Lemma [efr-TZ3Y]

Suppose \mathcal {C} admits weak conditionals, and let \mathcal {D} \to \mathcal {C}_\mathrm {det} be a fibration. Then two morphisms f_0,f_1: \bar {X} \to \bar {Y} in \overline {\mathcal {D}}, represented by commutative diagrams

as well as \phi _i: \bar {X}_{M_i} \to \bar {Y}_{M_i}, for i=0,1, are equal if and only if there exists a span M_0 \leftarrow K \to M_1 over X,Y, with a stochastic section X \to K lifting both the sections to M_0,M_1, so that the pullbacks of \phi _0,\phi _1 to K agree.

Proof

The relation here described clearly implies identity, and contains all the generating identities, so it suffices to show it is an equivalence relation. Reflexivity and symmetry are clear, so transitivity is the only issue. It suffices to show that, given a span X \leftarrow S \to Y and maps M_0 \to S \leftarrow M_1 so that the triangles commute, and so that the two induced sections X \to S agree, and a map \bar {X}_S \to \bar {Y}_S which pulls back to \phi _0,\phi _1, we can find K as above.

To do this, take K = M_0 \times _S M_1. Clearly the maps to M_0, M_1 are over X,Y, and by the existence of weak conditionals there exists a common lift of the sections to X \to K. By functoriality of pullbacks, the pullbacks of \phi _0,\phi _1 to K agree.

Context

Free Markov Prefibrations [efr-GO6R]

Because every map in Kl(\Delta )^\to factors into arrows which are "induced" from arrows in Kl(\Delta )^\to |_\mathrm {det} and the Markov prefibration property, it may initially be hoped that Kl(\Delta )^\to is in some sense "free" on the data of the fibration Kl(\Delta )^\to |_\mathrm {det} \to \mathsf {Set} and the inclusion \mathsf {Set} \to Kl(\Delta ). If that was true, we may further hope that taking the fiberwise opposite of the fibration and applying the same free generation principle would generate a good notion of stochastic lens.

Unfortunately, this is not the case. We will see that the free prefibration is given by gadgets which look a bit like an indexed version of optics, up to a sliding equivalence for deterministic maps on the residual. This prompts us to look for some extra structure on the fibration \mathcal {D}|_\mathrm {det} \to \mathcal {C}_\mathrm {det} which describes sliding equivalences for stochastic maps on the residual. In the next section, we will see that this is exactly the structure of an Eilenberg-Moore algebra for the free prefibration monad on \mathsf {Fib}(\mathcal {C}_\mathrm {det}).

In this section, we will give a description of the free Markov prefibration on a fibration \mathcal {D}_0 \to \mathcal {C}_{\mathrm {det}} (assuming \mathcal {C} is pullback positive). There is a fairly simple description of the hom-sets, but their composition is a bit tricky, and verifying associativity even more so. Hence we will employ a technical trick: by characterizing the hom-sets as "freely generated" in a certain sense from the hom-sets in \mathcal {D}_0, we can identify them with sets of natural transformations using a Yoneda-type argument, and infer composition and associativity from there.

Definition Indexed copresheaf [efr-001I]

Let p: \mathcal {D} \to \mathcal {C} be a functor. An indexed copresheaf on \mathcal {D} is a tuple (X \in \mathcal {C}, F: \mathcal {D} \to \mathsf {Set}, \alpha : F(-) \to \mathcal {C}(X,p(-))) consisting of a copresheaf, an object of \mathcal {C}, and a natural transformation \alpha as indicated. We say the indexed copresheaf is over X, and we will abuse the terminology by referring to F itself as an indexed copresheaf, leaving the transformation \alpha implicit (for example, "let F be an indexed copresheaf over X").

We denote the subset \alpha ^{-1}(\{f\}) \subseteq F(\bar {A}), for f: X \to p(\bar {A}) by F(\bar {A})_f.

A map of indexed copresheaves (X,F,\alpha ) \to (Y,G,\beta ) is a natural transformation F \to G and a map Y \to X \in \mathcal {C} so that the obvious square of natural transformations commutes. We denote the category of indexed copresheaves by \mathsf {IcoPSh}(\mathcal {C} / \mathcal {D}). Note that there is an obvious forgetful functor \mathsf {IcoPSh}(\mathcal {D} / \mathcal {C})^\mathrm {op} \to \mathcal {C}

Observe that for each object A \in \mathcal {D}, there is a corepresentable copresheaf (p(A), \mathcal {D}(A,-), p). Maps between these obey the Yoneda lemma, in the sense that they are in bijection with maps between the underlying objects in \mathcal {D}. This defines a fully faithful functor \mathcal {D} \to \mathsf {IcoPSh}(\mathcal {D} / \mathcal {C})^\mathrm {op} over \mathcal {C}.

Of course, there is a dual notion of indexed presheaf, but this will not interest us.

Proposition [efr-FOJT]

Let p: \mathcal {D} \to \mathcal {C} be any functor and let \mathcal {C}_0 \to \mathcal {C} be an identity-on-objects functor. Write \mathcal {D}_0 = \mathcal {D} \times _\mathcal {C} \mathcal {C}_0 for the pullback. If F: \mathcal {D} \to \mathsf {Set} is a copresheaf indexed over X \in \mathcal {C}, the pullback \bar {A} \mapsto F(\bar {A}) \times _{\mathcal {C}(X,p\bar {A})} \mathcal {C}_0(X,p\bar {A}) is a copresheaf on \mathcal {D}_0 indexed over X again in a unique way. This defines a functor \mathsf {IcoPSh}(\mathcal {D} / \mathcal {C}) \to \mathsf {IcoPSh}(\mathcal {D}_0 / \mathcal {C}_0). Moreover, this functor preserves the corepresentable copresheaves (since \mathcal {C}_0 \to \mathcal {C} is identity on objects, so is \mathcal {D}_0 \to \mathcal {D}, so this statement makes sense).

Corollary [efr-FRQG]

Let \mathcal {D}_0 \to \mathcal {C}_0 be a functor and let \mathcal {C}_0 \to \mathcal {C} be identity-on-objects and faithful. Then the pullback of the composite \mathcal {D}_0 \to \mathcal {C} and the inclusion \mathcal {C}_0 \to \mathcal {C} is identical to \mathcal {D}_0. Therefore, Proposition [efr-FOJT] gives a functor \mathsf {IcoPSh}(\mathcal {D}_0 / \mathcal {C}) \to \mathsf {IcoPSh}(\mathcal {D}_0 / \mathcal {C}_0).

The idea of our construction of the free Markov prefibration is to give a certain monad on \mathsf {IcoPSh}(\mathcal {D}_0 / \mathcal {C}) and consider the Kleisli maps between the representable copresheaves.

At this point, the notion of a deterministic map M \to X equipped with a stochastic (ie not necessarily deterministic) section begins playing a key role. The phrase "stochastic section" will always carry an implicit "of a deterministic map". In most cases the map that the section is a section of will be clear from the context.

Note that, given a stochastic section Y \to M and a deterministic map X \to Y, if \mathcal {C} is pullback-positive, there is a unique lifting of this to a section of the projection Y \times _X M \to Y. We will use this fact several times.

Definition Stochastic Module [efr-Y926]

Let \mathcal {C} be a Markov category, and let \mathcal {D}_0 \to \mathcal {C}_\mathrm {det} be a fibration. Then a stochastic module consists of

  1. An indexed copresheaf F = (A,F,\rho ) \in \mathsf {IcoPSh}(\mathcal {D}_0 / \mathcal {C})
  2. For each \bar {X} \in \mathcal {D}_0, Cartesian morphism \bar {a}: \bar {X}_M \to \bar {X} lying over a: M \to X, and stochastic section \alpha : X \to M, a function \alpha _*: F(\bar {X}) \to F(\bar {X}_M), which acts on the underlying morphisms in \mathcal {C} as composition with \alpha
  3. Satisfying, whenever given a commutative square of Cartesian morphisms:
    and stochastic sections \alpha ,\beta lying over a = p(\bar {a}), b = p(\bar {b}), so that we have a digram in \mathcal {C}:
    Where the maps except \alpha ,\beta are deterministic, \alpha and \beta are sections of a and b, and both the square of deterministic maps and the square involving \alpha ,\beta commute, the condition that the square
    commutes.
  4. And satisfying furthermore the equation, for every two stochastic sections \alpha :X \to M, \beta : M \to N, the equation (\beta \alpha )_* = \beta _*\alpha _* (note that this makes sense because pullbacks compose).

A morphism of stochastic modules is an indexed natural transformation which preserves the operations \alpha _*. The category of stochastic modules is denoted \mathsf {SMod}(\mathcal {D}_0 / \mathcal {C}). There is an apparent forgetful functor \mathsf {SMod}(\mathcal {D}_0 / \mathcal {C}) \to \mathsf {IcoPSh}(\mathcal {D}_0 / \mathcal {C}).

Remark [efr-GL6R]

Note that \alpha _* of course depends on f, not just \alpha .

If f: M \to X is a deterministic map, \alpha :X \to M is a stochastic section, and \bar {X}_M, \bar {X}_M' \to \bar {X} are two Cartesian lifts of f to \bar {X} \in \mathcal {D}_X, then applying the commutativity axiom for stochastic modules implies that the triangle

commutes. In what follows, we will simply write f^*\bar {X} for some choice of cartesian lift, and speak of \alpha _* : F(\bar {X}) \to F(f^*\bar {X}). The above shows that this is a harmless abuse---the actions \alpha _* are preserved by the identification of different Cartesian lifts. In particular, we will often make arguments as if pullbacks compose strictly, although in general they only compose up to isomorphism. The above triangle means this is harmless.

The term "stochastic module" is not very good, but this is mostly a nonce definition in any case, so we won't worry too much about it.

Stochastic modules over a given X \in \mathcal {C} can be seen to be monadic over the category of indexed copresheaves over that X. However, the compatibility of these local left adjoints with the structure of the rest of the category is somewhat subtle. However, we do have free stochastic modules on representable indexed copresheaves, as we will soon see.

Proposition [efr-2XPE]

Let \mathcal {D} \to \mathcal {C} be a Markov prefibration, and let \mathcal {D}_0 = \mathcal {D} \times _\mathcal {C} \mathcal {C}_\mathrm {det}. Then the corepresentable copresheaf \mathcal {D}(\bar {A},-), restricted to \mathcal {D}_0, (but not pulled back---that is, we remember the whole set \mathcal {D}(\bar {A},\bar {X}), even the part over stochastic f, but only the composition with maps in \mathcal {D}_0) is a stochastic module in a canonical way, with \alpha _*: F(\bar {X}) \to F(f^*\bar {X}) given by composition with the unique induced lift of \alpha . Moreover, any morphism of Markov prefibrations \phi : \mathcal {D} \to \mathcal {D}' induces a homomorphism of stochastic modules \mathcal {D}(\bar {A},-) \to \mathcal {D}'(\phi (\bar {A}),-)

Proof

Given f: M \to X and a stochastic section \alpha , it's clear that composition with the unique lift \bar {X} \to f^*\bar {X} is a map of the right type, so we just have to verify the equations. For the first equation (item 3 in the definition of stochastic module), we are comparing two maps F(g^*\bar {X}) \to F(b^*\bar {X}). These are given by composition with two maps, let's call them \\phi, \psi : g^*\bar {X} \to b^*\bar {X}. These two maps are lifts of f\alpha and \beta g, but by assumption these two are equal. Hence by the uniqueness property of Markov prefibrations, \phi = \psi , and we have our equation. The other equation follows in a completely analogous way.

To prove the homomorphism property, note that a morphism of prefibrations preserves Cartesian morphisms, and hence (by uniqueness) must preserve the unique lifts of stochastic sections. Then by functoriality it must preserve composition with these, which finishes the proof.

Lemma [efr-OH7U]
  1. Let \bar {X} \in \mathcal {D}_0 be an object. Then there is a free stochastic module T\mathcal {D}_0(\bar {X},-) on its representable copresheaf, in the sense that if F is another stochastic module, homomorphisms T\mathcal {D}_0(\bar {X},-) \to F are in bijection with indexed natural transformations \mathcal {D}_0(\bar {X},-) \to F
  2. The free stochastic module on a corepresentable presheaf is given as follows: an element of T\mathcal {D}_0(\bar {X},-)(\bar {Y}) consists of a diagram in \mathcal {C} of the form
    (where X = p(\bar {X}), Y = p(\bar {Y}),) where fs = 1_X and f,g are deterministic, plus a map f^*\bar {X} \to \bar {Y} lying over g. This is up to the equivalence relation which, given some other such tuple, identifies them whenever there exists deterministic h: N \to M as in this diagram:
    so that the two deterministic triangles commute, hs' = s, and so that the unique Cartesian map f'^*\bar {X} \to f^*\bar {X} over h forms a commutative triangle with the two maps to \bar {Y}. (Note that we do not claim the relation just described is inherently an equivalence relation, rather we form the equivalence relation generated by this). It is clear how a map \bar {Y} \to \bar {Z} acts on this to make it a copresheaf. It is indexed by taking a tuple as above to the composite X \to M \to Y. Given N \to Y with a stochastic section s': Y \to N,, and an element of T\mathcal {D}_0(\bar {X},-)(\bar {Y}), the induced element in T\mathcal {D}_0(\bar {X},-)(s'^*\bar {Y}) is given by forming the pullback M \times _Y N, taking the pullback of the map over g to one lying over the projection M \times _Y N \to N, and composing the section s with the induced lift M \to M \times _Y N
  3. The underlying copresheaf of this respects Cartesian maps in \mathcal {D}_0 \to \mathcal {C}_\mathrm {det}, in the sense that given a Cartesian map \bar {A} \to \bar {B} and an element \phi \in T\mathcal {D}_0(\bar {X},-)(\bar {B}) lying over a deterministic map X \to B, the natural map from lifts \psi \in T\mathcal {D}_0(\bar {X},-)(\bar {A}) to lifts X \to A is a bijection.

Proof

First, we have to verify the equations of a stochastic module for T\mathcal {D}_0(\overline {X},-). For the first case, suppose we are given a square

with all but the upwards maps stochastic. Now let \bar {Z} be some object over Z and suppose we are given an element of T\mathcal {D}_0(\bar {X}, \bar {Z}_Y), represented by a span X \leftarrow M \to Y, a section X \to M and a map \phi : \bar {X}_M \to \bar {Z}_M over M. Consider then the below diagram:
By definition, the first of the two possible elements of T\mathcal {D}_0(\bar {Z}_W) are given by either forming the pullback M \times _Y N, taking the lift of \alpha to M \to M \times _Y N and composing to get a section X \to M \times _Y N, and pulling back \phi along the projection to get a map \bar {X}_{M \times _Y N} \to \bar {Z}_{M \times _Y N}, then taking the span X \leftarrow M \times _Y N \to W. The second is given by first composing with the map Y \to Z, then applying the above procedure with the pullback M \times _Z W. The induced map M \times _Y N \to M \times _Z W exhibits the equality of these two under the equivalence relation defining T\mathcal {D}_0(\bar {X},-) (commutativity of the bottom-right square implies that triangle of stochastic sections commutes.)

Given some other stochastic module F over A with a map of indexed copresheaves \mathcal {D}_0(\bar {X},-) \to F, over A \to X, there is at most one extension to a map of stochastic presheaves T\mathcal {D}_0(\bar {X},-) \to F over A \to X---given an element with representative (s:X \to M, X \leftarrow M \to Y, \bar {X}_M \to \bar {Y}_M), it must go to the identity element of F(\bar {X}), acted on by the stochastic section s to produce an element of F(\bar {X}_M), followed by F applied to the map \bar {X}_M \to \bar {Y} over M \to Y.

But it is not hard to see that the equivalence relation imposed by T\mathcal {D}_0(\bar {X},-) is implied by the equations of a stochastic module, and so this map is well-defined, establishing the property.

Secondly, let \bar {A} \to \bar {B} be a Cartesian map over A \to B, and take a commutative triangle

of deterministic maps. Finally take an element of T\mathcal {D}_0(\bar {X},-)(\bar {B}) over the given map X \to B. We must show it has a unique lift to A over the given map X \to A. Let us take a representative given by a diagram:
First, note that the two maps M \to X \to B and M \to B do not necessarily agree. However, we can remedy this by replacing M by their equalizer---note that as we argued above, this gives an equivalent element of the stochastic module. (By writing their equalizer in \mathcal {C}_\mathrm {det} as the split pullback M \times _{B \times B} B, we can see that the section factors over this, even if \mathcal {C} does not have all equalizers in general). Hence we can assume the triangle formed by adding the dashed arrow commutes.

Since M \to B now factors over X, the lift X \to A gives a lift M \to A. Now by the Cartesian property of \bar {A} \to \bar {B}, there is a unique lift of p^*\bar {X} \to \bar {B} to this map (here we just use the fact that \mathcal {D}_0 is a fibration). This gives the desired lift.

Finally, given two distinct lifts (again, we can assume their maps M \to A factor over X,) clearly any map N \to M witnessing an identity between their composites \bar {X} \to \bar {Y} would likewise exhibit an identity between their lifts (since pullbacks compose). This proves uniqueness, and finishes the proof.

We are now ready to prove the main proposition of this section:

Proposition [efr-LGOA]

Let \mathcal {C} be a pullback-positive Markov category and let \mathcal {D}_0 \to \mathcal {C}_\mathrm {det} be a fibration. Consider the full subcategory of stochastic modules spanned by the free modules on the corepresentables. Denote the opposite of this category \bar {\mathcal {D}_0}. Clearly there is a commutative diagram

We claim:

  1. \bar {\mathcal {D}_0} \to \mathcal {C} is a Markov prefibration.
  2. There is a bijection \bar {\mathcal {D}_0}(\bar {A},-) \cong T(\mathcal {D}_0(\bar {A},-)). When the left-hand side is equipped with the canonical stochastic module structure, and the right is equipped with the free one, this is moreover a homomorphism (hence isomorphism) of stochastic modules.
  3. \bar {\mathcal {D}_0} \to \mathcal {C} is initial among Markov prefibrations receiving a map from \mathcal {D}_0. In other words, this construction gives a left adjoint to the pullback functor \mathsf {MarkPreFib}(\mathcal {C}) \to \mathsf {Fib}(\mathcal {C}_\mathrm {det})

Proof [efr-0BES]

First observe that, by Lemma [efr-OH7U], the pullback \bar {\mathcal {D}_0}\times _\mathcal {C} \mathcal {C}_\mathrm {det} \to \mathcal {C}_\mathrm {det} is indeed a fibration, with the image of the Cartesian lifts under the functor \mathcal {D}_0 \to \bar {\mathcal {D}_0} being Cartesian again. (This also establishes that \bar {\mathcal {D}_0} really does receive a map of fibrations from \mathcal {D}_0)

Given Cartesian f: \bar {A} \to \bar {B} \leftarrow \bar {C}, and a stochastic lift A \to C (= p\bar {A} \to p\bar {C}), consider the pullback A \times _B C, and the pullback of \bar {A} to it. There is a unique lift of A \to C to a section A \to A \times _B C, and this induces a unique lift \bar {A} \to (f\pi _1)^*\bar {A} using the stochastic module structure. The composite of this with the projection to B is a lift of \bar {A \to \bar {B}} over A \to C, as required by a Markov prefibration.

Analogously to the proof of Lemma [efr-OH7U], given some other lift A \leftarrow N \to C, h: A \to N, h^*\bar {A} \to \bar {C}, the fact this is a factorization implies the existence of some M with maps M \to A, M \to N and a lift A \to M of the section A \to N, so that the induced map between the pullbacks over M and N of \bar {A} makes the triangle into \bar {B} commute. But then since this is a triangle over deterministic bases, this implies the lifted triangle to \bar {C} also commutes, hence this M lifts to another representative of the lift we started with. But then it's not hard to see that this M maps to A \times _B C and exhibits an equation with the previously constructed "canonical" lift.

Hence \bar {\mathcal {D}_0} is a Markov prefibration, and by the above, the induced stochastic module structure on the corepresentable presheaves \bar {\mathcal {D}_0}(\bar {A},-) = T(\mathcal {D}_0(\bar {A},-)) is exactly the one given by T (in other words this equation is not merely a bijection of sets, but an isomorphism of stochastic modules).

Let \phi : \mathcal {D}_0 \to \mathcal {D}' be a functor over \mathcal {C}_\mathrm {det} \to \mathcal {C} into some other Markov prefibration which preserves Cartesian maps. Using the algebra structure on \mathcal {D}'(\phi \bar {B},-), we see there is a unique extension of \phi to \bar {\mathcal {D}_0}(\bar {X},-) which respects the stochastic module structure. By chasing the diagram around it's easy to see that this is functorial, and hence gives a map of Markov prefibrations---conversely, any such map extending \phi must be a stochastic module homomorphism. Thus there is a unique functor, proving initiality.

This characterization of the left adjoint makes it fairly easy to understand the induced monad on \mathsf {Fib}(\mathcal {C}_\mathrm {det}).

Remark [efr-ACIV]

We will sometimes refer to the morphisms of \overline {\mathcal {D}_0} as precharts. Taking the fibration \mathcal {C}^\to |_\mathrm {det} \to \mathcal {C}_\mathrm {det} as an example, it is not too hard to see that the precharts between X \otimes A \to X and B \otimes Y \to Y are representatives of co-optics {A \choose X} \rightrightarrows {B \choose Y}. (To see this, note that any prechart is equivalent to one where the apex of the span has the form M \otimes Y and the right leg is the projection to Y. Then the rest of the data is a map X \to M \otimes Y and a map M \otimes B \to A, since the X-coordiante of the latter map is determined by the span).

In fact their equivalence relation is given by sliding equivalence for deterministic maps (i.e morphisms in \mathsf {Optic}_{\mathcal {C}_\mathrm {det}}(\mathcal {C}_\mathrm {det}, \mathcal {C})). The precharts in \mathcal {D}_0^\mathrm {fop} will be called prelenses. We will speak of the tuple (M, p:M \to X, p':M \to Y, s: X \to M, \phi : p^*\bar {X} \to p^*\bar {Y}) representing a prechart just as a "decorated span (representing ...)". When part of the structure is understood, or can just be left abstracted, we will denote such a decorated span simply by (M,s,\phi ), or even just (M,\phi ). It will be clear from context which part of the structure is being specified.

Proposition [efr-A08L]

Let \mathcal {D}_0 be a fibration. Then the underlying fibration of the free Markov prefibration, \overline {\mathcal {D}_0}|_\mathrm {det}, has fiber over X \in \mathcal {C} given by

  1. Objects are simply objects of \mathcal {D}_{0,X}
  2. A morphism \bar {X} \to \bar {X}' consists of a deterministic f: M \to X, a stochastic section s: X \to M, and a map \phi : f^*\bar {X} \to f^*\bar {X}', up to the equivalence relation generated by, whenever g: N \to M is deterministic and s': X \to N is a factorization of s, identifying (M,f,s,\phi ) with (N,fg, s', g^*(\phi )).
  3. Given two such morphisms (M,f,s,\phi ), (N,f',s',\psi ), their composite is represented by M \times _X N \to X equipped with the section formed as the composite of X \to M and the lift of X \to N to the pullback, and the composite \pi _M^*(\phi )\pi _N^*(\psi ) \in \mathcal {D}_{0,M\times _X N}
  4. Given deterministic f: X \to Y, the pullback is given on such a map by taking the pullback M \times _Y X \to X, the induced section, and the pullback of the map \phi along the projection M \times _Y X \to M

It is immediately obvious that we have:

Corollary [efr-ILIQ]

The free Markov prefibration monad \overline {(-)}|_\mathrm {det} commutes with fiberwise opposites. In particular, algebra structures on \mathcal {D}_0 are in bijection with algebra structures on \mathcal {D}_0^\mathrm {fop}, and fiberwise opposites lifts to an involution of \mathsf {Alg}(\overline {(-)}|_\mathrm {det}).

Definition Stochastic Module Fibration [efr-OEFJ]

Let \mathcal {C} be a pullback-positive Markov category. The adjunction \overline {(-)} \dashv (-)|_\mathrm {det} induces a monad on \mathsf {Fib}(\mathcal {C}_\mathrm {det}). A module for this monad is called a stochastic module over \mathcal {C} (or, to distinguish it from the copresheaves of Definition [efr-Y926], a stochastic module fibration). The category of stochastic module fibrations is denoted \mathsf {SFib}(\mathcal {C})

Let us try to understand the structure of a stochastic module fibration. It is easiest to understand in the case of a projection map P \times X \to X. Suppose we have two objects A,B over X. Then we think of a map f: \pi _X^*A \to \pi _X^*B over P \times X as a map P \times A \to B, that is a map parameterized by P (this is literally the case for a codomain fibration). Given a stochastic section s of \pi _X, which amounts to a stochastic map X \to P, the stochastic module stucture picks out a new map s_*f, which is given over each point x \in X by choosing the parameter according to s, then applying f.

The definition of the composite in Proposition [efr-A08L], in these terms, tells us that given maps f: P \times A \to B, g: Q \times B \to C, and maps s: X \to P, t: X \to Q, the composite of s_*(f) and t_*(q) is equal to the map obtained by forming the parameterized composite Q \times P \times A \to C and applying the independent pairing \langle t,s \rangle : X \to Q \times P. This is of course how composition is supposed to work in a Markov category.

Lemma [efr-VF6V]

Let \mathcal {D} be a stochastic module over \mathcal {C}, and let

be given, so that every map except s,s',t is deterministic. Suppose the deterministic part of the diagram commutes, fs = g, s' is the induced section, and t is a section. Let \bar {A},\bar {B} be two objects over Z. Suppose given a map \phi : f^*\bar {A} \to f^*\bar {B}. Then t^*h^*(P) = (s')^*\pi _Y^*(P) : g^*\bar {A} \to g^*\bar {B}

In particular, this operation depends only on s. Moreover, it is functorial, in the sense that given a diagram

with the downwards maps deterministic, s^*t^* = (ts)^*

Remark [efr-U436]

The operations \alpha ^* associated to stochastic lifts are "functorial" in the sense that (\alpha \beta )^* = \beta ^*\alpha ^*. However they are not functorial in the sense that \alpha ^*(fg) = \alpha ^*(f)\alpha ^*(g)

To make sense of this, consider a simple case of a map m: I \to X in Kl(\Delta ). Given two objects over * (in Kl(\Delta )^\to ), a map \bar {A}_X \to \bar {B}_X is equivalent to a parametrized map X \times \bar {A} \to \bar {B}. The operation m^* consists in sampling this parameter according to the distribution m---but since composition in the fiber over X is defined by copying the parameter, but composition in the fiber over * (i.e just Kl(\Delta )) is defined by composing the kernels under conditional independence, these only agree if the distribution m is assumed to be deterministic.

Note that if either f or g is pulled back from a map \bar {A} \to \bar {B} (i.e, if they do not depend on the parameter X), the composition is preserved.

By construction, two morphisms in \overline {\mathcal {D}}_0 represented by spans with apex M,M', are identified if there exists a zig-zag M \to K_0 \leftarrow K_1 \to \cdots \leftarrow M' of spans (decorated with sections from the domain X and morphisms in the fiber, satisfying equations, etc). We will now prove a lemma that allows us to cut this down to a smaller set in many conditions. We will need the following hypothesis:

Definition Weak conditionals [efr-2MH5]

Let p: \mathcal {D} \to \mathcal {C} be a Markov prefibration. We say p (or, abusing notation, \mathcal {D}) admits weak conditionals if, given a Cartesian map \bar {Y} \to \bar {Z} and any map \bar {X} \to \bar {Z}, the existence part of the Cartesian condition holds---that is, every factorization p(\bar {X}) \to p(\bar {Y}) admits a lift, although not necessarily a unique one.

We say a Markov category \mathcal {C} admits weak conditionals if its codomain functor \mathcal {C}^\to \to \mathcal {C} is a prefibration which admits weak conditionals---this is equivalent to requiring that it is pullback-positive, and that all deterministic pullbacks are carried to weak pullbacks by the inclusion \mathcal {C}_\mathrm {det} \to \mathcal {C} (in other words, that they satisfy the existence part of the universal property even for pairs of nondeterministic maps).

Observe that, if \mathcal {C} admits conditionals, it certainly admits weak conditionals: given a pullback X \times _Z Y, and maps P \to X,Y, form a Bayesian inverse of Y \to Z with respect to the given measure, and use that to build a lifting X \to Y, which gives X \to X \times _Z Y---then a diagram chase verifies that this map has the desired properties.

Lemma [efr-TZ3Y]

Suppose \mathcal {C} admits weak conditionals, and let \mathcal {D} \to \mathcal {C}_\mathrm {det} be a fibration. Then two morphisms f_0,f_1: \bar {X} \to \bar {Y} in \overline {\mathcal {D}}, represented by commutative diagrams

as well as \phi _i: \bar {X}_{M_i} \to \bar {Y}_{M_i}, for i=0,1, are equal if and only if there exists a span M_0 \leftarrow K \to M_1 over X,Y, with a stochastic section X \to K lifting both the sections to M_0,M_1, so that the pullbacks of \phi _0,\phi _1 to K agree.

Proof

The relation here described clearly implies identity, and contains all the generating identities, so it suffices to show it is an equivalence relation. Reflexivity and symmetry are clear, so transitivity is the only issue. It suffices to show that, given a span X \leftarrow S \to Y and maps M_0 \to S \leftarrow M_1 so that the triangles commute, and so that the two induced sections X \to S agree, and a map \bar {X}_S \to \bar {Y}_S which pulls back to \phi _0,\phi _1, we can find K as above.

To do this, take K = M_0 \times _S M_1. Clearly the maps to M_0, M_1 are over X,Y, and by the existence of weak conditionals there exists a common lift of the sections to X \to K. By functoriality of pullbacks, the pullbacks of \phi _0,\phi _1 to K agree.