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Stream: learning: questions

Topic: Trying to classify self-adjoint functors

Jack Jia (Apr 10 2026 at 22:03):

There is a systematic way to produce a lot of self-adjoint functors (I don't count op as self-adjoint in general, so F:$\mathcal{C} \rightarrow \mathcal{C}$ must have source and target agree): We can use the colimit-diagonal-limit adjunction, and for each category where colimits and limits of a certain shape agree (for instance, an abelian category with finitary biproducts. e.g. $F:= \oplus \circ \Delta$ from the category of abelian groups to itself is self-adjoint).

Now I have not been able to come up with an example of a self-adjoint functor that does not come from this. Do all self-adjoint functors look like this?

Jack Jia (Apr 10 2026 at 23:32):

I guess I never clarified what "look like this" means: I mean if we have a pair of biadjoints $F \dashv G \dashv F$, then $F \circ G$ (or $G \circ F$) is self-adjoint. So is it true that every self-adjoint functor $H$ can be decomposed as $F \circ G$ for some $F$ and $G$ where $F \dashv G \dashv F$?

fosco (Apr 11 2026 at 09:28):

..and I guess you don't want to count $[-,X] : C \to C$ in a monoidal closed category $C$ because it's contravariant.

Jack Jia (Apr 11 2026 at 16:27):

fosco said:

..and I guess you don't want to count $[-,X] : C \to C$ in a monoidal closed category $C$ because it's contravariant.

Yes. I would like to say all functors are covariant and put op where I need them.

Jack Jia (Apr 11 2026 at 17:01):

Also I have worries about Fosco’s example, let me explain a bit: in the category of abelian groups, we have two functors both called $\text{Hom}(-,A)$, they are adjoints, but only on one side! Maybe Fosco’s example is fine though.

Todd Trimble (Apr 12 2026 at 03:33):

Any self-dual object in a monoidal category, i.e., an object $X$ equipped with maps $I \to X \otimes X$ and $X \otimes X \to I$ satisfying suitable triangular equations, should give an example. For example, in the category of sets and relations with the usual monoidal product given objectwise by taking cartesian product, there is a natural bijection between relations $X \times Y \nrightarrow Z$ and relations $Y \nrightarrow X \times Z$. Or even more simply, any finite-dimensional vector space can be equipped with a self-dual structure (using a nondegenerate bilinear form, inducing an isomorphism $V \cong V^\ast$), and then $V \otimes -$ is self-adjoint.

It seems to be we shouldn't expect such $X \otimes -$ to be describable as a composite coming $F \circ G$ from an ambidextrous adjunction $F \dashv G \dashv F$. That would be like saying that any self-dual object automatically carries a Frobenius monoid structure. Maybe someone can take that ball and run it into the endzone.

Todd Trimble (Apr 12 2026 at 17:58):

So one simple way to proceed would be to construct, explicitly, the free or initial compact closed category (let's say symmetric strict monoidal) generated by a self-dual object.

I like to think of this string-diagrammatically, and if you're used to this, you can probably do this yourself with no assistance from me. Formally, the objects are finite sets $[n] = \{1, \ldots, n\}$, and morphisms $[m] \to [n]$ are compact 1-manifolds whose boundary is the disjoint sum $[m] + [n]$, considered up to diffeomorphism rel boundary. Such a $1$-manifold is a finite union of circles or loops and line segments, so we could alternatively describe a morphism $[m] \to [n]$ as an ordered pair $(P, k)$ where $P$ is a partition of the disjoint sum $[m] +[n]$ into two-element subsets, each such subset being the boundary of a line segment, and where $k$ counts the number of loops.

Particularly, there is a morphism $u: [0] \to [2]$ consisting of a single line segment, and a morphism $c:[2] \to [0]$, again consisting of a single line segment.

Such morphisms can be composed in parallel (i.e., tensored by disjoint sum $+$) or in series. When composing in series, extra loops might be formed when we glue manifolds $[m] \to [n]$, $[n] \to [p]$ along $[n]$, so you need to keep track of these. Anyway, unless I've made some mistake, it should be true that this gives the free structure.

So $[1]$ is a self-dual object, with unit $u$ and counit $c$, and the functor $[1] + (-)$ will be self-adjoint. But this self-adjoint functor cannot be of the form $F \circ G$ for some ambidextrous adjunction $F \dashv G \dashv F$, for the simple reason that there is no monad structure on $[1] + (-)$. If there were, then you would have a multiplication $[2] \to [1]$. But there are no such maps in the free structure described above.