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.