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) but I still don't see how to write a proper proof.


2 Answers 2


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.


One way to interpret that coherence theorem in Mac Lane's book (Theorem XII.5.2) is Corollary 2.6 in the Joyal-Street paper that you cited: In the free braided monoidal category generated by a set $A$ of objects, a diagram commutes if and only if all legs with the same source and target have the same underlying braid. Using the terminology in Mac Lane's Theorem XII.5.2, if two composites of canonical maps acting on an $n$-fold product are equal for all braided monoidal categories, then in particular it is true in the free braided monoidal category generated by $n$ objects. By Joyal-Street Corollary 2.6, this happens if and only if those two composites have the same underlying braids.

In this context, there is nothing very special about braids and braided monoidal categories. More specifically, for an action operad $\mathsf{G}$, this coherence theorem is proved for $\mathsf{G}$-monoidal categories in Theorem 20.3.6 and its subsequent corollary in the book Infinity Operads and Monoidal Categories with Group Equivariance (arxiv version). Specifying to the braid operad $\mathsf{G} = \mathsf{B}$ recovers the aforementioned result of Joyal-Street; they are Theorem 21.3.3 and Corollary 21.3.4 in that book. The same coherence theorem for ribbon monoidal categories (for the ribbon operad) and coboundary monoidal categories (for the cactus operad) are in, respectively, Sections 22.3 and 23.4 in that book.

  • $\begingroup$ I understand the interpretation of the theorem but I have the impression that MacLane deduces it from theorem 1 above. I don’t understand how theorem 1 implies theorem 2 $\endgroup$
    – QGM
    Mar 20, 2022 at 22:23

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.