measure with given push-forwards Let $X,Y$ be locally compact spaces (in my specific case, they are locally compact groups). Suppose that we are given a measure $\mu$ on $X$ and a finite number of quotient maps $p_1,\ldots,p_n:Y\rightarrow X$. Under what conditions on $\mu, p_1,\ldots,p_n$ is there a measure $\eta$ on $Y$ such that the marginals $(p_1)_\ast\eta=\ldots=(p_n)_\ast\eta=\mu$?
EDIT clarification about the type of quotients needed.
In the specific problem I am trying to solve, the space $Y=G\times G\times G$ and $X=G\times G$ where the measure $\mu$ is a given Radon probability measure on $X$. The quotients are given by
$p_1(x,y,z)=(x,y)$
$p_2(x,y,z)=(z,y)$
$p_3(x,y,z)=(x,xyz^{-1})$
$p_4(x,y,z)=(z,xyz^{-1})$
So the crucial difficulty is that the quotient maps do not come from a product decomposition of $Y$.
I would like to have a lemma of the form:
lemma If the quotient maps $p_1,\ldots,p_n$ satisfy the following conditions:
$\ldots$
then for every probability measure $\mu$ on $X$, there exists a probability measure $\nu$ on $Y$ such that $(p_i)_\ast\nu=\mu$.
EDIT2 Changed the title to be more in line with common terminology, and clarified the question. Thanks to Benoît.
 A: This kind of problem has been studien in:
J. Hoffmann-Jørgensen The general marginal problem Functional Analysis II Lecture Notes in Mathematics Volume 1242, 1987, pp 77-367
A: This is not a complete answer, just a simple example showing what kind of things can go wrong.
Consider $Y=A^3$ and $X=A^2$ with $A$ a non-trivial measurable space, and let $p_1(x,y,z)=(x,y)$ and $p_2(x,y,z)=(y,z)$. Then when $\mu$ a probability on $X$ does not have equal marginals, there can be no $\nu$ on $Y$ such that $p_i(\nu)=\mu$.
A: I hope this helps.  For two projections $p_1$ and $p_2$, if there exists such an $\eta$ then $\eta( p_1(S_Y) ) = \eta( p_2( S_Y) )$ for all measurable sets $S_Y \subset Y$.  Upon setting $S_1 = p_1(S_Y)$ for an arbitrary $S_Y \subset Y$ we see that $\eta$ must satisfy $\eta( S_1) = \eta( p_2 \circ p_1^{-1}(S_1) )$ for any measure able set $S_1 \subset X$.  In otherwords $(p_2 \circ p_1^{-1})^* \eta = \eta$.  So we see that the answer to the question is yes if $(p_2 \circ p_1^{-1})^*$ has fixed points.  For $n$ projections we need to check $(p_i \circ p_j^{-1})^*$ for $i,j = 1,\dots,n$.  This is for a given fixed $\mu$ on $Y$.  If we want an $\eta$ for every $\mu$ my guess is that this happens if and only if $(p_i \circ p_j^{-1})^*$ is the identity.  However, this would mean that $p_i = p_j$ for all $i,j$.
  In the case presented here we find $p_3 \circ p_1^{-1}(x,y) = (x,G)$ while $p_1 \circ p_2^{-1}(x,y) = (G,y)$.  The first calculation tells us that $\eta$ must be independent of the second component in $G \times G$ while the second calculation tells us that $\eta$ must be independent of the first component of $G \times G$.  This $\eta$ must be very restricted, to say the least, and this restricts the class of possible $\mu$'s.
