# What are some examples of self-adjoint functors? Is this an example?

I've been trying to figure out what some examples of "self-adjoint" functors are, or when this even happens, since I've never seen this before. What I mean is if $$F: \mathcal{C} \to \mathcal{C}$$ is a functor, then $$F$$ is self-adjoint if we have the natural bijection $$\text{Hom}_{\mathcal{C}}(F(A), B) \cong \text{Hom}_{\mathcal{C}}(A, F(B))$$ for all objects $$A, B \in \mathcal{C}$$. In terms of the unit/counit language, the universal diagrams would be as below. (Forgive the abuse of $$f$$ and $$g$$.) The second half of this question is the following: Is $$(-)^{\text{op}}: \textbf{Cat} \to \textbf{Cat}$$ an example? I think so. This is because $$(\mathcal{A}^{\text{op}})^{\text{op}}= \mathcal{A}$$ for any category $$\mathcal{A}$$. Hence we can take the $$\eta_{\mathcal{A}} = \epsilon_{\mathcal{A}} = 1_{\mathcal{A}}$$, so the unit and counits are trivial. The unique existence of a functor completing the commutative triangles is just given by $$(-)^{\text{op}}$$, e.g., if we have a functor $$g: \mathcal{A} \to \mathcal{B}^\text{op}$$, then take $$f = g^{\text{op}}: \mathcal{B} \to \mathcal{A}^{\text{op}}$$. The diagram on the above left commutes, and similar reasoning gives us the counit diagram on the right. However, this example seems a bit trivial that happens to work out because $$(-)^{\text{op}}$$ is a nice functor; hence my question regarding more interesting examples.

• What software are you using to draw those diagrams so nicely? – StudySmarterNotHarder Aug 4 '20 at 20:44

This is a relatively uncommon scenario, but here are a few more examples.

• As you suggest, $$(-)^\mathrm{op} : \mathbf{Cat} \to \mathbf{Cat}$$ is self-adjoint. More generally, this should be true for the underlying category of any 2-category with a duality involution.
• If $$\mathscr C$$ has biproducts, so products coincide with coproducts, then the composite $$\oplus \circ \Delta_n$$ of the (discrete $$n$$-ary) diagonal functor $$\Delta_n$$ with the ($$n$$-ary) biproduct functor $$\oplus$$ is self-adjoint.
• For completeness, the identity functor is self-adjoint.

There are also similar examples that involve a variance change, i.e. functors $$F : \mathscr C^\mathrm{op} \to \mathscr{C}$$, such that $$F \dashv F^\mathrm{op}$$. These are called self-adjoint on the left. The name comes from the characterisation in terms of hom-sets of $$\mathscr C$$, i.e. we have natural isomorphisms $$\mathscr C(F(A), B) \cong \mathscr C(F(B), A)$$. Conversely, if $$F$$ is self-adjoint on the right, then we have natural isomorphisms $$\mathscr C(A, F(B)) \cong \mathscr C(B, F(A))$$.

• The contravariant powerset functor $$\mathcal P: \mathbf{Set} \to \mathbf{Set}^\mathrm{op}$$ is left-adjoint to $$\mathcal{P}^\mathrm{op} : \mathbf{Set}^\mathrm{op} \to \mathbf{Set}$$, i.e. self-adjoint on the right.
• More generally, in a symmetric monoidal closed category $$(\mathscr C, \otimes, I, \multimap)$$, for a fixed object $$A$$, the functor $$(-) \multimap A$$ is self-adjoint on the right.
• In a similar vein, functors self-adjoint on the right are used in Thielecke's Categorical Structure of Continuation Passing Style to describe the structure of CPS (see Example 4.3.2).
• I really like this answer, thank you! It seems that since this construction doesn't naturally arise, it's not actually that interesting. I guess this is a lesson in that "tools" should not always determine the problems of interest, but problems should guide the tools. – trujello Aug 3 '20 at 4:43
• @trujello The first example, of $\mathrm{op}$ arises very naturally. It also gives you self-adjoint functors on the category of groups, rings, monoids, etc. – Oscar Cunningham Aug 19 '20 at 10:58