# The Unapologetic Mathematician

## The General Associative Law

For any monoid object we have an associative law for the multiplication: $\mu\circ(\mu\otimes1_X)=\mu\circ(1_X\otimes\mu)$. This basically says that the two different ways of multiplying together three inputs to give one output are the same. Let’s call the result $\mu_3:X^{\otimes3}\rightarrow X$. In fact, we might go so far as to say $\mu_2=\mu:X^{\otimes2}\rightarrow X$, $\mu_1=1_X:X^{\otimes1}\rightarrow X$, and even $\mu_0=1:X^{\otimes0}\rightarrow X$.

This generalizes a lot. We want to say that there’s a unique way (called $\mu_n$) to multiply together $n$ inputs. The usual way is to pick some canonical form and show that everything can be reduced to that form. This ends up being a lot like the Coherence Theorem. In fact, if we take a monoid object in the category $\mathbf{Cat}$ of small categories, this is the Coherence Theorem for a strict monoidal category.

But there’s an easier way than walking through that big proof again, and it uses our diagrammatic approach! The first thing we need to realize is that if we can show this rule holds in $\mathrm{Th}(\mathbf{Mon})$, then it will hold for all monoid objects. That’s why the “theory of monoids” category is so nice — it exactly encodes the structure of a monoid. Anything that is true for all monoids can be proved by just looking at this category and proving it there!

So how do we show that the general associative law holds in $\mathrm{Th}(\mathbf{Mon})$? Now we need to notice that the functor that makes $\downarrow\otimes\uparrow$ into a monoid object is faithful. That is, if two Temperley-Lieb diagrams in the image are the same, then they must come from the same morphism in $\mathrm{Th}(\mathbf{Mon})$. But if two diagrams are equivalent they differ by either sliding loops arcs around in the plane — which uses the monoidal structure to pull cups or caps past each other — or by using the zig-zag identities — which encode the left and right identity laws. Thus any equalities that hold in the image of the functor must come from equalities already present in $\mathrm{Th}(\mathbf{Mon})$!

Now any way of multiplying together $n$ inputs to give one output is a morphism $f:M^{\otimes n}\rightarrow M$ in $\mathrm{Th}(\mathbf{Mon})$, which will be sent to a diagram $F(f):(\downarrow\otimes\uparrow)^{\otimes n}\rightarrow\downarrow\otimes\uparrow$ in $\mathcal{OTL}$. It’s not too hard to see that there’s really only one of these diagrams that could be in the image of the functor (up to equivalence of diagrams). So all such multiplications are sent to the same diagram. By the faithfulness above, this means that they were all the same morphism in $\mathrm{Th}(\mathbf{Mon})$ to begin with, and we’re done.

By the way, you should try playing around with the oriented Temperley-Lieb diagrams to verify the claim I made of uniqueness. Try to work out exactly what diagrams are in $\hom_\mathcal{OTL}((\downarrow\otimes\uparrow)^{\otimes n},\downarrow\otimes\uparrow)$, and then which ones can possibly be in the image of the functor. Playing with the diagrams like this should give you a much better intuition for how they work. If nothing else, drawing a bunch of pictures is a lot more fun than algebra homework from back in school.