There exists a (strong) symmetric monoidal functor \mathsf {Optic}(Kl(\Delta )) \to \mathsf {SLens}(Kl(\Delta )^\to ) which

1. Is fully faithful.
2. Preserves the good coproducts.
3. So that the image has all coproducts, and every object of the image is a coproduct of objects in \mathsf {Optic}(Kl(\Delta ))
4. And so that moreover the monoidal structure on the image is distributive

Markov categories generally do not have pullbacks, for the same reason that they usually don't have products. This issue generally hinders the construction of fibrations, in the ordinary sense, of Markov categories. However, we can go part of the way. The idea of the following definitions is that given a pullback in the deterministic category, say A \times _Y X, a map P \to A \times _Y X where the X-coordinate is deterministic should be uniquely determined by a choice of (deterministic) map P \to X and (stochastic) P \to A such that the square commutes---as we claimed above (and will see below), this holds for the Markov category of discrete probability Kl(\Delta ). The analogous statement for products---that a map P \to A \otimes X with deterministic X-component is uniquely determined by the projections (or marginals) P \to X, P \to A---is a consequence of positivity (see Proposition [efr-OYB6]), and hence holds in most Markov categories of interest.

Let \mathcal {C} be a Markov category, and let p: \mathcal {D} \to \mathcal {C} be a functor into it. Then we call p a Markov prefibration if the following two conditions hold:

1. The pullback \mathcal {D} \times _\mathcal {C} \mathcal {C}_\mathrm {det} \to \mathcal {C}_\mathrm {det} is a (Grothendieck) fibration
2. Given maps f: A \to C, g:B \to C in \mathcal {D}, such that p(f),p(g) are deterministic and f,g are Cartesian for the above fibration, p induces a bijection between maps h: A \to B \in \mathcal {C} such that gh = f, and maps h': p(A) \to p(B) so that p(g)h' = p(f). Note that when restricted to those maps where p(h) is deterministic, this being a bijection is the defining property of g being Cartesian (for any f, not necessarily a Cartesian one).

Given a Markov prefibration \mathcal {D}, we write \mathcal {D}|_\mathrm {det} for \mathcal {D} \times _\mathcal {C} \mathcal {C}_\mathrm {det}. We will refer to this as the deterministic part of \mathcal {D}---note that this does have the potential for confusion, as when \mathcal {D} is itself a Markov category, this is not necessarily the same as the deterministic subcategory of \mathcal {D}. When f \in \mathcal {D} lies inside \mathcal {D}|_\mathrm {det}, and is Cartesian for that fibration, we will simply refer to it as a Cartesian map in \mathcal {D} (there are no other types of Cartesian maps, so this should not lead to confusion). A morphism of Markov prefibrations is a functor \mathcal {D} \to \mathcal {D}' over \mathcal {C} which preserves Cartesian maps. The category of Markov prefibrations over \mathcal {C} thus defined is denoted \mathsf {MarkPreFib}(\mathcal {C}). Taking the deterministic part defines a functor (-)|_\mathrm {det}: \mathsf {MarkPreFib}(\mathcal {C}) \to \mathsf {Fib}(\mathcal {C}_\mathrm {det}).

Since we will shortly be working with a number of functors between categories whose objects are themselves categories with some structure, it may be thought that we should give some consideration to the strictness of our constructions---for example, we will shortly construct a left adjoint to (-)|_\mathrm {det}: \mathsf {MarkPreFib}(\mathcal {C}_\mathrm {det}) \to \mathsf {Fib}(\mathcal {C}), and it may well be asked how strict this adjoint is, whether we need to consider the definition of pseudomonad when we get so far, et cetera.

However, we can largely avoid this issue. The key observation is that none of our functors will alter the objects of the underlying category (since \mathcal {C}_\mathrm {det} \to \mathcal {C} is identity on objects,). Hence, all the natural transformations that we would ordinarily ask to be equivalences of categories will instead be isomorphisms, and we can largely ignore considerations of higher category theory---similarly, all our functors will be strictly functorial. As a simple example of this, observe that the pullback functor (-)|_\mathrm {det} is automatically strict---it simply consists in restriction to a subset of the morphisms in \mathcal {D} (which is automatically closed under composition), and thus clearly preserves composition strictly.

In a Markov prefibration p: \mathcal {D} \to \mathcal {C} (as previously noted), a morphism f: \bar {X} \to \bar {Y} in \mathcal {D} is called Cartesian if p(f) is deterministic and f is Cartesian in the fibration \mathcal {D}|_\mathrm {det} \to \mathcal {C}_\mathrm {det}. It is called vertical if p(f) is an identity. It is called a stochastic-Cartesian if there exists a Cartesian map r: \bar {Y} \to \bar {X} so that rf = 1_{\bar {X}} (recall that in this case f is uniquely determined by r and p(f)). Note that if f is stochastic-Cartesian and p(f) is deterministic, then f is Cartesian.

Let \mathcal {C} be a Markov category which is Cartesian (that is, one where all morphisms are deterministic). Then a Markov prefibration over \mathcal {C} is simply a Grothendieck fibration.

Let \mathcal {C} be a Markov category. Then \mathcal {C}^\to \to \mathcal {C} is a Markov prefibration with Cartesian maps given by pullback squares, if and only if \mathcal {C} is pullback-positive.

In a general Markov category, not every isomorphism is necessarily deterministic. This means that, in general, fibres over isomorphic objects in a Markov prefibration are not necessarily isomorphic or even equivalent as categories. This seemingly immoral situation is, in fact, in accordance with other results indicating that deterministic isomorphism is really the proper notion of identification in a Markov category. See eg Reference [rischel-fritz-infinite-products], Section 4, for further discussion of this point. (Since the basic idea of a Markov category involves objects equipped with some structure which is not preserved by all the morphisms, it is not so paradoxical that an isomorphism in this situation should be insufficient to render two objects identical).

Under very weak assumptions on the Markov category \mathcal {C}, such as positivity, all isomorphisms are deterministic. This implies that all Markov prefibrations over \mathcal {C} are isofibrations, and thus rules out any sort of behavior like the above. As noted, we are only concerned with positive Markov categories.

Relatedly, in the proof of Proposition [efr-0044], a careless prover may have erroneously concluded after the first step that all Cartesian lifts are deterministic---but since Cartesian lifts are characterized only up to isomorphism, not necessarily deterministic isomorphism, this does not automatically follow. (But of course replacing a nondeterministic lift with an isomorphic deterministic one in this situation cannot alter the unique existence of the factorization, so it does not matter).

Let p: \mathcal {D} \to \mathcal {C} be a Markov prefibration, let \mathcal {C}' \subseteq \mathcal {C}, \mathcal {D}' \subseteq \mathcal {D} be full subcategories so that p(\mathcal {D}') is contained in \mathcal {C}', and suppose \mathcal {C}' is a monoidal subcategory (which is then automatically a sub-Markov category). Suppose finally \mathcal {D}' is stable under pullback along deterministic morphisms in \mathcal {C}'. Then \mathcal {D}' \to \mathcal {C}' is again a Markov prefibration.

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.

The codomain functor \mathsf {BorelStoch}^\to \to \mathsf {BorelStoch} is a Markov prefibration.

\mathsf {FinStoch}^\to \to \mathsf {FinStoch} is a Markov prefibration.

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.

Let \mathcal {C} be a positive Markov category with supports. Then \mathsf {Optic}(\mathcal {C}) \to \mathcal {C} is a Markov prefibration.

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}.

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).

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).

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}).

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

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}),-)

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

• March 27, 2025
• Eigil Fjeldgren Rischel

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.

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

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})

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.

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

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}).

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

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

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.

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).

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

Consider the adjunction |-|: \mathsf {Mon} \leftrightarrows \mathsf {Set} : (-)^* between the category of monoids and the category of sets. Given a monoid M, |M|^* consists of lists of elements in M, and ||M|^*|^* consists of lists of such lists. The two maps ||M|^*|^* \to |M|^* consist in either concatenating the lists, or replacing each list with its product. Clearly these two maps are coequalized by the product map |M|^* \to M. Moreover it's clear that two lists have the same product if and only if they are identified in this coequalizer (simply consider a singleton list-of-lists, which identifies any given list with the singleton corresponding to its product).

Let \mathsf {GrpTop} be the category of topological groups, and let R: \mathsf {GrpTop} \to \mathsf {Set} forget both the group structure and the topology. Clearly this is right adjoint to the free group in the discrete topology, and the monad of this adjunction is the free group monad, which we write RL for now. The canonical comparison functor \mathsf {GrpTop} \to \mathsf {Grp} just forgets the topology. Given a (non-topological) group, described by a map RLG \to G, we can form the diagram of topological groups LRLG \rightrightarrows LG. Here LG is the free, discrete group on G and LRLG is the free discrete group on the underlying set of LG. Their coequalizer is simply G equipped with the discrete topology, which is indeed the left adjoint to \mathsf {GrpTop} \to \mathsf {Grp}

If \mathcal {D} is a Markov prefibration, we call it a Markov fibration if the diagram \overline {\operatorname {Free}(\mathcal {D}|_\mathrm {det})} \rightrightarrows \overline {\mathcal {D}|_\mathrm {det}} \to \mathcal {D} is a coequalizer in \mathsf {Cat}_{/\mathcal {C}}.

Given an algebra \alpha : \operatorname {Free}(\mathcal {D}_0) \to \mathcal {D}_0 of the free Markov prefibration monad, let \overline {\operatorname {Free}(\mathcal {D}_0)} \rightrightarrows \overline {\mathcal {D}_0} \in \mathsf {MarkPreFib} be as above. We say \alpha presents a Markov fibration if the coequalizer of these maps in \mathsf {Cat}_{/\mathcal {C}} is a Markov prefibration.

Let \alpha : \operatorname {Free}(\mathcal {D}_0) \to \mathcal {D}_0 be stochastic module which presents a Markov fibration. Then the underlying algebra of the Markov prefibration obtained as the coequalizer of \overline {\operatorname {Free}(\mathcal {D}_0)} \rightrightarrows \bar {\mathcal {D}_0} is isomorphic to \alpha , and in particular this prefibration is a Markov fibration.

This correspondence determines an equivalence of categories between the full subcategory \mathsf {MarkFib}(\mathcal {C}) of \mathsf {MarkPreFib}(\mathcal {C}) spanned by the Markov fibrations, and the full subcategory \mathsf {SFib}(\mathcal {C})^p \subseteq \mathsf {SFib}(\mathcal {C}) spanned by those algebras which present a Markov fibration.

Let \mathcal {D}_0 \to \mathcal {C}_\mathrm {det} be a stochastic module fibration.

Let \alpha : X \to M be a stochastic section, let \bar {X} \in \mathcal {D}_{0,X} be an object, and let \phi : \bar {X}_M \to \bar {X}_M \in \mathcal {D}_{0,M} be an endomorphism of its pullback. Observe that if there exists f: N \to M so that f^*(\phi ) = 1 and \alpha factors over f, then \alpha _*(\phi ) = 1. We say \mathcal {D}_0 has weak supports if this implication is an equivalence

Let F,G: \mathcal {D} \rightrightarrows \mathcal {D}' be a parallel pair in \mathsf {Cat}_{/\mathcal {C}}, and suppose both are identity on objects. Suppose moreover this is a reflexive pair, i.e there is some S: \mathcal {D}' \to \mathcal {D} so that FS = GS = 1_{\mathcal {D}'}. Then the coequalizer in \mathsf {Cat}_{/\mathcal {C}} is again identity on objects, and is given on hom-sets simply by the coequalizer of the parallel pair \mathcal {D}(x,y) \rightrightarrows \mathcal {D}'(x,y)

Let \mathcal {D}_0 be a fibration equipped with a stochastic module structure. Consider the equivalence relation on \overline {\mathcal {D}_0}(\bar {X},\bar {Y}) which identifies two precharts (M, \phi ), (N, \phi ') if there exists a map f: N \to M over X,Y and a stochastic section s of f which preserves the section from X, so that s^*\phi ' = \phi (note that this makes sense because pullbacks compose).

1. This equivalence relation respects composition, and so defines a category which we denote \mathsf {SChart}(\mathcal {D}_0)
2. In \mathsf {Cat}_{/\mathcal {C}}, \overline {\operatorname {Free}(\mathcal {D}_0)} \rightrightarrows \overline {\mathcal {D}_0} \to \mathsf {SChart}(\mathcal {D}_0) is a coequalizer diagram. In particular, \mathcal {D}_0 presents a Markov fibration if and only if \mathsf {SChart}(\mathcal {D}_0) is a Markov prefibration (in which case \mathsf {SChart}(\mathcal {D}_0) is the fibration it presents)
3. If \mathcal {D}_0 has weak supports, \mathsf {SChart}(\mathcal {D}_0) is a prefibration

Proof

• April 5, 2025
• Eigil Fjeldgren Rischel

Recall that the fibration \operatorname {Free}(\mathcal {D}_0) has fibers whose morphisms \bar {X} \to \bar {X}' are given by tuples s: X \to M: p, p^*\bar {X} \to p^*\bar {X}' (up to a certain equivalence relation). Forming the free Markov prefibration \overline {\operatorname {Free}(\mathcal {D}_0)} on this fibration, we find that the morphisms are given by diagrams

equipped with a map \phi : p^*q^*\bar {X} \to p^*r^*\bar {Y}. The two maps \overline {\operatorname {Free}(\mathcal {D}_0)} \to \overline {\mathcal {D}_0} carry such a thing to first, the map resulting from forgetting M and just composing q,r with p to get a span (and composing the sections to get a new section), and secondly, the map with apex M obtained by using the stochastic module structure to push \phi down into a map over M. It is clear that this is equivalently the equation described in the theorem. This establishes points 1. and 2., since by Lemma [efr-L7L2] we can compute such coequalizers hom-set by hom-set.

Now we wish to prove that \mathsf {SChart}(\mathcal {D}_0) is a Markov prefibration given weak supports. Since by the coequalizer presentation, its deterministic part is isomorphic to \mathcal {D}_0, the fibration property is automatic. It remains to verify that, given a triangle

in \mathcal {C} with the vertical and horizontal maps deterministic, an object \bar {A} \in \mathcal {D}_{0,Y} and Cartesian maps \bar {A}_X \to \bar {A} \leftarrow \bar {A}_Z, there exists a unique lift \bar {A}_X \to \bar {A}_Z in \mathsf {SChart}(\mathcal {D}_0).

Such a lift is given by a diagram

equipped with p^*\bar {A}_X \to q^*\bar {A}_Z. By taking the equalizer of the two maps M \to Y (the section factors over this), we may assume these two are equal, which implies that the pulled-back objects are equal---denote this object \bar {A}_M. Now the hypothesis is that after postcomposing with the Cartesian map \bar {A}_Z \to \bar {A}, this gives the map \bar {A}_X \to \bar {A}. This postcomposition is given simply by postcomposing the leg M \to Z with the map Z \to Y (and observing that, by functoriality of pullbacks, this does not alter the pulled-back object). Then the claim is that integrating this map \bar {A}_M \to \bar {A}_M down into a map \bar {A}_X \to \bar {A}_X, it gives the identity. But by assumption this means we can pull back to some object M' \to M (lifting the section from X) where the two maps are already equal to the identity. But this pull-back can be applied to the original map \bar {A}_X \to \bar {A}_Z as well. But this implies every such map is equal to the one represented by the diagram

and the identity on \bar {A}_{X \otimes Y}, with the map M' \to X \otimes Y giving the witness, since identities pull back. This map only depends on the underlying X \to Y, hence \mathsf {SChart}(\mathcal {D}_0) is indeed a prefibration.

Let (M,\phi ) and (N,\psi ) be two representatives of charts. Given some possibly stochastic map f: M \to N over X and Y, recall (Lemma [efr-VF6V]) that we can define f^*\psi , regardless of whether f is deterministic or the section of a deterministic map. If there exists such a map f, we can always factor it over the pullback M \times _{X \times Y} N as a section followed by a deterministic map. Hence the equivalence relation defining \mathsf {SChart} is equivalent to the relation identifying two representatives whenever there exists such an f with f^*\psi = \phi

By construction, for each Markov prefibration \mathcal {D}, there is a canonical functor \mathsf {SChart}(\mathcal {D}|_\mathrm {det}) \to \mathcal {D} over \mathcal {C}, which restricts to an isomorphism on the deterministic part (and in particular preserves Cartesian morphisms). \mathcal {D} is a Markov fibration if and only if this is an isomorphism.

Also by construction, given a morphism of stochastic modules F: \mathcal {D}_0 \to \mathcal {D}_0', there is an induced functor \mathsf {SChart}(\mathcal {D}) \to \mathsf {SChart}(\mathcal {D}') over \mathcal {C}. This restricts to F on the deterministic part and in particular preserves Cartesian morphisms.

Let \mathcal {D}_0 be a stochastic module fibration. Then \mathcal {D}_0 presents a Markov fibration if and only if \mathcal {D}_0^\mathrm {fop} does it.

Proof

• April 27, 2025
• Eigil Fjeldgren Rischel

Consider a triangle of this for in \mathcal {C}:

where the maps to Z are deterministic. Suppose given Cartesian lifts \bar {X} \to \bar {Z}, \bar {Y} \to \bar {Z} of the cospan. These are the same in both cases, coming from Cartesian maps in \mathcal {D}_0 \mathcal {D}_0^\mathrm {fop} (which are the same). We must show that there is a unique lift of f to a chart in \mathsf {SChart}(\mathcal {D}_0) if and only if there is a unique lift to a chart in \mathsf {SChart}(\mathcal {D}_0^\mathrm {fop}). Clearly it suffices to prove the "only if" implication, so suppose \mathsf {SChart}(\mathcal {D}_0) is a prefibration.

By passing to the equalizer as in the proof that \overline {\mathcal {D}_0} is a prefibration, we may assume that any such lift is represented by a diagram

where the outer square and the triangle X \to M \to Y commutes, and f' is a section.

Take such a diagram and let \phi : \bar {Z}_M \to \bar {Z}_M be the map representing a chart. Then the claim is there exists some zig-zag of chart equivalences identifying (M,\phi ) with (M,1). But clearly this is invariant under passing to the fiberwise opposite, and so \mathcal {D}^\mathrm {fop} is also a prefibration.

We refer to \mathsf {SChart}(\mathcal {D}_0) as the category of stochastic charts in \mathcal {D}_0. We refer to \mathsf {SChart}(\mathcal {D}_0^\mathrm {fop}) as stochastic lenses and denote it also \mathsf {SLens}(\mathcal {D}_0)

Kl(\Delta )^\to is a Markov fibration. We have already seen that it is a Markov prefibration, and that the map from the coreflection is full. So it suffices to prove faithfulness. Consider a map in {Kl(\Delta )^\to |_\mathrm {det}}, given by a diagram

We can factor the section X \to M as X \to X \times Y \to M, where the first map is just the pairing and the second is a conditional distribution. This induces a factorization of the lift \bar {X} \to M \times _X \bar {X} over \bar {X} \to \bar {X} \times Y. By composing the map M \times _X \bar {X} \to \bar {Y} with this factorization to build the map \bar {X} \times Y \to \bar {Y}, we have found a new representative for the same map.

Hence every map over X \to Y has a representative where the residual is X \times Y. We would like to argue that, since the map \bar {X} \times Y \to \bar {Y} is given by the conditional distribution of the composite map \bar {X} \to \bar {Y}, it is uniquely determined by it, and thus if two distinct maps in \overline {Kl(\Delta )^\to |_\mathrm {det}} have the same underlying map in Kl(\Delta )^\to , they must have equal representatives of this form, and so be identified in the coreflection (which must therefore be isomorphic to Kl(\Delta )^\to ). But of course, the two maps may only be almost certainly equal.

In this case, there is a simple fix: instead of taking X \times Y as the residual, take the subset S given by those pairs (x,y) where y has positive probability given x. The pairing factors over this, of course, and two maps \bar {X} \times _X S \to \bar {Y} which give the same distribution \bar {X} \to \bar {Y} really must have the same value on every point. This proves that \mathsf {SChart}(Kl(\Delta )^\to ) \to (Kl(\Delta ))^\to is faithful and hence an isomorphism.

\mathsf {BorelStoch}^\to , as we have noted, is a Markov prefibration, and hence induces a stochastic module structure on \mathsf {BorelStoch}^\to |_\mathrm {det}. This structure does not present a Markov fibration. To see this, note that in that case its fiberwise opposite would also present a Markov fibration. Then this fibration, \mathsf {SLens}(\mathsf {BorelStoch}^\to |_\mathrm {det}), would be a prefibration whose deterministic part was \mathsf {BorelStoch}^\to |_\mathrm {det}^\mathrm {fop}. But Example [efr-8B5X] shows that this is impossible.

Let \mathcal {C} be a Markov category with supports. Then the stochastic module induced by \mathcal {C}^\to \to \mathcal {C} presents a Markov fibration. If \mathcal {C} has conditionals, this Markov fibration is isomorphic to \mathcal {C}^\to .

Proof

• April 6, 2025
• Eigil Fjeldgren Rischel

Given a section s: X \to M:p and \phi : A \times _X M \to A \times _X A, simply take the pullback to the support of s. It must be the case that \phi (a,m) = (a,m) s-almost surely, which implies strict equality on the support. Hence by Proposition [efr-TBZZ], \mathcal {C}^\to presents a Markov fibration.

There is an induced map \mathsf {SChart}(\mathcal {C}^\to |_\mathrm {det}) \to \mathcal {C}^\to , which we claim is an isomorphism. So consider a map in \mathcal {C}^\to :

This map is in the image of

where the map \bar {X} \times Y \to \bar {Y} is taken to be a conditional. Note that every map in \mathsf {SChart}(\mathcal {C}^\to |_\mathrm {det}) can be represented in this form, by taking a conditional of M given Y to build a section to M \to X \otimes Y. Since conditionals are almost-surely equal, by restricting to the support of X \to X \otimes Y, we can find a representative which only depends on the overall map \bar {X} \to \bar {Y}, which proves that the map from \mathsf {SChart} is faithful, concluding the proof.

If \mathcal {C} is Cartesian (even if it does not admit pullbacks), the definition of Markov fibration over \mathcal {C} still makes sense, the Markov fibrations are exactly the Grothendieck fibrations, and their fiberwise opposites are simply their fiberwise opposites in the usual sense.

A Markov category is said to be an extensive Markov category if it admits finite coproducts, whose injections are deterministic, and which satisfy the following equivalent conditions:

1. If we let \mathcal {C}_{/a}^\mathrm {det} refer to the full subcategory of the slice spanned by the deterministic morphism x \to a, we have an equivalence of categories \mathcal {C}_{/a}^\mathrm {det} \times \mathcal {C}_{/b}^\mathrm {det} \cong \mathcal {C}_{/a + b}^\mathrm {det}, given by taking coproducts
2. \mathcal {C}_\mathrm {det} is an extensive category in the usual sense and the inclusion \mathcal {C}_\mathrm {det} \to \mathcal {C} preserves pullbacks along coproduct inclusions.

More generally, for an infinite regular cardinal \kappa , we say that \mathcal {C} is \kappa -extensive if \mathcal {C}_\mathrm {det} is a \kappa -extensive category in the ordinary sense and both coproduct inclusions and pullbacks along them are preserved by the functor \mathcal {C}_\mathrm {det} \to \mathcal {C}.

Let \mathcal {C} be an extensive Markov category, let \mathcal {D}_0 \to \mathcal {C}_\mathrm {det} be a fibration which satisfies \mathcal {D}_{0,X+Y} = \mathcal {D}_{0,X} \times \mathcal {D}_{0,Y}. Note that this implies \mathcal {D}_0 admits finite coproducts, and they're given exactly by this pairing. Suppose \mathcal {D}_0 is equipped with a stochastic module structure. Then \mathcal {D}_0 \hookrightarrow \mathsf {SChart}(\mathcal {D}_0) preserves the finite coproducts. In particular, \mathcal {D}_0^\mathrm {fop} has the same coproducts as \mathcal {D}_0, and \mathcal {D}_0^\mathrm {fop} \to \mathsf {SLens}(\mathcal {D}_0) preserves them as well.

Suppose \mathcal {D} is a Markov fibration so that each pullback functor f^*: \mathcal {D}_Y \to \mathcal {D}_X for f: X \to Y \in \mathcal {C}_\mathrm {det} can be taken to be bijective on objects, and that these can furthermore be chosen strictly functorial (so that (fg)^* = g^*f^*). Then, writing objects \bar {X} \in \mathcal {D}_X as \binom {A \in \mathcal {D}_*}{X \in \mathcal {C}}, where A is the unique object in \mathcal {D}_* which pulls back to \bar {X} under the deletion X \to *, (note that this means f^*\binom {A}{Y} = \binom {A}{X}) we may characterize the fiberwise dual as having hom-sets \mathcal {D}^{\mathrm {fop}}(\binom {A}{X},\binom {B}{Y}) = \mathcal {D}(\binom {B}{X},\binom {A}{Y})

Let \mathcal {D} \to \mathcal {C} be a Markov prefibration, and let F: \mathcal {C}' \to \mathcal {C} be any functor from another Markov category which preserves deterministic maps. Then the pullback \mathcal {D}' = \mathcal {D} \times _{\mathcal {C}} \mathcal {C}' \to \mathcal {C}' is again a Markov prefibration, and the functor \mathcal {D}' \to \mathcal {D} preserves Cartesian maps.

Proof

• April 9, 2025
• Eigil Fjeldgren Rischel

Since pullbacks compose, \mathcal {D}'|_\mathrm {det} \to \mathcal {C}'_\mathrm {det} is the pullback of \mathcal {D}|_\mathrm {det} along \mathcal {C}'_\mathrm {det} \to \mathcal {C}_\mathrm {det}. Since fibrations are stable under pullback, this is a fibration.

Now consider a triangle in \mathcal {C}':

with X, A \to B deterministic, and let \bar {B}_X \to \bar {B} \leftarrow \bar {B}_A be Cartesian maps lying over these. We must show there is a unique lift \bar {f}: \bar {B}_X \to \bar {B}_A rendering the lifted triangle commutative. By definition, to give such a morphism is to given one over F(f), and the triangle commutes if and only if its image in \mathcal {D} commutes.

The triangle in \mathcal {C}' goes to a triangle of the same class in \mathcal {C}, and the Cartesian maps go to Cartesian maps of the same type. Hence there is a unique lift of F(f) of the given type, which is exactly what we needed to show.

Given any functor F: \mathcal {C} \to \mathcal {C}' between Markov categories, we may attempt to define an oplax monoidal structure F(X \otimes Y) \to F(X) \otimes F(Y) by pairing the projections.

This is not necessarily a natural transformation. However, if it is, F it automatically equips F with the structure of an oplax monoidal functor. Recall that oplax monoidal functors carry comonoids to comonoids. An oplax monoidal functor between Markov categories preserves the given comonoids if and only if it is induced like this.

Hence, there is at most one way to equip a functor between Markov categories with such a structure---it is a property, not extra structure. Call such a functor an oplax Markov functor. Note that oplax Markov functors preserve deterministic morphisms.

(Fritz Reference [fritz-synthetic-markov-cats] defines a Markov functor to be a strong monoidal functor which preserves the comonoids. Clearly this is a proper subset of our oplax markov functors.)

We will denote by \mathsf {MarkPreFib} the category whose objects are Markov prefibrations \mathcal {D} \to \mathcal {C}, and whose functors are commutative squares

We will let \mathsf {Markov}^\mathrm {oplax} denote the category of Markov categories and oplax Markov functors. Note that by Lemma [efr-9VAH], the forgetful functor \mathsf {MarkPreFib} \to \mathsf {Markov}^\mathrm {oplax} is a fibration.

Let \mathsf {Markov}_s \subseteq \mathsf {Markov}^\mathrm {oplax} denote the subcategory of strictly monoidal Markov functors (i.e those where the oplaxator F(X \otimes Y) \to F(X) \otimes F(Y) is the identity).

1. \mathsf {Markov}_s and \mathsf {Markov}^\mathrm {oplax} admit all products, computed simply as products in \mathsf {Cat}.
2. \mathsf {Markov}_s admits all finite limits, and these are preserved by the inclusion into \mathsf {Markov}^\mathrm {oplax}

The category \mathsf {MarkPreFib}(\mathcal {C}) admits all products, and pullbacks along isofibrations. These are simply computed as limits in \mathsf {Cat}_{/\mathcal {C}}

The functor \overline {(-)}: \mathsf {Fib} \times _\mathsf {Cat} \mathsf {Markov}^\mathrm {oplax} \to \mathsf {MarkPreFib} preserves products.

For each Markov category \mathcal {C} with weak conditionals, \overline {(-)}: \mathsf {Fib}(\mathcal {C}_\mathrm {det}) \to \mathsf {MarkPreFib}(\mathcal {C}) preserves the terminal object, and pullbacks along isofibrations.

For any (pullback-positive) \mathcal {C}, the monad \overline {(-)}|_\mathrm {det} on \mathsf {Fib}(\mathcal {C}_\mathrm {det}) preserves products.

The property of having weak supports is stable under equalizers in stochastic module fibrations over \mathcal {C}. If \mathcal {C} has weak conditionals, it is also stable under finite products (hence all finite limits).

A monoidal Markov prefibration is a markov prefibration p: \mathcal {D} \to \mathcal {C} equipped with a monoidal category structure on \mathcal {D} so that p is strict monoidal and so that the underlying fibration is a monoidal fibration (i.e so that p preserves Cartesian lifts).

A braided or symmetric monoidal Markov prefibration is a monoidal prefibration equipped with a braiding or symmetry on \mathcal {D} so that p is moreover a braided monoidal functor.

Let F,G: \mathcal {C}' \rightrightarrows \mathcal {C}, S: \mathcal {C} \to \mathcal {C}' be a reflexive pair of identity-on-objects, strict monoidal functors. Let E: \mathcal {C} \to \mathcal {D} be the coequalizer in \mathsf {Cat}. Then \mathcal {D} inherits a monoidal structure making E strict monoidal. Moreover, if \mathcal {C} is symmetric or braided, \mathcal {D} inherits this structure making E a braided functor.

1. The free prefibration monad on \mathsf {Fib}(\mathcal {C}_\mathrm {det}) has a canonical lifting to \mathsf {MonFib}(\mathcal {C}_\mathrm {det})
2. Given a monoidal prefibration, its underlying stochastic module acquires the structure of an algebra of this lifted monad.
3. Given an algebra for the lifted monad \mathcal {D}_0, \mathsf {SChart}(\mathcal {D}_0) acquires a monoidal structure so that \mathsf {SChart}(\mathcal {D}_0) \to \mathcal {C} is a strict monoidal functor.
4. If \mathcal {D}_0 moreover has weak supports, this forgetful functor is a monoidal prefibration.

Every statement holds also for braided or symmetric fibrations.

Proof

• April 25, 2025
• Eigil Fjeldgren Rischel

By Reference [moeller-vasilakopoulou], \mathsf {MonFib}(\mathcal {C}_\mathrm {det}) is equivalent to the category of pseudomonoids in \mathsf {Fib}(\mathcal {C}_det). Since the monad preserves products, it must preserve pseudomonoids, which is all we need.

Now suppose \mathcal {D} \to \mathcal {C} is a monoidal prefibration. Then its underlying fibration is a monoidal fibration, hence an object of \mathsf {MonFib}(\mathcal {C}_\mathrm {det}). The claim is that the functor \overline {\mathcal {D}|_\mathrm {det}}|_\mathrm {det} \to \mathcal {D}|_\mathrm {det} is monoidal. The induced monoidal structure is given on objects by \otimes _\mathcal {D} and takes a pair of morphisms in the fiber represented by sections (s: X \to M, \phi : \bar {X_0}_M) \to \bar {X_1}_M and s': X \to M', \phi ': \bar {X'_0}_M \to \bar {X'_1}_M to \langle s,s' \rangle : X \to M \times _X M', \pi _M^*\phi \otimes _\mathcal {D} \pi _{M'}^*\phi '. Recalling that the algebra structure is defined by taking (s,\phi ) to the composite \bar {X_0} \to \bar {X_0}_M \xrightarrow {\phi } \bar {X_1}_M \to \bar {X_1}, and chasing the below diagram around, it is apparent that the algebra structure preserves the monoidal structure.

Now let \mathcal {D}_0 be a monoidal stochastic module in this sense. It suffices to show that the free prefibration \overline {\mathcal {D}_0} is a monoidal prefibration, by Lemma [efr-7GH5], and the induced functor \mathsf {SChart}(\mathcal {D}_0) \to \mathcal {C} will clearly be strict monoidal if \overline {\mathcal {D}_0} \to \mathcal {C} is.

To construct this monoidal structure on \mathcal {D}_0, simply note that since \overline {(-)} preserves global limits as well, there is an induced monoidal structure on \overline {\mathcal {D}_0} so that the forgetful functor is strict monoidal. Recalling that the Cartesian lifts of f: X \to Y \in \mathcal {C} to \overline {\mathcal {D}_0} are given by the span X = X \to Y and the morphism 1_{f^*\bar {Y}}, it is easy to see by unwinding the definition that these are stable under tensor.

If \mathcal {D}_0 has weak supports, we have already proven that \mathsf {SChart}(\mathcal {D}_0) \to \mathcal {C} is a strict monoidal functor, and weak supports are equivalent to the claim that it is a prefibration. Since the Cartesian lifts are just the equivalence classes of the Cartesian lifts in \overline {\mathcal {D}_0}, the preceding claim that they are stable under tensor implies the same for \mathsf {SChart}(\mathcal {D}_0), finishing the proof.

The last point is mostly trivial. The product preservation still establishes the lifting to \mathsf {BrMonFib}(\mathcal {C}_\mathrm {det}) and \mathsf {SymMonFib}(\mathcal {C}_\mathrm {det}). Given a braided or symmetric monoidal prefibration, the braidings are Cartesian and in the deterministic part, so the underlying fibration is braided/symmetric and they are preserves by the stochastic module structure. The braiding/symmetry on \mathsf {SChart} follows again from Lemma [efr-7GH5], and there is nothing to show for the last point.

A monoidal Markov fibration is a monoidal prefibration \mathcal {D} \to \mathcal {C} which is a Markov fibration. It is braided or symmetric if it is braided or symmetric as a prefibration

Since \overline {(-)} preserves both global products and fiberwise ones, it induces both a fiberwise monoidal structure and a "global" monoidal structure on \overline {\mathcal {D}_0}. Here we only use the global one. If \mathcal {C} were Cartesian, the global one would be induced from the local one by, given maps \bar {X_1} \to \bar {X_2}, \bar {Y_1} \to \bar {Y_2}, pulling each of them back along the square

Let \mathcal {D} be a monoidal stochastic module fibration over \mathcal {C} and suppose each fiber \mathcal {D}_X is a Markov category, and this structure is preserved by the pullback functors f^*. Then \mathsf {SChart}(\mathcal {D}) carries the structure of a Markov category, so that \mathsf {SChart}(\mathcal {D}) \to \mathcal {C} is a Markov functor.

If \mathcal {D} \to \mathcal {C} is a Markov prefibration which is also a strict Markov functor, the induced monoidal structure on \mathcal {D}|_\mathrm {det} acquires a fiberwise Markov structure.

Let \mathcal {D} be a stochastic module fibration, and suppose each fiber has coproducts, and these are preserved by the pullback functors. Then \mathsf {SLens}(\mathcal {D}) is a Markov category, with monoidal structure given by \binom {\bar {X}}{X} \& \binom {\bar {Y}}{Y} = \binom {\pi _X^*\bar {X} + \pi _Y^*\bar {Y}}{X \otimes Y}

Let \mathcal {C} act on \mathcal {D}. Then \mathsf {Optic}_\mathcal {C}(\mathcal {D}) := \mathsf {Optic}_\mathcal {C}(\mathcal {C},\mathcal {D}) has a functor to \mathcal {C}. The deterministic part \mathsf {Optic}_\mathcal {C}(\mathcal {D})|_\mathrm {det} admits the structure of a stochastic module fibration. There is an isomorphism \mathsf {Optic}_\mathcal {C}(\mathcal {D}) \to \mathsf {SChart}(\mathsf {Optic}_\mathcal {C}(\mathcal {D})|_\mathrm {det}).

If \mathcal {D} is itself symmetric monoidal and the action is symmetric (meaning it is given by M \cdot A = F(M) \otimes A for some symmetric monoidal functor F: \mathcal {C} \to \mathcal {D}, see Reference [actegories-amthematician-capucci-gavranovic] 5.4.3 and 5.5.12), this stochastic module is symmetric monoidal and the isomorphism is an isomorphism of symmetric monoidal categories.

Let \mathcal {C} be any pullback-positive Markov category. Then \mathcal {C}^\to \to \mathcal {C} is a Markov prefibration which thus induces a stochastic module structure on \mathcal {C}^\to |_\mathrm {det}. Writing simply \mathsf {SChart}(\mathcal {C}), \mathsf {SLens}(\mathcal {C}) for \mathsf {SChart}(\mathcal {C}^\to |_\mathrm {det}), \mathsf {SLens}(\mathcal {C}^\to |_\mathrm {det}), we have:

1. There is a functor \mathsf {Optic}(\mathcal {C}) \to \mathsf {SLens}(\mathcal {C}), which is fully faithful. Dually there is a functor \mathsf {coOptic}(\mathcal {C}) \to \mathsf {SChart}(\mathcal {C}) which is fully faithful.
2. \mathsf {SLens}(\mathcal {C}) and \mathsf {SChart}(\mathcal {C}) both admit symmetric monoidal structures, which make the functors \mathsf {SChart}(\mathcal {C}), \mathsf {SLens}(\mathcal {C}) \to \mathcal {C} strict symmetric monoidal, as well as the functors \mathsf {Optic}(\mathcal {C}) \to \mathsf {SLens}(\mathcal {C}), \mathsf {coOptic}(\mathcal {C}) \to \mathsf {SChart}(\mathcal {C}) strong symmetric monoidal.
3. If \mathcal {C} is extensive, this functor preserves the coproducts {A \choose X} + {A \choose Y} = {A \choose X+Y}, and \mathsf {SChart}(\mathcal {C}),\mathsf {SLens}(\mathcal {C}) both admit all finite coproducts.
4. If \mathcal {C} moreover has conditionals and supports, \mathsf {SChart}(\mathcal {C}) = \mathcal {C}^\to

Given a compact Hausdorff space X, a Banach space bundle is a space over X, V \to X, equipped with a fiberwise (complex) vector space structure +: V \times _X V \to V, \cdot : \mathbb {C} \times V \to V, so that there exists a cover \{U_i\} of X so that for each U_i, there exists a Banach space V_i and a homeomorphism V \times _X U_i =: V_{U_i} \cong V_i \times U_i over U_i, which is moreover linear in each fiber, where V_i is equipped with the norm topology.

Note that this determines a local norm on each V_{U_i} (and in particular each V_x) up to equivalence (but no stricter than that). In particular each V_x is a Banach space.

A morphism of Banach space bundles is a continuous map f: V \to W over X which is linear on each bundle. Note that this implies that on a suitable cover U_i, the maps f: V_{U_i} \to W_{U_i} obey \left \lVert f(v) \right \rVert \leq C_i \left \lVert v \right \rVert for some C_i \in \mathbb {R}, for any local norms inducing the topologies, and hence by compactness there exists some C so that \left \lVert f(v) \right \rVert \leq C \left \lVert v \right \rVert for each v.

If X \to Y is a continuous map, there is a pullback functor \mathsf {Ban}_Y \to \mathsf {Ban}_X. The Grothendieck construction of this gives a fibration \mathsf {BanBun} \to \mathsf {CHaus}

Note that Tychonoff spaces include all compact Hausdorff spaces. Therfore consider the full subcategory \mathsf {CHausStoch} \hookrightarrow \mathsf {TychStoch} spanned by these. The fibration \mathsf {BanBun} admits the structure of a stochastic module: given M \to X, s: X \to M a kernel, and a linear continuous map f: V \times _X M \to W \times _X M, given v \in V_x, there is an induced function M_x \to W_x given by f(v,-). Since this is bounded (being continuous on a compact space) it is (Bochner) integrable, define s^*f(v) to be this integral.

This example is analogous to optics for the action of categories of markov kernels on categories of vector spaces (as in Proposition [efr-A6YR]). Note that we do not expect this type of example to present a Markov fibration. The reason is simply that, given a parametrized linear map M \times \mathbb {R}^n \to \mathbb {R}^n and a measure on M, the fact that the expectation map \mathbb {R}^n \to \mathbb {R}^n is the identity by no means implies that the original map is almost surely the identity or anything like that. If (m,e_0) \mapsto e_1 and m has positive probability, this can be canceled out by (m',e_0) \mapsto -e_1. This is impossible for probability kernels.

If we add an assumption of positivity, it seems plausible that examples of this type will present Markov fibrations---but of course, that brings us quite close to categories of Markov kernels in any case.

Let \mathcal {C} be a representable, positive Markov category so that \mathcal {C}_\mathrm {det} admits intersections and the probability monad P preserves them. Then each slice (\mathcal {C}_\mathrm {det})_{/X} inherits a monad structure given by P_X(B \to X) = P(B) \times _{P(X)} X. This is pseudofunctorial in X. Moreover, the stochastic module structure on \mathcal {C}^\to |_\mathrm {det} extends to a stochastic module structure on the fibration representing the pseudofunctor X \mapsto \mathsf {Alg}(P_X).

Proof

• May 2, 2025
• Eigil Fjeldgren Rischel

The monad is induced by the adjunction \mathcal {C}_\mathrm {det}{/X} \leftrightarrows \mathsf {Alg}(P)_{PX}. Let f: X \to Y (deterministic). For abstract reasons there is a natural transformation P_Yf^* \to f^*P_X. Writing this out, we find P(A \times _Y X) \times _{PX} X \to P(A) \times _{PY} X

By representability, the unit X \to PX is a monomorphism. Hence a map into P(A \times _Y X) \times _{PX} X is precisely a map in \mathcal {C} into A \times _Y X so that the marginal on X is deterministic. But by pullback-positivity this is precisely a map (in \mathcal {C}_\mathrm {det}) into P(A) \times _{PY} X.

Let M \to X, two P_X-algebras A, B, and a map M \times _X A \to M \times _X B which is a homomorphism for the induced P_M-algebras, and a map X \to PM be given. Note that by the above, P_M(M \times _X A) \cong M \times _X P_XA. Then the induced map A \to B is given by A \to PM \times _{PX} A \cong P_X(M) \times _X A \hookrightarrow P_X(M \times _X A) \to P_X(B) \to B

We must show this is a P_X-homomorphism. Let us simplify by working internally to (\mathcal {C}_{\mathrm {det}})_{/X}---thus we have a Cartesian category equipped with a strong commutative monad P, a map * \to PM, and a map M \times A \to B which is a parametrized algebra homomorphism, in the sense that the diagram

commutes. Now we must show the map A \to PM \times A \to P(M \times A) \to PB \to B is a P-homomorphism. Write E_A,E_B for the structure maps of the two algebras. Consider this diagram:

The triangle at the top left commutes because P is strong. The square to the right does not commute in general---however, since P is commutative, the composite maps PM \times PA \to P^2(M \times A) \xrightarrow {\mu } P(X \times A) agree. Since the map P^2(M \times A) \to B factors over this, we may replace one edge of this square with another. The square to the right of that is simply P(-) applied to the previous diagram, and so commutes by assumption. The "triangle" under that is just two copies of the same maps, so commutes. The square on the left of the diagram commutes by functoriality of product. The square to the right of that commutes again because P is strong. Hence the outer square commutes, which is precisely the homomorphism property we wanted.

It is apparent that, if A, B = P_XA', P_B' are free algebras, this restricts to the stochastic module structure of \mathcal {C}^\to |_\mathrm {det} (viewing \mathcal {C} as the Kleisli category of P). But since every algebra is a coequalizer of free algebras, it follows that the action on general algebras is determined uniquely by this. This implies the equations of a stochastic module.