Let $X, Y, Z$ be measurable spaces with measures $\mu\_X, \mu\_Y, \mu\_Z$ respectively. Let $\pi\_Y : Y \times Z \rightarrow Y$ be the projection on $Y$ and $\pi\_Z : Y \times Z \rightarrow Z$ the projection on $Z$. Let $\psi : X \rightarrow Y \times Z$ be a **surjective** map such that $(\pi\_Y \circ \psi)\_{\*} \mu\_X = \mu\_Y$ and $(\pi\_Z \circ \psi)\_{\*} \mu\_X = \mu\_Z$. Can we deduce that $\psi\_{\*}\mu\_X = \mu\_Y \otimes \mu\_Z$ (this is a notation for the product measure of $\mu\_Y$ and $\mu\_Z$)?
If yes how? If not, what other conditions are required?


In case the answer to the above question is no, the following is a more precise account of my problem: Let $G$ be a Lie group and $H\_i, K\_i \leq G$ Lie subgroups for $i = 1,2,3$. Let $X\_i = K\_i \setminus G /H\_i$ be the double coset space given by quotienting from the right by $H\_i$ and from the left by $K\_i$,i.e. an element of $X\_i$ is a double coset of the form $K\_igH\_i$. Endow $X\_i$ with the unique $G$ invariant measure and denote this measure by $\mu\_{X\_i}$. I have a map $\psi: X\_1 \rightarrow X\_2 \times X\_3$ which satisfies the properties above. Namely, $\psi$ is surjective and $(\pi\_{X\_i} \circ \psi)\_{\*}\mu\_{X\_1} = \mu\_{X\_i}$ for $i =2,3$. I would like to show that $\psi\_{\*}\mu\_{X\_1} = \mu\_{X\_2} \otimes \mu\_{X\_3}$. Thank you in advance for your answer.



Pushforward of measure under surjective map
The answer to the first part is no. The point is that surjectivity is not as strong a condition in this context as one might wish for.


Let $X=Y=Z=[0,1]$ with the Borel $\sigma$-algebra and $\mu\_X=\mu\_Y=\mu\_Z$ be the uniform distribution. Consider the non-surjective function $x\mapsto (x,x)$. It's push-forward is clearly not $\mu\_Y\otimes\mu\_Z$, it is the uniform distribution on the diagonal. But you can take an uncountable subset $N$ of $X=[0,1]$ of measure zero (the Cantor ternary set will do). Then $N$ will actually be measurably isomorphic to the set $\{(x,x)\mid x\in N\}\cup\{(x,y)\mid x,y\in [0,1], x\neq y\}$ by Kuratowski's Borel isomorphism and the fact that all uncountable Borel sets in Polish spaces have the cardinality of the continuum. Let $\phi$ be such a measurable isomorphism. Now define $\psi:X\to Y\times Z$ by $\psi(x)=(x,x)$ for $x\notin N$ and $\psi(x)=\phi(x)$ for $x\in N$. The pushforward of $\mu\_X$ under the surjection $\phi$ is the uniform distribution on the diagonal.


