# Pushforward of measure under surjective map

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.

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.