It's straightforward to check that if F admits any oplax Markov structure, then it preserves deterministic maps and this restricts to an oplax monoidal structure on F_\mathrm {det} (the fact that the oplaxator itself must be deterministic follows from the associativity). But functors between Cartesian categories always admit a unique oplax monoidal structure. Hence any two oplax structures on F agree. The only obstruction to this oplax functor extending to a full Markov structure is that it may not be natural with respect to nondeterministic morphisms. Given X \to X', Y \to Y', the tensor X \otimes Y \to X' \otimes Y' is equal to the independent pairing of X \otimes Y \to X \to X' and the analagous map to Y'. Applying the assumptiong that F preserves independent pairings to this gives naturality. The converse is clear. Finally F is strong if F(X \otimes Y) \to F(X) \otimes F(Y) is an isomorphism, but this is exactly the claim that F_\mathrm {det} preserves products.

Note first that \mathsf {Borel} is countably extensive and Boolean, by Reference [chen-universal-stdborel-2019] theorem 1.1. Hence it suffices to observe that the Giry monad preserves pullbacks along coproduct inclusions. Consider X \to A + B. The claim is that the square

is a pullback, where X_A is the preimage of A inside X. But this is clear: a probability measure on A is exactly a probability measure on A + B which happens to be concentrated on A, and a probability measure on X has an image in G(A + B) of this form if and only if it is concentrated on X_A (and is such equivalent to a measure on X_A).

Let m_1,m_2 denote two coinflip structures. Let \mu _1 = m_1(\pi _1, \pi _2) : A \otimes A \to A, analogously \mu _2 (we condense the notation by writing \mu _1,\mu _2 regardless of the object in question). Note m_1(a,b) = \mu _1\langle a,b \rangle , where \langle a,b \rangle is any pairing of a,b, by naturality.

First consider m_1(m_2(a,b),m_2(c,d)). By the above this is \mu _1 \langle \mu _2 \langle a,b \rangle , \mu _2 \langle c,d \rangle \rangle . Applying naturality to the morphism \mu _2, we can also rewrite this as \mu _2 \circ (m_1(\langle a,c \rangle ,\langle b,d \rangle ))

Now take c = b, d = a. By symmetry for m_2 we can rewrite the left-hand side as m_1(m_2(a,b),m_2(a,b)). By idempotency this is simply m_2(a,b). The other expression now becomes \mu _2 \circ (m_1(\langle a,b \rangle ,\langle b,a \rangle )). Now consider the first projection of the map m_1(\langle a,b \rangle ,\langle b,a \rangle ). By naturality it is equal to m_1(a,b). By symmetry, so is the second projection. Hence this is a pairing of m_1(a,b) with itself. Hence the composite is equal to m_2(m_1(a,b),m_1(a,b)) = m_1(a,b). This concludes the proof.

Let f be a coinflip. Consider the map 2 \xrightarrow {\oplus f} 2 given by flipping the coin, applying \neg : 2 \to 2 in one branch and doing nothing in the other. We claim this is equal to deletion followed by the coinflip. This can be checked on each summand (since 2 is a coproduct), where it follows by symmetry.

Now let f_1, f_2 be two coinflips and consider the operation given by 1 \xrightarrow {f_1 \otimes f_2} 2 \otimes 2 \xrightarrow {\oplus }. This is easily seen to be equal both to f_1 ; (\oplus f_2) and f_2 ; (\oplus f_1). But these are equal to f_2 and f_1 respectively by the above.

Let A be an iterable midpoint algebra. Note that any map f: [0,1] \to A must be a homomorphism for the infinitary midpoint operator M. Hence given a_0,a_1 \in A, the unique f so that f(0) = a_0, f(1) = a_1 is given by f(0.b_1b_2 \dots ) = M(a_{b_1}, \dots ). This is straightforwardly seen to be well-defined and a midpoint homomorphism.

We identify the elements of \bar {\Delta }(X) with countably supported probability measures on X. Observe that \bar {\Delta } X is an iterable midpoint algebra (with the infinitary choice operation given by the sum M((a_i)) = \sum _i 1/2^i a_i). By the preceding, every finitely supported measure can be obtained by iterating M, and every countably supported measure can be obtained by applying M to a sequence of these (half of the probability measure must be concentrated in some finite subset, then 1/4 of the remainder again in some finite subset, and so on). Hence there is at most one midpoint homomorphism \bar {\Delta }X \to Y extending any given function f: X \to Y, for any iterable midpoint algebra Y.

Now we must show that one exists. By a finitary dyadic measure, we mean one which is finitely supported and where each weight a_x has the form k/2^n. Note that these are exactly those that can be written using just the binary choice operator m(x,y). Also note that any two such terms are equal in a generic midpoint algebra if and only if they represent the same such measure. Hence given f: X \to Y, there is a unique and well-defined f(a) defined on the finitary dyadic measures, which is a midpoint homomorphism.

Now for any probability measure \mu , we can write it as \mu = M(a_1,\dots ), where each a_i is a finitary dyadic measure. We define our operation as f(\mu ) = M(f(a_1), \dots ), with f(a_i) defined as above. It suffices to show that this is well-defined, since f(m(\mu ^1,\mu ^2)) = f(m(M(a_1^1,\dots ),M(a_1^2,\dots ))) = f(M(m(a^1_1,a^2_1), \dots )) = M(m(f(a_1^1),f(a_1^2)),\dots ) = m(f(\mu ^1),f(\mu ^2))

Now let M(a_1,\dots ) = M(b_1,\dots ) = \mu be two representations in this form of the same measure. Then there exists some smallest N so that there exists a finitary dyadic probability measure c_1 = \sum _i \frac {k_i}{2^N} \delta _{x_i} so that k_i/2^N < 2 \mu (x_i) whenever this is positive - that is, so that c/2 is strictly less than \mu everywhere on the support.

Now in the sum \sum _i 1/2^i a_i must exceed c/2 by some finite partial sum, say M. Then \sum ^M_{i=1} 1/2^i a_i is a finitary dyadic measure. Hence a finite manipulation using the properties of midpoint algebras can rewrite the term M(a_1,\dots ) into M(c_1,a_2',\dots ,a_M',a_{M+1}). Iterating this we build c_2,\dots so that M(a_1,\dots ) = M(c_1,\dots ). Since these only depended on the measure \mu , we similarly find M(b_1,\dots ) = M(c_1,\dots ). Since this argument used only the axioms of an iterable midpoint algebra, it follows that f is well-defined.

Let s: X \to A \otimes X be a map, and suppose u_1,u_2 both satisfy the equation u_i = m(\pi _A s, u_i \pi _X s). Let a_n : X \to A denote the map defined inductively by a_1 = \pi _A s, a_{n+1} = a_n \pi _X s. Let b_i^{n} be defined inductively by b_i^1 = u_i, b_i^{n+1} = b_i^n \pi _X s. Then we see that b_i^n = m(a_n, b_i^{n+1}) for both i=1,2. Therefore they must agree, and in particular u_1 = u_2. This concludes the proof.

By Reference [escardo-simpson-universal-interval-2001], a midpoint algebra (A,m) in \mathsf {Set} is iterable if and only if is satisfies these conditions:

1. There exists an operator M: A^\omega \to A so that M(a_1,a_2, \dots ) = m(a_1, M(a_2, a_3, \dots ))
2. Given two sequences a_i, b_i so that b_i = m(a_i, b_{i+1}) for all i, then b_1 = M(a_1, a_2, \dots )

Let a countable family f_i : A \to B be given. Consider the operation \sum ^\omega _{i=0} A \to B \times \sum ^\omega A given by the copairing of all the f_i paired with the map \sum ^\omega _{i=0} A \to \sum ^\omega _{i=0} A which maps each summand to the next one (i.e (i,a) \mapsto (i+1,a)). Now an iteration of this map u: \sum _i A \to B is equivalently a sequence of maps g_i so that g_i = m(f_i,g_{i+1}). Hence the existence and uniqueness of such an operator precisely corresponds to points 1 and 2 above.

Since \mathsf {Borel} is Cartesian, any such Markov functor must factor over \mathcal {C}_\mathrm {det}. \mathcal {C}_\mathrm {det} has countable limits (it has countable products by Lemma [efr-OH6B] and pullbacks along monomorphisms, which along with finite products suffices to build equalizers) and is countably extensive and Boolean by assumption. The assumptions on i are equivalent to the factorization \mathsf {Borel} \to \mathcal {C}_\mathrm {det} preserving countable limits and countable coproducts. Hence this uniqueness is exactly Reference [chen-universal-stdborel-2019] theorem 1.1

Let r_1: 2^\omega \to 2 be the indicator of the subset consisting of those streams which have only finitely many 1s. Observe that for any finite N, r_1 factors over the second coordinate of the decomposition 2^\omega \to 2^N \times 2^\omega .

As a result, if we form the joint I \to 2^\omega \otimes 2 given by sampling the uniform distribution and computing r_1, the second coordinate is independent of any finite prefix of the first. Hence, by the abstract Kolmogorov zero-one law (Reference [rischel-fritz-infinite-products], theorem 5.3), the composite r_1 c^\omega : I \to I+I is deterministic.

Hence it classifies some subterminal object q. By symmetry, r_0 c^\omega is the indicator of q as well, where r_0 is the indicator of sequences with only finite many 0s. But these two subterminal objects must be disjoint, hence they must both be empty. (This step uses the fact that pullbacks along monomorphisms in \mathcal {C}_\mathrm {det} are still pullbacks in \mathcal {C})

So r_0c^\omega , r_1c^\omega are both equal to 0 : I \to 2. It follows that c^\omega factors over the subobject 2^\omega \setminus \{0\}^\omega \hookrightarrow 2^\omega (since any sequence with finitely many zeroes certainly has at least one 1).

Let F: BT(A)^{\omega \times \omega } \times (2^\omega \setminus 0)^2 \times (2^\omega )^\omega \to A be a parametrization of the sampling map. Specifically, the first two bitstreams select the indexes (i,j) into the matrix of finitary distributions, after which the ith of the remaining bitstreams is used to choose a branch.

Let us adopt the convention that the first index chooses the column and the second index chooses the row when describing the manipulations below.

Given a sequence (d_{ij}) \in BT(A)^{\omega \times \omega }, construct r_1(d)_{ij} as follows:

1. r_1(d)_{1j} = g(d_{11}, g(d_{12}, d_{21}))
2. r_1(d)_{2j} = g(d_{1(j+2)},d_{2(j+1)})
3. r_1(d)_{ij} = d_ij for i \geq 3

First note that clearly r_1(d) has the same image in \Delta (A) as d—the first column of r_1(d) has the same sum as the three most probably indices of d, and the second column of r_1(d) has the same sum as the remainder of the first two columns of d. Also note that there exists a permutation \rho _1 : (2^\omega \setminus 0 )^2 \times (2^\omega )^omega \to (2^\omega \setminus 0)^2 \times (2^\omega )^omega so that F(r_1(d),\rho _1(b)) = F(d,b).

\rho _1(b_1,b_2,(b_3^i)) can be chosen as follows: it does not modify b_3^i for i>2 (since these bits are only used for choices in the first column). Omitting therefore these strings from the notation below, we define \rho _1(1.c,1.r,s^1,s^2) = (1.c,r,1.s^1,s^2) \rho _1(1.c,01.r,s^1,s^2) = (1.c, r,01.s^1,s^2) \rho _1(01.c,1.r,s^1,s^2) = (1.c,r,10.s^2, s^1) \rho _1(1.c, 00.r, s^1,s^2) = (01.c, r, s^2, 1.s^1) \rho _1(01.c, 0.r, s^1,s^2) = (01.c, r, s^1, 0.s^2) \rho _1(00.c,r,s^1,s^2) = (00.c,r,s^1,s^2)

It is straightforward to verify that \rho _1 is well-defined, that F(r_1(d),\rho _1(b)) = F(d,b) as required. Moreover we see that every finite projection of the output of \rho _1 factors over some finite prefix of the input, and that the resulting maps 2^N \to 2^M carry the uniform distribution to the uniform distribution. This implies that \rho _1 c^\omega = c^\omega .

r_1 "flattens" the first column of d. Now we can apply r_1 again to the remaining columns to flatten the second column, and similarly apply \rho _1 ignoring the first bit of c and s^1 to obtain a corresponding permutation. Let r_N, \rho _N be the map an permutation that flatten the first N columns. Note that by construction each output bit of r_N is eventually stable for high N, and the same is true for \rho _N. Let r_\infty , \rho _\infty be the limiting maps.

Then when the first 1 in the part of \rho _\infty (b) which selects the i appears before index N, F(r_\infty (d),\rho _\infty (b)) depends only on the first N rows, hence is equal to F(r_N(d),\rho _\infty (b)), which is then equal to F(d,b). Since \rho _\infty is the limit of permutations, it preserves the uniform measure, and so the composite F(d,c^\omega ) = F(r_\infty (d), \rho _\infty (c^\omega )) = F(r_\infty (d), c^\omega ).

This proves that r_\infty preserves the sampling map. But clearly r_\infty factors over the inclusion of BT(A)^\omega \hookrightarrow BT(A)^{\omega \times \omega } as the "flat" sequences of sequences, and equally clearly this inclusion preserves the sampling map. This concludes the proof.

By Lemma [efr-X0VA], we see that we can immediately flatten any nested sequence in BT(A)^{\omega ^n} into a sequence of dyadic binary trees in BT(A)^\omega , without altering either the sampling map or the represented distribution in \bar {\Delta (A)}. Hence it suffices to prove that the sampling map BT(A)^\omega \to A factors over \bar {\Delta (A)}.

It is clear that the sum map BT(A)^\omega \to \bar {\Delta }(A) is surjective. We will begin by choosing a particular splitting of this map, assigning to each countably supported measure a "normal form" in BT(A)^\omega

Fix an arbitrary total ordering on A. If A is countable there is no difficulty doing this, if A = 2^\omega , we may use the lexical ordering.

Given a finitary distribution \mu , we compute its normal form as follows:

1. Let N_0 be the smallest natural so that there exists a dyadic distribution \nu _0 = \sum _i \frac {k_i}{2^N} a_i so that \frac {1}{2}\nu _0 < \mu - that is, for every element of A, the probability under \frac {1}{2}\nu _0 is strictly less than the probability under \mu .
2. Let \nu _0 be a dyadic distribution as above, choosing the lowest possible a_i according to the chosen ordering. Represent \nu _0 as a balanced binary tree of minimal depth, with the leaves increasing left to right.
3. Repeat these steps with \mu - \frac {1}{2}\nu _0 to construct \nu _1, \nu _2, \dots .

It is clear that this defines a measurable map \Delta (A) \to BT(A)^\omega , and that this is a section of the map \sum _i 1/2^i \nu _i in the other direction.

1. Let S: BT(A)^\omega \times (2^{\omega } \setminus 0) \times (2^\omega )^\omega \to A be the sampling map, which uses the first bitstream to choose the index i from the sequence, then the ith one of the remaining streams to sample from the given binary tree.
2. Note that by construction, if \nu _1 is the first element in the normal form of \mu , and \lambda _i is another sequence of binary trees so that \sum _i 1/2^i \lambda _i is equal to \mu , then there exists some finite prefix so that 1/2 \nu _0 < \sum _{i=0}^N 1/2^i \lambda _i. Then there exists some rearrangement of this prefix \lambda _i' so that \lambda _1' = \nu _1 (and \lambda '_i = \lambda _i when i>N). Let \mathrm {nm}_k: BT(A)^\omega \to BT(A)^\omega be the map which puts the first k elements into the normal form like this. Note that for each k there exists some permutation \sigma _k : BT(A)^\omega \times (2^{\omega } \setminus 0) \times (2^\omega )^\omega \to (2^{\omega } \setminus 0) \times (2^\omega )^\omega which, for each possible sequence, witnesses this rearrangement (in the same way as Lemma [efr-X0VA]).
3. By construction the nth coordinate of \mathrm {nm}_k(a) is equal for all k > n, and so there is a well-defined limiting map \mathrm {nm}_\infty : BT(A)^\omega \to BT(A)^\omega , which (by construction) carries a sequence to its normal form.
4. For \sigma _k, similarly note that the first n bits (deciding the index into the sequence) are fixed for k > n, and the same is true for the n first streams used to sample from the given binary trees.
5. Let \mathrm {nm}_\infty and \sigma _\infty be the limiting maps.

Now observe that F(a,b) = F(\mathrm {nm}_\infty (a),\sigma _\infty (a,b)), and that \sigma _\infty (a,b) for each a preseves the uniform measure. Hence just as before, this proves that the sampling map commutes with taking the normal form, as we wanted.

Observe that (for any monad), the Kleisli category is presented by adding morphisms s_A: TA \to A for each A, subject to the equations s_A\eta _A = 1_A and s_A s_{TA} = s_A \mu _A. We have just constructed the s_A, so we now have to prove that they satisfy these equations. Unitality is clear. The multiplicativity condition follows by observing that any finitary distribution on \Delta (A) can be decomposed as an iterated sum coming from \Delta (A)^{\omega ^N}, at which point this reduces to the mentioned \omega -midpoint homomorphism property.

We proceed by induction on k, the case k=0 being trivial.

First, consider a kernel \alpha : A \to (2^\omega )^k which, on the first N bits in the first coordinate, is concentrated on a given string a = a_1,\dots a_N which contains at least one 1. This kernel factors over the inclusion of this component: (2^\omega )^k = \{a\} \otimes 2^{\{N+1, \dots \}} \otimes (2^\omega )^{k-1} \sqcup (2^N \setminus \{a\}) \otimes 2^{\{N+1, \dots \}} \otimes (2^\omega )^{k-1}

This factorization is preserved by F (which preserves coproducts). Note that \vee ^k, when restricted to this summand, factors over the projection (2^\omega )^k \to (2^\omega )^{k-1}. So we can write F(\vee ^k)\bar {F}(\alpha ) = F((1,\vee ^{k-1}) \pi )\bar {F}(\alpha ). It is clear that F(\pi )\bar {F}(\alpha ) = \bar {F}(\pi \alpha ), and so by induction this means F(\vee ^k)\bar {F}(\alpha ) = F(\vee ^k \alpha ).

By an analogous argument (without passing through the finite truncation), if \alpha is concentrated on the string 0 \in 2^\omega in the first coordinate, again F(\vee ^k)\bar {F}(\alpha ) = F(\vee ^k \alpha )

Now consider the generic kernel \phi of the lemma. Let A = A_0 \sqcup A_1, where A_0 consists of those a for which the probability that the first coordinate is zero is \geq 1/2, and A_1 is the complement, consisting of those for which at least one 1 has probability >1/2. Note that by \sigma -continuity, there exists for each a \in A_1 some N_a so that there is a probability \geq 1/2 of an 1 in the first N_a coordinates. Note that the least such N_a is a measurable function of a, and so A_1 decomposes into a sum A_1 = A_1^1 \sqcup A_1^2 \dots , where on A_1^N there is a \geq 1/2 probability of an 1 in the first N coordinates.

Now let \mu _1 be the kernel defined as follows:

1. On A_0, it is deterministically 0 in the first coordinate, and given by the conditional distribution given this in the last k coordinates.
2. For a \in A_1^N, choose some distribution \nu on 2^N so that \nu / 2 is less than the marginals of \phi , and so that \nu is concentrated away from 0 (this is possible by construction of A_1^N). Then for each b \in 2^N, let \xi _b be the measure which is concentrated on that bitstring on the first N bits, and given by the conditional distribution according to \phi everywhere else. Let \mu _1(a) = \sum _{b\in 2^N} \nu (b) \xi _b

Observe that by construction, \mu _{1/2} \leq \phi . Now we can continue the recursion, defining \mu _2, \mu _3, \dots , and we must necessarily have \sum _i 1/2^i \mu _i = \phi . This sum must be preserved by \bar {F} (where the meaning of this sum in \mathcal {C} is given by the canonical sampling morphism X^\omega \to X, or equivalently by the dual I \to \omega ). Therefore we have F(\vee ^k)\bar {F}(\phi ) = F(\vee ^k)(\sum _i 1/2^i \bar {F}(\mu _i)) = \sum _i 1/2^i F(\vee ^k) \bar {F}(\mu _i) = \sum _i 1/2^i F(\vee ^k \mu _i) = F(\vee ^k \phi ), finishing the proof.

Using Lemma [efr-VH33], there is a unique extension F: \mathsf {Borel}_{\bar {\Delta }} \to \mathcal {C}. There is a faithful (identity on objects) functor \mathsf {Borel}_{\bar {\Delta }} \to \mathsf {BorelStoch} - the only case where this is not full is \operatorname {\mathrm {Hom}}(A,2^\omega ).

It is apparent that we must extend this functor by defining \bar {F}(\phi : A \to 2^\omega ) = \lim _n F(\pi _{2^n}\phi ), using the universal property of F(2^\omega ) = F(2)^\omega . It is apparent that this preserves independent pairings—the only question is whether this extension is actually functorial. By the limit property it suffices to prove that it is functorial for any composable pair A \xrightarrow {\phi } 2^\omega \xrightarrow {\psi } K with K finite.

Let a function f: 2^\omega \to K be good if \bar {F}(\phi ) ; F(f) = F(\phi ; f) for all kernels \phi : A \to 2^\omega (for all A). Let an algebra of sets \mathbb {A} \subseteq \mathcal {B}(2^\omega ) (i.e a collection of subsets stable under finite unions and complements) be called good if every \mathbb {A}-measurable map 2^\omega \to K to a finite set is good. Now we claim:

1. The class \mathbb {A}_0 of sets of the form V \times 2^\omega for V \subseteq 2^N, N finite, is a good algebra.
2. If \mathbb {A} is a good algebra, let \mathbb {A}^+ be the smallest algebra containing all countable unions of sets in \mathbb {A}. Then \mathbb {A}^+ is again a good algebra.
3. Any directed union of good algebras \mathbb {A}_0 \subseteq \mathbb {A}_1 \dots \subseteq A_\alpha \subseteq is again a good algebra.

First note that, by Zorn's lemma, this straightforwardly implies the full Borel \sigma -algebra is a good algebra, which in turn concludes the proof.

Point 3 holds, since both being an algebra and being good are finitary properties (any map to a finite set which is measurable for the union must be measurable at some finite stage). Point 1 is a straightforward consequence of the fact that the projections 2^\omega \to 2^N are good by construction. So we are left with point 2.

Let \mathbb {A} be a good algebra. Let us identify sets by their indicators 2^\omega \to 2. Then we can write any set in \mathbb {A}^+ as g(\vee (f_1^i(-)), \vee f_2^i(-), \dots , \vee f_k^i(-)), where g: 2^k \to 2 is some function, \vee : 2^\omega \to 2 is the indicator of the sequences with at least one 1, and f_j^i, 0 \leq j \leq k, 0 \leq i < \infty are a bunch of \mathbb {A}-measurable functions.

By assumption the pairing of all the f's, 2^\omega \to (2^\omega )^k is good, and so it suffices to show that the mapping (\vee )^k : (2^\omega )^k \to 2^k is good. By induction we may suppose this holds for (\vee )^{k-1}, since it is trivial for k=0.

Now let \phi : A \to (2^\omega )^k be some kernel. Write A = A_0 + A_1, where A_0 is the subset where the probability of only zeroes in the first stream is at least 1/2, and A_1 is the complement, where the probability of at least one 1 in the first stream is >1/2. Observe that by \sigma -continuity, for each a there must be some N so that the probability of at least one 1 in the first N elements of the first stream is at least 1/2. Decompose A_1 into A_1^1 + A_1^2 + \cdots , where for a \in A_1^N there is at least a 1/2 probability of the first 1 being before N. Let \psi _1 : A \to (2^\omega )^k be a kernel which on A_0 is given by (0, \otimes \mu _{2\dots k}) where \mu _{2,\dots k} is the kernel giving the conditional distribution of the remaining streams if the first stream is 0. On A_1^N, it is given by a linear combination \sum _{s \in 2^N \setminus 0} \delta _{s} \otimes \mu _s, where \mu _s is the conditional distribution of the rest of the stream and the remaining streams - we choose this linear combination so that 1/2 of it is \leq the actual probability of each of those prefixes.

Thus we obtain a decomposition of \phi as \sum _i \frac {1}{2^i} \phi _i where for each \phi _i, the distribution on the first coordinate is either concentrated on 0 \in 2^\omega , or concentrated away from 0 on some finite prefix. Each of these satisfy F(\vee ^k)\bar {F}(\phi _i) = F(\vee ^k \phi _i). Hence, since F(\vee ^k) must preserve the infinite linear combination, \vee ^k must be good as desired.

Now let f: 2^\omega \to B be a kernel, and let \phi : A \to 2^\omega again be a kernel. We can factor f as a measurable map \bar {f} : 2^\omega \times 2^\omega \to B, composed with the uniform c^\omega : I \to 2^\omega . By the above we have F(f) = F(\bar {f})\circ (F(c^\omega ) \otimes 2^\omega ). By monoidality, it follows that \bar {F}(f)\bar {F}(\phi ) = F(f\phi ) as desired.

Subject to these assumptions, \mathsf {Borel}_T is a Markov category which satisfies the hypotheses of Theorem [efr-TVLB]. Hence there is a strong monoidal, identity-on-objects functor \mathsf {BorelStoch} = \mathsf {Borel}_G \to \mathsf {Borel}_T. Such functors correspond bijectively to morphisms of monads G \to T (see eg Reference [barr-wells-triples-2005], Theorem 3.6.3), finishing the proof.