Given $\Omega = [0,1] \times [0,1]$ endowed with the borel sigma algebra. Can we characterize all random variables $Z:\Omega \to [0,1]$ s.t. for any probability distribution $P=P_1 \times P_2$ on $\Omega$ we have $\forall A \in \sigma(\pi_1), B \in \sigma(\pi_2): P(A \cap B|Z) = P(A|Z)P(B|Z)$ a.s.? I.e. for any probability distribution that respects the independence of the projections, they remain independent after conditioning on $Z$. In the remaining document I refer to this as 'the property'.
Here is approximately what I've figured out so far:
My conjecture is that $Z$ satisfies this if and only if $\forall z:\{Z=z\} \in \mathbb{B} \times \mathbb{B}$, i.e. all atoms of $\sigma(Z)$ are rectangles.
It is easy to see that this is a necessary condition:
If there is a $z$ where $\{Z=z\}$ is not a rectangle, then there are $(a_1,a_2),(b_1,b_2) \in \{Z=z\}$ with $(a_1,b_2) \notin \{Z=z\}.$ We can then define
$$P_1=\frac{1}{2} \epsilon_{a_1} + \frac{1}{2} \epsilon_{a_2}$$
$$P_2=\frac{1}{2} \epsilon_{b_1} + \frac{1}{2} \epsilon_{b_2}$$
$$P = P_1 \times P_2$$
Now since $(a_1,a_2), (b_1,b_2) \in \{Z=z\}$ and $(a_1,b_2) \notin \{Z=z\}$,
we have $P(Z=z)$, $P(\pi_1=a_1, Z=z)$ , $P(\pi_2=b_1, Z=z) \notin \{0,1\}$,
but $P(\pi_1=a_1, \pi_2=a_2, Z=z) = 0$
giving: $P(\pi_1=a_1\mid Z=z)P(\pi_2=a_2\mid Z=z) \neq P(\pi_1=a_1,\pi_2=a_2\mid Z=z)$
Furthermore, the conjecture is true for $Z(\Omega)$ countable, since if $\{Z=z\}=C_1 \times C_2$ is not a nullset we have for $A=A_1 \times [0,1] \in \sigma(\pi_1)$, $B = [0,1] \times B_2 \in \sigma(\pi_2)$ $$P(A|Z=z)P(B|Z=z)$$ $$=\frac{P(A,Z=z)}{P(Z=z)}\frac{P(B,Z=z)}{P(Z=z)}$$ $$=\frac{P((A_1 \cap C_1)\times C_2)}{P(Z=z)}\frac{P(C_1\times (B_2 \cap C_2))}{P(Z=z)}$$ $$=\frac{P_1(A_1 \cap C_1)P_2(C_2)}{P_1(C_1)P_2(C_2)}\frac{P_1(C_1)P_2(B_2 \cap C_2)}{P_1(C_1)P_2(C_2)}$$ $$=\frac{P_1(A_1 \cap C_1)}{P_1(C_1)}\frac{P_2(B_2 \cap C_2)}{P_2(C_2)}$$ $$=\frac{P(A \cap B \cap \{Z=z\})}{P(Z=z}$$ $$=P(A \cap B \mid Z=z)$$
One can show with levy's upward theorem that if there is a sequence of partitions of [0,1] that generate $\sigma(Z)$, s.t. for each atom there is a point from which on the part it is in are rectangles, then $Z$ fulfills the desired property. However there are $Z$ which aren't like that, but still fulfill the property, so I think this is a dead end.
It also seems like functions of this form are 'rare'. If we instead let $Z: \Omega \to \Omega$, which is equivalent since polish spaces are measurably isomorphic, then one can show that if $$ Z = \sum_{n \in \mathbb{N}} 1_{A_n}(f_n(\pi_1),g_n(\pi_2)), \text{where } A_n \in \sigma(Z) $$ then $Z$ fulfills the property. In fact I have not been able to construct any $Z$ which is not equivalent to some random variable with this property. So one way to prove the conjecture would be to show that any random variable with rectangle atoms is in fact generated by disjoint union of restricted $\sigma$-algebras that are generated by rectangles, i.e. Let $(A_n)_n$ be a countable partition of $\Omega$, s.t. $A_n \in \mathbb{B}$ and for each $n \in \mathbb{N}$ let $\mathfrak{A}_n$ be a sigma algebra on $A_n$ defined by $$\mathfrak A_n = \sigma(\sigma(B_1) \times \sigma(B_2))|_{A_n} \text{where } B_1,B_2 \subset \mathbb{B}$$ Then if $\sigma(Z)=\sigma(\mathfrak{A}_n:n \in \mathbb{N})$, $Z$ fulfills the property. If any sigma-algebra that is induce by a random variable whose atoms are rectangle is generated in this way, the conjecture follows.
I would be glad if someone has any ideas how to proceed, or how to construct a nontrivial function whose sigma algebra has rectangle atoms.