Is the identity our old friend \$$\mathrm{hom}\$$? Or rather, in this case, \$$\mathrm{hom}\_{\mathbf{Bool}}\$$?

Edit: I ask because it has the same typing constraints,

\$\mathrm{hom} : X^{op} \times X \to \mathbf{Set} \\\\ \mathrm{hom}\_{\mathbf{Bool}} : X^{op} \times X \to \mathbf{Bool}. \$