Adjoint functor of power-set functor

Question

I want to ask a maybe stupid question. Power-set functor $P: \mathbf{Set}^{op} \rightarrow \mathbf{Set}$ is a well-known contravariant functor which sends each set to its powser set. A natural question is which functor is right(left) adjoint to it and I guess its oppostie functor $P^{op}: \mathbf{Set} \rightarrow \mathbf{Set}^{op}$ is right adjoint, is it true?

Answer 1

Although the question has an easy answer (given in my comment under the question), it may be well to expand that answer by pointing out some related facts.

So in the first place, we can say that in any cartesian closed category $\mathbf{C}$, or even in any symmetric monoidal closed category meaning there is a family of adjunctions $- \otimes A \dashv [A, -]$ parametrized over objects $A$, every object $D$ induces an adjunction where $[-, D]^{op}: \mathbf{C} \to \mathbf{C}^{op}$ is left adjoint to $[-, D]: \mathbf{C}^{op} \to \mathbf{C}$. For there are natural bijections

$$\hom(A, [B, D]) \cong \hom(A \otimes B, D) \cong \hom(B \otimes A, D) \cong \hom(B, [A, D])$$

where the middle isomorphism uses symmetry. This may be rewritten as a natural bijection

$$\hom_{\mathbf{C}^{op}}([B, D], A) \cong \hom_{\mathbf{C}}(B, [A, D])$$

which expresses the adjunction $[-, D]^{op} \dashv [-, D]$.

A deeper fact is that in the special case $\mathbf{C} = Set$ and $D = 2 = \{0, 1\}$, the functor

$$P = [-, 2]: Set^{op} \to Set$$

is monadic. That is to say: $Set^{op}$ is equivalent to a category of algebras $\mathbf{A}$ of an algebraic theory (identifying $P$ with the underlying set-functor $U: \mathbf{A} \to Set$). In this case, $\mathbf{A}$ is the category of complete atomic Boolean algebras (CABAs), or equivalently the category of completely distributive complete Boolean algebras. This is saying in particular that every CABA is a power set Boolean algebra $P(A)$, and conversely. (Contrast with the case of complete Boolean algebras, where in fact the underlying set-functor has no left adjoint.)

One of the big theorems of elementary topos theory (due to Robert Paré) is the corresponding generalization, that if $\Omega$ is the subobject classifier of a topos $\mathbf{E}$, then

$$P(-) = [-, \Omega]: \mathbf{E}^{op} \to \mathbf{E}$$

is also monadic. Since monadic functors reflect limits, this implies that finite limits exist in $\mathbf{E}^{op}$ since they exist in $\mathbf{E}$, and therefore a topos is finitely cocomplete.

Answer 2

It does not have a left adjoint, because it is essentially the forgetful functor from complete and atomic boolean algebras to the category of sets.

See https://ncatlab.org/nlab/show/complete+Boolean+algebra.