Let $C$ be a coupling between two measures, $C= \mu^1 \mbox{ } t \mbox{ } \mu^2$ ($t$ is the symbol of binary operator of the coupling (I can't find a more proper symbol here)). The measures are both defined on a product space $X = \prod_{s \in S} \Omega_s$. How can I prove that,

$$(\mu^1 \circ \pi^{-1}_s) \mbox{ } t \mbox{ } (\mu^2 \circ \pi_s^{-1} ) = (\mu^1 \mbox{ } t \mbox{ } \mu^2) \circ \pi_s^{-1},$$ that means that the projection on a coordinate $s$ of the coupling between two measures is the coupling of the projections of those measures on the same coordinate?

P.S. the coupling is not necessarily the product between the two measures. The coupling $C = \mu^1 \mbox{ } t \mbox{ } \mu^2$ is any measure $C$ acting on the $\sigma$-algebra of subsets of $X^1 \times X^2$, with $X^i = X$, which has as marginals $\mu^1$ and $\mu^2$, i.e. $\mu^1 = C \circ \pi^{-1}_1$, $\mu^2 = C \circ \pi^{-1}_2$, where $\pi_i$ is the projection $\pi_i : X^1 \times X^2 \longrightarrow X^i$.

thecoupling of the projections"? Do you mean isacoupling of the projections? – Anthony Quas Nov 23 '12 at 20:10