# What kind of natural isomorphisms are alpha, lambda, and rho of a monoidal category

I've read about monoidal categories, and I understand them, however I'm still confused about calling the associator, left unitor, and right unitor natural isomorphisms.

Natural isomorphisms are just natural transformations, where every component has an inverse in the category itself.

But I wonder what kind of natural transformation they are? What are their domain and codomain? E.g. the left unitor:

$$\lambda: (I ~\otimes ~ -) \cong -$$

A natural transformation has a component for every object in the domain. I.e. it should also have a component for $(c,c)$ for any object $c$. But what would $\lambda(c ~\otimes ~ c)$ be? just an identity function?

The same applies to the associator, what e.g. would $\alpha(I \otimes I)$ be?

• $\lambda$ is an isomorphism from the functor $I\otimes -$ to $1$. It means that you have $\lambda_A: I\otimes A \to A$ for every $A$ Dec 11 '17 at 14:41
• For the associator I think this is appropriate reading: math.stackexchange.com/questions/440758/… Dec 11 '17 at 14:44
• But what kind of functor is $1$? If $I \otimes -$ maps $(I,c)$ to $I \otimes c$, does $1$ map $(1,c)$ map to $c$? Why is it 1? It implies some kind of identity? Dec 11 '17 at 14:59
• $1$ is the identity functor, which is a very simple functor. It takes everything to itself. $\lambda$ takes $I\otimes A$ to $A$ so that the element $I$ behaves like the unit of a group or a monoid under the product $\otimes$. (hence the name monoidal category!) Dec 11 '17 at 15:17
• Ahh, I get it, once I thought about $I \otimes -$ being an endofunctor, it clicked. Dec 11 '17 at 15:21

Compiled from my comments above:

$\lambda$ is an isomorphism of the endofunctors $I\otimes$ and $1$, where $1$ is the identity functor. The map $\lambda_A$ goes from $I\otimes A$ to $A$ and tells us that $I\otimes A \cong A$.

This is just like the identity element of a monoid, which is where the name comes from.

Correspondingly, the other isomorphism tells us that $A\otimes I\cong A$.

The associator, which tells us that $\otimes$ is associative, is a bit more complicated, and if you want details about it you can see this question for instance: Understanding associators as natural transformations

• I think in the second paragraph you mean "$I\otimes A \simeq A$", right?
– Pece
Dec 11 '17 at 17:00
• @Pece I mean whatever symbol you think means isomorphism. Dec 11 '17 at 17:13
• I was trying to point out a typo, the $I$ and $A$ should be swapped when talking about the left unitor.
– Pece
Dec 11 '17 at 22:00
• @Pece with "other isomorphism" I'm pretty sure he's referring to the right unit or Dec 11 '17 at 23:25
• @hgiesel I was talking about the first occurence, but Berci edited it since. (Maybe I should have done that directly.)
– Pece
Dec 12 '17 at 6:10