Whenever a category $\mathcal{C}$ is being viewed a subcategory of a dagger category $\mathcal{D}$, define that the dagger category generated by $\mathcal{C}$ is the least subcategory of $\mathcal{D}$ that is closed under the dagger operation and includes $\mathcal{C}$.
Question. Viewing $\mathcal{C}=\mathsf{Set}$ as a subcategory of $\mathcal{D}=\mathsf{Rel}$, what is the dagger category generated by $\mathcal{C}$? I'm especially interested to see whether all of $\mathcal{D}$ is generated in this case.
I think that indeed all of $\mathsf{Rel}$ is generated by the following argument:
Consider a relation $R\subseteq X\times Y$, then we have natural maps $i\colon R\to X\times Y$, $\pi_1\colon X\times Y\to X$ and $\pi_2\colon X\times Y\to Y$. Define $R^\prime = (\pi_2\circ i)\circ(\pi_1\circ i)^\dagger$.
Now $x R^\prime y$ iff $\exists z\in R\colon x(\pi_2\circ i) z\mbox{ and }z(\pi_1\circ i)^\dagger y$, which is true iff $\exists z\in R\colon x = \pi_2(i(z))\mbox{ and }y=\pi_1(i(z))$, which in turn is equivalent to $(x,y)\in R$ or $xRy$. Hence $R=R^\prime$.
Since $\pi_2\circ i$ is in $\mathsf{Set}$ and $(\pi_1\circ i)^\dagger$ is in $\mathsf{Set}^\dagger$, this proves that $R$ is in the dagger category generated by $\mathsf{Set}$.