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Let $\mathcal{A}$ and $\mathcal{B}$ be two $2$-categories and $F : \mathcal{A} \to \mathcal{B}$ be a lax $2$-functor. Given $1$-cells $(f_{i})_{0 \leq i \leq n}$ of $\mathcal{A}$ such that the composition $f_{n} \circ f_{n-1} \circ \cdots \circ f_{0}$ makes sense, this data together with the structural $2$-cells of $F$ give many paths of $2$-cells going from $F(f_{n}) \circ F(f_{n-1}) \circ \cdots \circ F(f_{0})$ to $F(f_{n} \circ f_{n-1} \circ \cdots \circ f_{0})$, for instance $$ F(f_{n}) \circ F(f_{n-1}) \circ \cdots \circ F(f_{0}) \Rightarrow F(f_{n} \circ f_{n-1}) \circ F(f_{n-2}) \circ \cdots \circ F(f_{0}) \Rightarrow \cdots $$ $$\Rightarrow F(f_{n} \circ f_{n-1} \circ \cdots \circ f_{0}) $$ and $$ F(f_{n}) \circ F(f_{n-1}) \circ \cdots \circ F(f_{0}) \Rightarrow F(f_{n}) \circ F(f_{n-1}) \circ \cdots \circ F(f_{1} \circ f_{0}) \Rightarrow \cdots $$ $$\Rightarrow F(f_{n} \circ f_{n-1} \circ \cdots \circ f_{0}) $$ which correspond to what one gets by "parenthesizing on the left" and "parenthesizing on the right" respectively. It seems to seem obvious that it follows from the definition of lax functor that the $C_{n}$ ways to parenthesize the left hand side all give the same $2$-cell $$ F(f_{n}) \circ F(f_{n-1}) \circ \cdots \circ F(f_{0}) \Rightarrow F(f_{n} \circ f_{n-1} \circ \cdots \circ f_{0}) $$ Since I need this property for a text I am writing, I would like to provide a reference. My question is the following:

Where is this result rigorously stated, and where is it rigorously proved? Hopefully, the two references will be the same.

Edit: I am aware that this result is "obvious". In addition, it is certainly classical, by which I mean that all the people working with lax functors use it routinely. However, if one wants to state it and prove it, the question arises as to what is the best way to state the result, which I think turns out not to be completely trivial. Furthermore, writing a rigorous proof certainly does require some work. I am sure there are some people here who have already used this result. How do they state it? To which reference do they point? Or is the reader assumed to find this fact so obvious that no one ever cares to provide a proof or a reference?

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I don't understand why my LaTeX is botched up every time I copy-paste it, while I don't have any trouble to compile it with TeXShop. I'll try to fix the mess. Any answer to the question in this comment is welcome – Jonathan Chiche Jan 7 '12 at 8:35
Many thanks to Alain Valette for having cleaned up the mess, about which I am sorry. I notice only now that the "How to write math" part recommends "putting backticks around any math that contains underscores". – Jonathan Chiche Jan 7 '12 at 11:14
It would be useful for me if a new tag "2-category-theory" (where specific questions arise which belong neither to "higher algebra" nor "category theory" as most people understand these terms) was created. I think some other people may use it sometimes too. But I have not been able to find rules to abide by when creating a new tag. – Jonathan Chiche Jan 9 '12 at 8:45
A 2-category-theory tag might be useful. Then again, if people don't think that 2-category theory is part of higher category theory, then I would say that they have an odd view of what higher category theory is. – Tom Leinster Jan 9 '12 at 16:33

The closest statement I know is Theorem 1.6 in Gordon-Power-Street "Coherence for tricategories." It actually deals with pseudofunctors instead of lax functors, but I believe it can easily be modified to cover lax functors. The proof is the same proof of Theorem 1.7 in Joyal-Street "Braided tensor categories."

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Ah, thanks! I'm abroad for a few weeks and cannot look it up now. I'll check that when I'm back. Besides, before seeing your answer, I thought I would answer my question myself because Bénabou told me last month that this coherence result was a special case of a quite general result in his thesis, which should have been published but, for some reason unknown to him, was not. I'll write about that as soon as I find the proper reference. – Jonathan Chiche Feb 3 '12 at 7:44

For composable couple of morphisms $g\circ f$ let

$T(g, f): F(g)\circ F(f) \Rightarrow F(g\circ f)$ the canonical cell.

For a triple of composable $h\circ g\circ f$ let $T_r(h, g, f):=T(T(h, g),f)$ (A short way to write $T(hg, f)\ast T(h, g)f$)

and $T_l(h,g,f):= T(h,T(g, f))$ for the axiom of coherence $T'_3=T''_3$.

If we have composable $n$ morphisms $f_n\circ \ldots\circ f_1$ for the various (associativity) coherent disposition of parenthesis (as in a no necessarily associative binary composition) we have a cell $F(f_n)\circ \ldots F(f_1) \to F(f_n\circ \ldots f_1)$, we want to prove that all these cells are equal, for $n=3$ this is true. For induction we suppose the assert for 3, ..., n and indicate with $T^k$ ($3\leq n\leq n$) the unique composition cell $F(f_k)\circ \ldots F(f_1) \to F(f_k\circ \ldots f_1)$. Let $T_{n+1}: F(f_{n+1})\circ \ldots F(f_1)\xrightarrow{1\circ T^n} F(f_{n+1})\circ F(f_n\circ \ldots f_1)\xrightarrow{T} F(f_{n+1}\circ f_n\circ \ldots f_1) $, let $T': F(f_{n+1})\circ \ldots F(f_1)\to F(f_{n+1}\circ f_n\circ \ldots f_1) $ associated to some coherent parenthesis disposition. Now if $f_{n+1}$ is not inside a parenthesis (do not allow the total parenthesis on the whole string) we have that $T'(f_{n+1}, \ldots f_1): F(f_{n+1})\circ F(f_n)\circ \ldots F(f_1)\xrightarrow{1\circ T} F(f_{n+1})\circ F(f_n\circ \ldots f_1)$ and this is $T_{n+1}$, if $f_{n+1}$ is inside a parenthesis consider the maximal parenthesis containing $f_{n+1}$ and let $f_{n+1},\ldots f_{i+1}$ the part of string contained in this parenthesis, the case $i=n-1$ follow as above,

if $i< n-1$ then

$T'(f_{n+1}, \ldots f_1)= T(T^{n-i}(f_{n+1},\ldots f_i), T^{i}(f_i,\ldots f_1))=$

$T(T((f_{n+1},T^{n-i-1}(f_n, \ldots f_i)), T^{i}(f_i,\ldots f_1))=$

$T((f_{n+1},T^n(f_n, \ldots f_1))=$

$T_{n+1}(f_{n+1}\circ \ldots f_1) $.

This is only a traslation from the classical algebra theorem.

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Thanks. I have not looked at the details, mainly because I was in fact not asking for a proof, but rather a reference. I knew that the classical proof that, say, arbitrary multiplication in a monoid is well-defined, could be adapted to show the coherence result here. The fact which I find disturbing is that there does not seem to be any published text containing both a rigorous statement and a rigorous proof. – Jonathan Chiche Nov 10 '12 at 18:19
I think that adding this proof on your thesis or article fill this lack. P.S. do you understand italian? I was looking for your Email (no found it), mine is – Buschi Sergio Nov 11 '12 at 8:57

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