Let $\mathcal A$ be a monoidal category. We know that $\mathcal A$ is monoidally equivalent to a strict monoidal category $\mathcal A^{\mathrm{str}}$. In many books/papers it is assumed without loss of generality that $\mathcal A$ is strict to make things easier and not have to worry about associator and unitor isomorphisms. But how do I know that things I prove about $\mathcal A$ in the strict case will also hold if $\mathcal A$ is not strict?

Here is an explicit example of what I am talking about. Assume $\mathcal A$ is a braided monoidal category and suppose $(C, \Delta, \epsilon), (C', \Delta' , \epsilon')$ are coalgebras in $\mathcal A$. This means $\Delta \colon C \to C \otimes C$ is a morphism and $\epsilon \colon C \to I$ is a morphism such that $(1 \otimes \Delta)\Delta = \alpha_{C,C,C}(\Delta \otimes 1)\Delta$ and $1_C = \rho_C (1 \otimes \epsilon)\Delta = \lambda_C(\epsilon \otimes 1) \Delta$ (and similarly for $C'$). I want to show that $C \otimes C'$ becomes a coalgebra. It does, and the comultiplication $\tilde \Delta$ is given by $$ C \otimes C' \xrightarrow{\Delta \otimes \Delta'} (C \otimes C) \otimes (C' \otimes C') \xrightarrow{\alpha} C \otimes (C \otimes (C' \otimes C') \xrightarrow{1 \otimes \alpha} C \otimes ((C \otimes C') \otimes C') \xrightarrow{1 \otimes(\sigma_{C,C} \otimes 1)} C \otimes ((C' \otimes C) \otimes C') \xrightarrow{1 \otimes \alpha^{-1}} C \otimes ( C' \otimes (C \otimes C'))\xrightarrow{\alpha^{-1} } (C \otimes C) \otimes (C' \otimes C') $$ However, this is absolutely terrible and checking that it does give a comultiplication is excruciating. In the strict case it is much easier since all the $\alpha$'s will disappear. How can I be sure (in general) that if a diagram commutes in the strict case then it will commute in the non-strict case with all the extra isomorphisms?

  • 1
    $\begingroup$ The coherence theorem can also be interpreted as, roughly speaking, "everything commutes". This is how you justify assuming strictness. $\endgroup$
    – Zhen Lin
    Nov 16, 2012 at 19:08

1 Answer 1


As Zhen Lin noted, the coherence theorem is basically equivalent to stating that, for every path $\langle f_1,..,f_n\rangle$ of 1-cells, between any 2 parenthesized composition of the path (e.g. between $\left(\big((f_1f_2)f_3\big)..f_n\right)$ and $f_1\left(\big( f_2 f_3\big)f_4... \right) $ ), there is a unique isomorphism built up by the associator and unit isomorphisms. Formally, there are of course, more such isomorphism, but the theorem just states that they all coincide.

Tom Leinster has another approach, so called unbiased monoidal (or bi-) categories, which written in simplified words, considers not only binary products a priori, but an $n$-fold product (without parenthesis) for all paths of length $n$, as basic operation. Then, the coherence axiom becomes requiring that, for any path of 1-cells, if we 'put 2 pairs of (disjoint) parentheses', then it doesn't matter, which one is evaluated first -i.e. these are commuting squares, for example $f_1f_2f_3f_4\to (f_1f_2)f_3f_4 \to (f_1f_2)(f_3f_4)$ and $f_1f_2f_3f_4\to f_1f_2(f_3f_4) \to (f_1f_2)(f_3f_4)$ must coincide. (Note that the middle terms are 3-fold compositions.)

Note also that allowing paths of length $0$ (as being the vertices) just defines the unit 1-cell(s) and provides the unit coherence isomorphisms..

From this, the coherence theorem is very easily seen, by simple induction. (However, to prove that we get the 'same kind' of monoidal (or bi-) category concept, we need to use the original coherence theorem..)


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