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Stream: learning: questions

Topic: Do all adjunctions in Sets arise from product ⊣ exponential?

Ignat Insarov (Apr 15 2025 at 22:00):

The adjunction between product and exponential functors, from which the Cartesian closed structure arises, is famous. (Which is to say, the only one I know, sad laugh.) It is also special because it involves two endofunctors — it is not a «free-forgetful» adjunction. Are there any other adjunctions that completely fit into the category of sets and functions like this, and that are not constructed on top of product and exponential, but are entirely independent? Is there any kind of a classification of adjunctions between endofunctors?

John Baez (Apr 15 2025 at 22:21):

Are you familiar with the adjunction between the binary product $\times: \mathsf{Set} \times \mathsf{Set} \to \mathsf{Set}$ and the diagonal functor $\Delta : \mathsf{Set} \to \mathsf{Set} \times \mathsf{Set}$, or the adjunction between the binary coproduct $+ : \mathsf{Set} \times \mathsf{Set} \to \mathsf{Set}$ and the diagonal functor?

Taken together with the one you mentioned, these are nice because we can start with the diagonal functor and then build up addition (binary coproduct), multiplication (binary product) and exponentiation by taking adjoints.

Ignat Insarov (Apr 15 2025 at 22:23):

Yes, I am more or less familiar with them, but they are not endofunctors!

John Baez (Apr 15 2025 at 22:26):

Oh, you need endofunctors?

John Baez (Apr 15 2025 at 22:26):

Okay.

John Baez (Apr 15 2025 at 22:37):

I wonder if I know any "left adjoint monads" on Set. These are endofunctors that happen to be monads and happen to be left adjoints . They have right adjoints that are comonads.

John Baez (Apr 15 2025 at 23:11):

I guess the example you mentioned, $S \times - : \mathsf{Set} \to \mathsf{Set}$, becomes a left adjoint monad when the set $S$ is a monoid, but I'm wondering if there are other examples.

Mike Shulman (Apr 15 2025 at 23:17):

If $F:\rm Set \to Set$ is a left adjoint, then in particular it is cocontinuous. Since Set is the free cocompletion of a point, this means that $F$ is uniquely determined, up to unique isomorphism, by its value on a one-element set. If $F(1) = S$, then the unique cocontinuous extension of this is $F(X) = S\times X$. So the answer is no! There are no other endo-adjunctions of Set.

John Baez (Apr 15 2025 at 23:47):

Darn. I should have seen that.

The good news is that you've now got the complete classification of all adjoint endofunctors on $\mathsf{Set}$, @Ignat Insarov.

Kevin Carlson (Apr 16 2025 at 01:20):

So the category of left adjoint, equivalently right adjoint, endofunctors of Set is not so huge, being just Set itself. But there are vastly more if you loosen up a bit. For instance parametric right adjoints (which preserve just the connected limits, or roughly speaking “all limits except products and the terminal object”) are [[polynomial functors]], which are all the functors of the form $X\mapsto A X^B+C X^D+\cdots$. In particular, they subsume both the left and the right adjoint endofunctors that were listed above!

Ignat Insarov (Apr 16 2025 at 11:02):

I am very proud that I can understand what Mike just said!

Madeleine Birchfield (Apr 16 2025 at 12:37):

It's also true in $(\infty,1)$-category theory for left adjoint endofunctors on $\infty\mathrm{Grpd}$ being the product functor $X \mapsto S \times X$

Jencel Panic (Apr 24 2025 at 07:04):

Mike Shulman said:

If $F:\rm Set \to Set$ is a left adjoint, then in particular it is cocontinuous. Since Set is the free cocompletion of a point, this means that $F$ is uniquely determined, up to unique isomorphism, by its value on a one-element set. If $F(1) = S$, then the unique cocontinuous extension of this is $F(X) = S\times X$. So the answer is no! There are no other endo-adjunctions of Set.

Wow, this comment should be a paper.

Kevin Carlson (Apr 24 2025 at 23:44):

It’s not a new result.