The following is usually called "Gluing Lemma":
Given two measures $\gamma^+ \in \prod(\mu,\rho)$ and $\gamma^- \in \prod(\rho,\nu)$, there is at least one measure $\sigma\in \mathcal P(\Omega \times \Omega \times \Omega)$ such that $(\pi_{x,y})_\# \sigma = \gamma^+$ and $(\pi_{y,z})_\# \sigma = \gamma^-$,where $\pi_{x,y}$ and
$\pi_{y,z}$ denote the projections on the two first and two last variables, respectively.
Note that $\gamma^+ \in\prod(\mu,\rho)$ means that $\gamma ^+$ is a coupling of measures $\mu$ and $\rho$, in other words, the marginals of $\gamma^+$ are $\mu$ and $\rho$, respectively.
Now, I was wondering if there exists some sort of "Stronger Version" of the Gluing Lemma… In this original formulation, we need that both $\gamma^+$ and $\gamma^-$ to share the $\rho$ as marginal. But perhaps there are weaker conditions for the existance of such $\sigma$.
Here is an example of what I was looking for:
(Fictitious Lemma)
Given two measures $\gamma^+ \mathcal P(\Omega\times \Omega)$ and $\gamma^- \in \mathcal P(\Omega\times \Omega)$. If the support of $(\pi_y)_\#\gamma^+ = (\pi_y)_\#\gamma^-$ , then there is at least one measure $\sigma\in \mathcal P(\Omega \times \Omega \times \Omega)$ such that $(\pi_{x,y})_\# \sigma = \gamma^+$ and $(\pi_{y,z})_\# \sigma = \gamma^-$,where $\pi_{x,y}$ and
$\pi_{y,z}$ denote the projections on the two first and two last variables, respectively.
Best Answer
No, the condition $(\pi_y)_\#\gamma^+=(\pi_y)_\#\gamma^-$ is necessary: If you take the pushforward under $\pi_y$ of the identities $(\pi_{x,y})_\# \sigma = \gamma^+$ and $(\pi_{y,z})_\# \sigma = \gamma^-$, you get $(\pi_y\circ \pi_{x,y})_\#\sigma=(\pi_y)_\#\gamma^+$ and $(\pi_y\circ \pi_{y,z})_\#\sigma=(\pi_y)_\#\gamma^-$. But $\pi_y\circ\pi_{x,y}=\pi_y\circ\pi_{y,z}=\pi_y$, so that the two left sides coincide.