Example \mathsf {Stoch}^\to as Markov prefibration [efr-NC7D]

Recall that the deterministic maps in \mathsf {Stoch} are those kernels valued in \{0,1\}-valued measures. (And that these are not the same thing as the measurable maps). Given a pair f:X \to Z, g: Y \to Z of such maps, we claim that the subset of X \times Y given by those points where f(x) = g(y) as measures on Z, equipped with the subset \sigma -algebra and its inclusions into X and Y, is a pullback in \mathsf {Stoch}_\mathrm {det}.

To see this, observe that \mathsf {Stoch}_\mathrm {det} is the Kleisli category for the submonad G_{0,1} of the Giry monad which takes a space to the subspace of 0,1-valued probability measures on it. Note that by the general theory of Markov categories this has products given by the products in \mathsf {Meas}. It hence suffices to prove that the pullback of f \times g: X \times Y \to Z \times Z along the diagonal Z \to Z \times Z exists---if it does, it has the universal property of the product X \times _Y Z.

To see that it does (and that it's given by the space described above), note that the diagonal is a split monomorphism, so it suffices to show that h: P \to X \times Y \in \mathsf {Stoch}_\mathrm {det} factors over X \times _Y Z if and only if the composite to Z \times Z lifts over the diagonal.

Now this lift over Z exists if and only if the probability of the diagonal \operatorname {im}(Z) \subseteq Z \times Z under the composite is 1. By definition, it is \int f(\operatorname {im}(Z) \mid x) g(\operatorname {im}(Z) \mid y) h(dx,y \mid p) Since the function being integrated is an indicator, this is simply the measure of the set \{(x,y) : \mathbb {P}(\operatorname {im}(Z) \mid x,y) = 1\}. If the probability of the diagonal is one, clearly the two marginals agree. Conversely, since \mathsf {Stoch}_\mathrm {det} is Cartesian, it must be the case that if the marginals agree the probability of the diagonal is 1. Therefore this is equal to the subset X \times _Z Y \subseteq X \times Y described at the start. But for this to have measure 1 for all p \in P is equivalent to the desired lifting property. This finishes the argument.

Moreover, this argument relied only on the 0,1-valuedness of the maps f,g, not h: P \to X \times Y or its marginals. Hence this also proves that the pullback along the diagonal is preserved by the inclusion \mathsf {Stoch}_\mathrm {det} \to \mathsf {Stoch}. Since \mathsf {Stoch} is known to be positive, this implies \mathsf {Stoch}^\to \to \mathsf {Stoch} is a Markov prefibration.