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. enter image description here

(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.

  • $\begingroup$ What software are you using to draw those diagrams so nicely? $\endgroup$ Aug 4, 2020 at 20:44

1 Answer 1


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).
  • 2
    $\begingroup$ 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. $\endgroup$
    – trujello
    Aug 3, 2020 at 4:43
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    $\begingroup$ @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. $\endgroup$ Aug 19, 2020 at 10:58

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