Example [efr-8B5X]

The functor \mathsf {Optic}(\mathsf {BorelStoch}) \to \mathsf {BorelStoch} is not a Markov prefibration, although its pullback over \mathsf {BorelStoch}_\mathrm {det} is a Grothendieck fibration.

To see this, first consider the deterministic pullback. An optic \binom {A}{X} \to \binom {B}{Y} with deterministic base can be identified with a map X \otimes B \to A (and the base deterministic map X \to Y). To see this, first observe that the subset of \mathsf {BorelStoch}(X, Y \otimes M) with the marginal X \to Y deterministic is in bijection with \mathsf {BorelStoch}_\mathrm {det}(X,Y) \times \mathsf {BorelStoch}(X,M), since \mathsf {BorelStoch} is positive. Hence we can calculate \int ^M \mathsf {BorelStoch}_\mathrm {det}(X,Y) \times \mathsf {BorelStoch}(X,M) \times \mathsf {BorelStoch}(M \otimes B, A) \cong \mathsf {BorelStoch}_\mathrm {det}(X,Y) \times \mathsf {BorelStoch}(X \times B, A), using the ninja yoneda lemma as in Proposition [efr-M19V]. Hence this part is a fibration with the fiber over X being the coKleisli category of the X \times - monad, and the pullback functors given by reindexing these parametrized maps. The Cartesian lift of a map X \to Y at \binom {B}{Y} is given by the optic \binom {B}{X} \to \binom {B}{Y} with unit residual and identity backwards component.

Now, let g: I \to \mathbb {R} denote the standard Gaussian distribution, let f: \mathbb {R} \otimes \mathbb {R} \to \mathbb {R} be the function given by f(x,y) = 0 if x=y and y otherwise, and consider the two optics \binom {\mathbb {R}}{*} \to \binom {\mathbb {R}}{\mathbb {R}} given by (I, g: I \to \mathbb {R}, 1_\mathbb {R}: \mathbb {R} \to \mathbb {R}), (\mathbb {R}, \mathrm {copy}_\mathbb {R} g : I \to \mathbb {R} \otimes \mathbb {R}, f: \mathbb {R} \otimes \mathbb {R} \to \mathbb {R}) (where we recall that the first argument is the residual). They cannot be equal, as postcomposition with the optic \binom {\mathbb {R}}{\mathbb {R}} \to \binom {*}{*} given by the identity \mathbb {R} \to \mathbb {R} yields, for the former, the standard Gaussian g: I \to \mathbb {R}, and for the latter, the constant zero map. But postcomposition with the projection \binom {\mathbb {R}}{\mathbb {R}} \to \binom {\mathbb {R}}{*} does give the same optic (the identity), because, for every fixed y \in \mathbb {R}, when x is normally distributed, f(x,y) = y with probability one. Hence the unique lifting of Cartesian maps over Cartesian maps cannot hold.