Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Let $\mathcal{C}$ be a monoidal category. By Mac Lane's coherence theorem for monoidal categories, there are strong monoidal functors $F : \mathcal{C} \to \mathcal{C}_s$ and $G : \mathcal{C}_s \to \mathcal{C}$ such that $\mathcal{C}_s$ is a strict monoidal category and $(F, G)$ form an equivalence of categories. [CWM, Ch. XI, §2] Informally, this tells us that we may pretend that the associators $\alpha$ and unitors $\lambda, \rho$ are identity arrows rather than mere isomorphisms.

My question concerns the analogue for symmetric monoidal categories: suppose $\mathcal{C}$ has a braiding $$\gamma_{a, b} : a \otimes b \to b \otimes a$$ such that $\gamma_{b, a} \circ \gamma_{a, b} = \textrm{id}$, and suppose it is compatible with the monoidal structure making $\mathcal{C}$ into a symmetric monoidal category.

  1. Does $F \gamma$ make $\mathcal{C}_s$ into a symmetric monoidal category?

  2. Assuming that (1) is true, do "all" diagrams involving $\gamma$ with atomic subscripts commute in $\mathcal{C}_s$? More precisely, let $\sigma$ be a permutation on $k$ letters, and suppose that $\sigma = \tau_1 \cdots \tau_m = \tilde{\tau}_1 \cdots \tilde{\tau}_n$ for some transpositions $\tau_i$, $\tilde{\tau}_j$ of adjacent pairs; does the equation $\tau_1 \cdots \tau_m = \tilde{\tau}_1 \cdots \tilde{\tau}_n$ still hold true when I interpret $\tau_i$ and $\tilde{\tau}_j$ as the corresponding braiding operation?

    For example, $$(1 \, 2 \, 3) = (1 \, 2) (2 \, 3) = (2 \, 3) (1 \, 2) (2 \, 3) (1 \, 2)$$ so I expect that $$(\gamma_{a, c} \otimes \textrm{id}_b) \circ (\textrm{id}_a \otimes \gamma_{b, c}) = (\textrm{id}_a \otimes \gamma_{b, c}) \circ (\gamma_{a, b} \otimes \textrm{id}_c) \circ (\textrm{id}_a \otimes \gamma_{b, c}) \circ (\gamma_{a, b} \otimes \textrm{id}_c)$$ which is indeed correct.

  3. Let $a = a_1 \otimes \cdots \otimes a_n$, $b = b_1 \otimes \cdots \otimes b_m$. Let $\sigma$ be the permutation $$(a_1, \ldots, a_n, b_1, \ldots, b_m) \mapsto (b_1, \ldots, b_m, a_1, \ldots, a_n)$$ Assuming (2) holds, let $\gamma_\sigma : a \otimes b \to b \otimes a$ be the morphism corresponding to the permutation $\sigma$. Then, is it true that $\gamma_{b, a} = \gamma_\sigma$?

    For example, take $n = 2$, $m = 1$. I expect it to be true that $$\gamma_{a \otimes b, c} = (\gamma_{a, c} \otimes \textrm{id}_b) \circ (\textrm{id}_a \otimes \gamma_{b, c})$$ and this turns out to be an instance of the hexagon axiom, in the special case of a strict monoidal category. But what if $n > 2$ or $m > 1$?

  4. Does this imply that "all" diagrams involving $\alpha, \lambda, \rho, \gamma$ commute in $\mathcal{C}$?

I suspect that the answers to all the questions are yes, at least if I understood Mac Lane [CWM, Ch. XI, §3] correctly, but I haven't found an explicit statement of the implied coherence theorem for symmetric monoidal categories. (There is an obvious way of defining a "free symmetric strict monoidal category" generated by another category, and I imagine there is some way of defining a "free symmetric monoidal category"; the coherence theorem ought to state that the two are equivalent via strong braided monoidal functors.)

share|cite|improve this question
Coherence is well-known to be true for symmetric monoidal categories. Have you googled some time? – Martin Brandenburg Mar 2 '12 at 8:53
up vote 2 down vote accepted
  1. Yes. By functoriality and strong monoidality, $F\gamma$ must satisfy the symmetry axioms.
  2. This is precisely the statement of coherence for SMC's. Mac Lane gives the braided version at the end of XI.5. Note, however, that this does not imply that "all" diagrams of the correct type commute. Only those corresponding to the same element of the symmetric group.
  3. The fact that you say "the" symmetry map corresponding the given permutation is already a consequence of coherence for SMC's. All symmetry maps corresponding to the given permutation (including $\gamma_{a,b}$) are equal.
  4. Nope, c.f. #2. In fact, the "all diagrams commute"-type coherence theorem is quite atypical. Most monoidal categories with extra structure have a coherence theorem more along the lines of "this is how to (efficiently) decide which diagrams commute: (...)".
share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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