Disintegration-like theorem $\def\b{\mathcal B}\def\p{\mathcal P}\def\d{\mathrm d}$
Let $X$ be a (standard) Borel space: a topological space isomorphic to a Borel set of a complete separable metric space. Denote by $\b(X)$ the Borel $\sigma$-algebra of $X$, by $\p(X)$ the set of Borel probability measures on $X$ endowed with a topology of weak convergence.
Let $Y$ be another Borel space and let $f:X\to Y$ be a Borel map. Consider a measure $\alpha\in \p(X)$ and a stochastic kernel $P:X\to \p(X)$. Let us denote by $Q:X\to\p(Y)$ the pushforward kernel:
$$
  Q(B|x) := P(f^{-1}(B)|x) \qquad \text{ for all }x\in X,B\in \b(Y).
$$
Does there exist a stochastic kernel $R:X\times Y \to \p(X)$ satisfying the following conditions:


*

*$R(f^{-1}(y)|x,y) = 1$ for $\alpha\otimes Q$-almost all $(x,y)\in X\times Y$, that is
$$
  \int_{X\times Y}R\left(\left.X\setminus f^{-1}(y)\right|x,y\right)\;Q(\d y|x)\;\alpha(\d y) = 0
$$

*for all $A\in \b(X)$ and $\alpha$-almost all $x\in X$ it holds that
$$
  P(A|x) = \int_Y R(A|x,y)\; Q(\d y|x).
$$

The statement sounds very much related to the Disintegration theorem, however the current version is seemingly more general and I was not able to transform it as a special case of the original formulation. 

Regarding the assumptions on the measurable spaces involved. In the formulation I have linked the Radon spaces are required (I don't know whether each Borel space is Radon), whereas in "Conditioning and Disintegration" of Kallenberg (Ch. 5 in 1st ed., Ch.6 in 2nd ed.) the disintegration theorem only requires the existence of regular conditional distributions. As in my case everything shall only hold almost surely, I guess that the result I am looking for may hold for general measurable spaces just based on the existence of conditional expectation for general measurable spaces. Although I know that there exists a connection between the conditional expectation and the disintegration, I'm not sure whether it can help me here.
 A: I think the following affirmative answer is correct. However, it makes use of fancy machinery and it will probably take some effort to turn it into a pedestrian and rigorous proof. The basic idea is using the fact that a measure and a kernel give rise to a product distribution, and up to null sets, only one kernel does so.
To make things easier to see, take $X_1=X_2=X$. We have a measure $\alpha$ on $X_1$, a kernel $P$ from $X_1$ to $X_2$ and a function $f$ from $X_2$ to $Y$. Applying generalized abstract nonsense, there is a category of kernels and measurable spaces, the category of probabilistic mappings. We can identify probability measures with kernels from a terminal one-point measurable space and measurable functions with Dirac-measure-valued kernels. So we have
$$1\xrightarrow{~~~\alpha~~~} X_1\xrightarrow{~~~P~~~} X_2\xrightarrow{~~~f~~~} Y$$
and $Q=f\circ P$. We do get a resulting measure $\mu$ on $X_1\times X_2\times Y$. We can apply a result for product regular conditional probabilities to get a kernel $R:X_1\times Y\to X_2$ that gives us $\mu$ when combined with $\text{proj}_{X_1\times Y}(\mu)$. Moreover, such kernels are unique up to $\text{proj}_{X_1\times Y}(\mu)$-zero sets, so it is sufficient to work with $R$.

*

*The $X_2\times Y$-marginal of $\mu$ is supported on the graph of $f$, since $(P\circ\alpha,f\circ P\circ\alpha)=(P\circ\alpha,Q\circ\alpha)$ is. But the complement of the graph would have positive probability if $R$ would not be $Q\circ\alpha$-almost surely be supported on the graph of $f$, by integrating over the measure of the sections, Fubini-style.


*When we endow $X_1$ with $\alpha$, there are two ways to get the correct joint distribution on $X_1\times X_2$: First, we can apply $(\text{id}_{X_1},P)$, second, we can apply $(\text{id}_{X_1},R\circ (\text{id}_{X_1},Q))$. By the uniqueness of RCP, they must coincide $\alpha$-almost surely.
A: I don't see any difficulty after rephrasing the question in the language of probability theory. Let $(A_1, A_2)$ be a couple of random variables such as $A_1 \sim \alpha$ and $(A_2 | A_1=x) \sim P(\cdot | x)$. Then set $B_2=f(A_2)$, so that $(B_2 | A_1=x) \sim Q(\cdot |x)$. Take a regular version $R(\cdot | x,y)$ for the conditional law ${\cal L}(A_2 | A_1=x, B_2=y)$, provided by the disintegration theorem.  
Your first condition is
$$
1 = \Pr(f(A_2)=B_2) = \mathbb{E}\left[\Pr(f(A_2)=B_2 | A_1, B_2) \right] 
$$
and your second condition is 
$$
\Pr(A_2 \in A | A_1) = \mathbb{E}\left[\Pr(A_2 \in A| A_1, B_2) \big\vert A_1 \right]. 
$$
