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Stream: learning: questions

Topic: ✔ Whence «strong» functors and «commutative» monads?

Ignat Insarov (Jun 18 2025 at 03:08):

Thank you @John Baez, this makes a lot of sense!

Notification Bot (Jun 18 2025 at 03:09):

Ignat Insarov has marked this topic as resolved.

John Baez (Jun 18 2025 at 07:14):

Great!

Aaron David Fairbanks (Jul 02 2025 at 04:39):

Too late to tag you @Oscar Cunningham ?

Ignat Insarov (Jul 02 2025 at 08:12):

I should rejoice to see those diagrams too.

Oscar Cunningham (Jul 02 2025 at 10:10):

Oh, thanks for the reminder!

So, suppose we have a monoid $M$ with multiplication $m$ and unit $e$. We can draw the multiplication $m:M\times M\to M$ as a diagram with two inputs and one output.
(1)

Likewise the unit can be drawn as a diagram with no inputs and one output.
(2)

The associativity axiom is drawn as follows:
(3)

And the identity property like so:
(4)

If $M$ is a commutative monoid, then it satisfies $m(x,y)=m(y,x)$, which we draw as an axiom saying that you can swap the two wires:
(5)

Note that we need to draw a 'braiding' to swap the legs representing $x$ and $y$. This makes sense for monoids in $\mathbf{Set}$ because for any sets $X$ and $Y$ there is a natural transformation $X\times Y \to Y\times X$ sending $(x,y)$ to $(y,x)$.

Now suppose we have a monad $(T,\mu,\eta)$ on some category $\mathbf{C}$. We can draw the natural transformation $\mu$ like so:
(6)

The red areas represent the category $\mathbf{C}$. The top edge represents the functor $T\circ T$. The bottom edge represents the functor $T$, and the interior of the diagram represents the natural transformation $\mu:T^2\to T$. We can represent $\eta:\mathrm{Id}_\mathbf{C}\to T$ similarly:
(7)

These natural transformations obey associativity and identity axioms that look just the same as (3) and (4) except in red boxes. This is what is meant by 'a monad is just a monoid in the category of endofunctors'.

Now, can we define what it means for $T$ to be commutative? We would want to draw a diagram like (5). But there is no way to ascribe a meaning to the 'braiding' in that diagram. If $X$ and $Y$ are endofunctors of $\mathbf{C}$ there is no canonical natural transformation $X\circ Y\to Y\circ X$. So we can't define 'commutative monad' as simply a commutative monoid in the category of endofunctors.

Now suppose that we are given a monoidal structure on the category $\mathbf{C}$. This has a multiplication map $\otimes:\mathbf{C}\times\mathbf{C}\to\mathbf{C}$. We can draw a diagram for this in three dimensions. It gives a way to take two of the red sheets representing $\mathbf{C}$ and join them together.
(8)

A strong monad is a monad $T$ equipped with a right-strength and a left-strength. These strengths are natural transformations that allow us to move the wire representing $T$ through $\otimes$ onto either of the two sheets.
(9)

(The strengths also have to obey some axioms which I can't be bothered to draw.) A commutative monad is then defined as a strong monad obeying the following axiom:
(10)

Note that on the left-hand-side of this axiom the wire from the left side of $\mu$ ends up on the front sheet, and the wire from the right side of $\mu$ ends up on the rear sheet. Whereas on the right-hand-side of this axiom the wire from the left side of $\mu$ ends up on the rear sheet, and the wire from the right side of $\mu$ ends up on the front sheet.

So this axiom says that a commutative monad is precisely a monad where you can swap the two wires!

Oscar Cunningham (Jul 02 2025 at 10:14):

@Ignat Insarov @Aaron David Fairbanks

Ignat Insarov (Jul 04 2025 at 08:10):

@Oscar Cunningham   This is lovely! You should make a blog post out of it or something like that.

Oscar Cunningham (Jul 04 2025 at 08:19):

Yeah, it would be good to also draw the diagrams for monoidal monads and explain why they're equivalent. I'll put it on my stack of unwritten blog posts.