Strictification and Mac Lanes Coherence Theorem - Examples

One route around the pentagon is the composite $((A\otimes B)\otimes C)\otimes D\stackrel{{\alpha }_{A\otimes B,C,D}}{\to }(A\otimes B)\otimes (C\otimes D)\stackrel{{\alpha }_{A,B,C\otimes D}}{\to }A\otimes (B\otimes (C\otimes D))$.

The other route is the composite $((A\otimes B)\otimes C)\otimes D\stackrel{{\alpha }_{A,B,C}\otimes 1_D}{\to }(A\otimes (B\otimes C))\otimes D\stackrel{{\alpha }_{A,B\otimes C,D}}{\to }A\otimes ((B\otimes C)\otimes D)\stackrel{1_A\otimes {\alpha }_{B,C,D}}{\to }A\otimes (B\otimes (C\otimes D))$.

Both composites are canonical morphisms with the same source and target, so [L1] makes them equal.

Apply the outer associator to obtain $((A\otimes (B\otimes C))\otimes (D\otimes E))\stackrel{{\alpha }_{A,B\otimes C,D\otimes E}}{\to }A\otimes ((B\otimes C)\otimes (D\otimes E))$.

Apply $1_A\otimes {\alpha }_{B,C,D\otimes E}$ to the result of step 1.1 and get $A\otimes ((B\otimes C)\otimes (D\otimes E))\to A\otimes (B\otimes (C\otimes (D\otimes E)))=v$.

The composite of steps 1.1 and 2.1 is therefore a canonical map $u\to v$. Any other chain of associators with the same endpoints is equal to it by [L1].

A unit-free length-three word must be obtained by combining exactly three copies of $(-)$ with two binary operations. Up to the placement of brackets, the only possibilities are $((-\mathrm{\square }-)\mathrm{\square }-)$ and $(-\mathrm{\square }(-\mathrm{\square }-))$. Words of length three containing one or more copies of $e_0$ also exist, but are outside this unit-free subcollection.

These two words have the same length, so there is exactly one morphism between them in each direction. In the monoidal structure of [L2], those are the associator and its inverse at the generator.

Thus the unit-free part of the length-three slice of $W$ is the basic coherence picture: two bracketings and one canonical comparison each way.

By [L1], the tensor product is $\times$, so the strictification formula from [L2] becomes $L(X)(Y)=X\times Y$. Its right-module structure map at $(Y,Z)$ is the cartesian associator $(X\times Y)\times Z\to X\times (Y\times Z)$.

The binary comparison $L(X)\otimes L(Y)\Rightarrow L(X\times Y)$ therefore has component the inverse reassociation $X\times (Y\times Z)\to (X\times Y)\times Z$, and the unit comparison is $Z\to 1\times Z$.

So in the cartesian case the abstract strictification is completely explicit: it packages ordinary reassociation maps of products into a strict monoidal category of endofunctors.

By [L1], the two formal vertices $x$ and $(x\otimes \mathbf{1})$ both evaluate to the single object $\ast$ of $\mathcal{M}$.

Consider the square all of whose vertices are $\ast$, whose top edge is $u$, and whose other three edges are $1$. The clockwise composite is $u$, while the counterclockwise composite is $1$, so the square does not commute because $u\mathrm{\ne }1$.

This realizes the mechanism described in [L2]: formally distinct words can collapse to one object, and a noncanonical endomorphism then breaks the blanket slogan. Hence the slogan is false.

The explicit associativity axiom is $\mu \circ (\mu \otimes 1_M)=\mu \circ (1_M\otimes \mu )\circ {\alpha }_{M,M,M}$, and the unit axioms are $\mu \circ (\eta \otimes 1_M)={\lambda }_M$ and $\mu \circ (1_M\otimes \eta )={\rho }_M$.

By [L2], the same relations may be written as $\mu (\mu \otimes 1_M)=\mu (1_M\otimes \mu )$ and $\mu (\eta \otimes 1_M)=1_M=\mu (1_M\otimes \eta )$, with the bracket suppressions understood canonically.

Thus the two notations express the same monoid-object structure; the second is shorter only because coherence has already absorbed the bracket changes.