second theorem (coherence theorem) claiming that each composite of canonical maps in $M$ induces a braiding (element of the braid group), and that two such composites are equal for all $M$ if and only they induce the same element braiding.


I understand how to construct the braiding from a given composition of canonical morphisms, but I fail to write it properly using the previous theorem and I don't know how to prove that two such maps are equal if and only if they induce the same braiding.


I also read the original preprint of Joyal and Street which appears in the references given by MacLane ([http://maths.mq.edu.au/~street/JS1.pdf](http://maths.mq.edu.au/%7Estreet/JS1.pdf)) but I still don't see how to write a proper proof.



Coherence theorem in braided monoidal categories
Theorem 1 says precisely that for every braided monoidal category $M$ and every object $V \in M$, any isotopy class of braid on $n$ strands induces an isomorphism
$$
V^{\otimes n}\longrightarrow V^{\otimes n}.
$$
In particular this theorem also says this isomorphism is well defined, i.e. it does not depend one the particular braid one choose in the equivalence class. In other words, it does indeed says that if two braids are representatives of the same element in the braid group $B\_n$, then they define the same morphism for all pair $(M,V)$. Conversely, applying this to the case $M=B$ and $V=\bullet$, i.e. to the identity functor $B\rightarrow B$, one sees that this is an "only if": if two braids are **not** isotopic, then there exists at least one braided monoidal category (e.g. $B$ itself) for which these two braids induce non-equal morphisms.


So to me theorem 2 is really just a slightly weaker formulation of theorem 1 but they're still pretty much saying the same thing..


Also as Donald says, theorem 1 stats precisely that $B$ is the free braided monoidal category on one object. As a useful analogy, if $A$ is an abelian group and $A\_0$ its underlying set, the statement that there is a natural bijection
$$
Hom\_{Ab}(\mathbb{Z},A)\cong A\_0$$
states precisely that $\mathbb{Z}$ is the free abelian group on one generator.


