A Negative Answer to Wolansky's Question on Mutually Dominating Multiphase Measure Spaces

In an Oberwolfach report of 2017, Wolansky introduced a domination order between vector-valued measure spaces (X, σ) and (Y, η) with k components, defined by one Markov kernel that transports every σ_i to η_i, and asked whether two atomless spaces that dominate each other must admit deterministic maps T : X → Y and S : Y → X with T_#σ_i = η_i and S_#η_i = σ_i. We show that the answer is no as posed, already for k = 2 and compact atomless spaces. Mutual domination is equality of the colour laws (Blackwell equivalence), and a deterministic transport has to respect colours, so it must carry colour fibres onto colour fibres. This gives explicit examples on [0,1] and [0,1]² in which only one of the two maps exists, and an example on [0,1] in which neither exists. On the positive side, for standard Borel spaces with equal colour laws, T exists whenever the colour fibres of X are atomless, for instance for atomless spaces with k = 1 or with finitely many colours. We give necessary conditions in general and an exact classification for finite spaces. The results are consequences of the classical comparison of experiments (Blackwell, Le Cam, Torgersen), and we claim no novelty beyond answering this question with explicit atomless examples and the atomless-fibre positive case. This is an unrefereed note.