# Coherence in pseudo.monoids

In the article of A. Joyal and R. Street:

Braided tensor categories, (Adv. Math. N.102 (1993), no. 1, 20-78)

they define a tensor object (or pseudo.monoids) in a monoidal 2-category $\mathcal{C}$ (a monoidal structure on the underling category where all data are 2-natural too) as a 6-pla $(A, m , e, \alpha, \lambda, \rho)$ where $A\in\mathcal{C}$, $m: A \otimes A \to A$, $e: I \to A$, and 2-isomorphisms:

$\alpha : m \ast (m \otimes 1) \Rightarrow m \ast (1 \otimes m) : A \otimes A \otimes A\to A$, $\rho: m\ast Ae \Rightarrow 1_A$, $\lambda: m\ast eA \Rightarrow 1_A$ where $Ae: A \cong A \otimes I \xrightarrow{1\otimes e} A\otimes A$ similarly for $eA$ and $AeA$ ecc. and we leave out canonical isomorphisms of $\mathcal{C}$.

With the following axioms:

MA) the composition:

$m\circ (m\otimes 1)\circ (m\otimes 1\otimes 1) \xrightarrow{m\circ( \alpha\otimes 1)} m\circ (m\otimes 1)\circ (1\otimes m\otimes 1) \xrightarrow{\alpha\otimes 1} m\circ (1\otimes m)\circ (1\otimes m\otimes 1) \xrightarrow{m\circ (1\otimes \alpha)}m\circ (1\otimes m)\circ (1\otimes 1\otimes m)$

is equal to:

$m\circ (m\otimes 1)\circ (m\otimes 1\otimes 1) \xrightarrow{\alpha\otimes 1} m\circ (1\otimes m)\circ (m\otimes 1\otimes 1) \xrightarrow{=} m\circ (m\otimes 1)\circ (1\otimes 1\otimes m) \xrightarrow{\alpha\otimes 1} m\circ (1\otimes m)\circ (1\otimes 1\otimes m)$

MU) the composition:

$m\circ(m\otimes 1)\circ (AeA) \xrightarrow{m\circ(\rho^{-1}\otimes 1)} m \xrightarrow{m\circ \lambda} m\circ(1\otimes m)\circ AeA$ is equal to: $m\circ(m\otimes 1)\circ (AeA) \xrightarrow{\alpha\circ AeA} m\circ(1\otimes m)\circ AeA$

If we consider a pseudo-monoid in the monoidal 2-category $Cat$ (with cartesian monoidal structure) we get a usual (small) monoidal category, and by G. Kelly theorems all diagrams of canonical isomorphisms are commutative, but this is like saying that all the (two dimensional) diagram built by canonical 2-cells are commutative (or coherent).

I ask: Is this true in general? Are all the cell diagrams of a pseudomponoids (built using only the canonical $\alpha, \rho, \lambda$ the tensor composition, and the canonical isomorphisms of $\mathcal{C}$) coherent?

I just know that a pseudo-monoid is a particular tricategory (with the theorem of triequivalence to a Gray-enriched category), but in this full-generality we have only a "weak coherence" (no all diagrams commute).

EDIT: Edit: Given a diagram of canonical cells on $\mathcal{C}$ one could think to use some "kind of Yoneda lemma" for traslate the question on $Cat$, the fault is that $[-, X]: \mathcal{C}^{op}\to Cat$ don't preserve (neither has a some kind of correlation to) the monoidal product $\otimes$. But if $X$ is a $\otimes$-monoid then
$[-, X]: \mathcal{C}^{op}\to Cat$ is a monoidal functors

(from the morphisms-dual (the some monoidal product) of $\mathscr{C}$ to the monoidal cartesian structure of $Cat$)

and if applied on a pseudo-monoid of $\mathcal{C}$ maps it in a pseudo-monoid of $Cat$ (diagrams (MF1), (MF2), (MF3) p. 473 of Kelly, Eilenberg "Closed CAtegories", then if applied to a cells diagrams (relative to a pseudo-monoid) we get a similar diagram (relative to pseudo-monoid image)

(the some if $X$ is a comonoid and consider $[X,-]: \mathcal{C} \to Cat$, if $X$ is a pseudo-monoid $[-, X]$ is a pseudo.monoidal functors i.e. the diagrams (MF1), (MF2), (MF3) commuting but 2-isomorphisms).

Now the functors $[X, -]$ ($X\in \mathcal{C}$ simply) are useful because they are collectively faithful (locally too) (then if their images is ever a coherent diagram the initial diagram is coherent), but in our situation we have to use only monoids $X$ (no simple objects).

Then the second question is: Has a monoidal 2-category $\mathcal{C}$ enough-monoids (i.e. the family of $[-, X]$ for $X$ monoid, is (locally) faithful) ?

If we request $\mathcal{C}$ representable as 2-category (i.e. with comma objects) has it enough-monoids?

EDIT:

I did this partial answer (enough for many results):

Given a 2-category $\mathscr{A}$ and a object $G \in C$ we define the category $\mathscr{A} \searrow G$ with object of type $(A, a)$, $a: A \to G$ and morphisms of type $(f, \phi ): (A, a) \to (B, b)$, $f: A \to B$, $\phi : a \Rightarrow b \circ f$ and natural composition and unities. Let $(G, \alpha, \lambda , \rho, m, e)$ a pseudomonoid of a monoidal 2-category $\mathscr{C}$, and consider the full subcategory $\widetilde{G} \subset \mathscr{C} \searrow G$ where objets are defined by induction as:

1) $(I, e)$ and $(G, 1_G)$ are object of $\widetilde{G}$.

2) if $(X, x), (Y, y)$ are object of $\widetilde{G}$ then so is also their monoidal product $(X, x) \otimes (Y, y)$ defined as $(X \otimes Y, m\circ (x \otimes y))$.

Now $\widetilde{G}$ has a monoidal structure, with $(I, e)$ as unity object, monoidal product defined as above, with canonical isomorphism defined by induction using the elements $\alpha, \lambda , \rho, m, e$ (I seems that work, it is tedious but no hard).

Then we have a partial results: all diagram of cells (having a final vertex $G$ corresponding to canonical diagrams of $\widetilde{G}$) are coherent.

If we suppose that $\mathscr{C}$ has a symmetry, can generalize a bit because a monoidal product of copies of $G$ or $I$ become the monoidal product of the pseudo.monoids $G$ or $I$, and a reasoning similar to above (make a new induction building, that start from a object already built with the prior induction) concludes the proof.

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The answer is "yes, certainly", but I'd have to think about where this might be in the literature. Drawing a blank... –  Todd Trimble Jul 24 '13 at 21:28

The fact that “all diagrams commute” in a general pseudomonoid follows easily from Chapters 5 and 6 of my PhD thesis, which shows how both the statement and the proof of Mac Lane’s coherence theorem for monoidal categories can be reinterpreted as being about pseudomonoids in an arbitrary monoidal bicategory.

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Very intersting arguments, just what I wish (a traslation of the linguage of the coherence in monoidal category to a more general setting) may be need a bit of time for read it. –  Buschi Sergio Jul 25 '13 at 13:11
Thank you, Robin! (And welcome to MO!) –  Todd Trimble Jul 25 '13 at 14:15