Is there an extension of the covariant powerset functor to $\mathbf{Rel}$ that both agrees with its definition on $\mathbf{Set}$, and preserves the dagger involution on $\mathbf{Rel}$?
Relations are like functions, but more symmetrical: they behave the same with respect to inputs and outputs. I'm trying to understand whether there is a relational extension of the idea of "mapping a function over a set" that is symmetrical in this way. I'm using "mapping" here in the programming sense of applying something to every element of a set.
We have the covariant powerset functor $P: \mathbf{Set} \rightarrow \mathbf{Set}$, which maps a function over sets: For $f: X \rightarrow Y$,
$$P(f): P(X) \rightarrow P(Y)$$ $$P(f)(A) = \{f(a) \,|\, a \in A\}$$
We could also say that $P(f)$ sends sets to their images under $f$. The idea of images is not exclusive to functions, it also applies to relations; so we can define another functor $P_{RF}: \mathbf{Rel} \rightarrow \mathbf{Set}$ that sends sets to their images under a relation. For $R \subseteq X \times Y$,
$$P_{RF}(R): P(X) \rightarrow P(Y)$$
$$P_{RF}(R) (A) = \{y \; | \; \exists a \in A, a R y \}$$
However, this functor treats inputs and outputs differently: it's a functor to $\mathbf{Set}$, not $\mathbf{Rel}$. Even if we think of $\mathbf{Set}$ as a subcategory of $\mathbf{Rel}$ using the functor $Graph: \mathbf{Set} \hookrightarrow \mathbf{Rel}$, we have that
$$Graph(P_{RF}(R^\dagger)) \neq Graph(P_{RF}(R))^\dagger$$
What I'm wondering is: is there a functor $P_{RR} : \mathbf{Rel} \rightarrow \mathbf{Rel}$ such that
$$P_{RR} \circ Graph = Graph \circ P$$ and $$P_{RR}(R^\dagger) = P_{RR}(R)^\dagger$$ ?