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