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Let $X,Y,Z$ be metric spaces. Let $\mu,\nu,\omega$ be the probability measures on $X,Y,Z$, respectively. Moreover, assume all three measures vanish on small sets. Assume $T:X\to Y$ is the optimal transport map from $\mu$ to $\nu$, i.e., $\nu=T\#\mu$; and $S:Y\to Z$ is the optimal transport map from $\nu$ to $\omega$.

I wonder if $T\circ S$ is the optimal transport map from $\mu$ to $\omega$. I know the statement is true when $X=Y=Z=\mathbb{R}$ where the optimal transport map between distributions of two 1-d random variables can be analytically expressed. Moreover, the statement is false if the measures are discrete in $\mathbb{R}^d$ for $d\geq 2$. But, how about the general case? Is there somehow a gluing lemma here?

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  • $\begingroup$ Why the differential geometry tag? $\endgroup$ Commented Nov 9, 2021 at 21:33
  • $\begingroup$ OT is connected to maps in over manifolds, so I used that tag. Is there any issue with that? $\endgroup$ Commented Nov 11, 2021 at 9:02
  • $\begingroup$ Up to now, I have only a partial answer to this. In Wasserstein-2 space, such measures are called compatible measures. That is, if the transport maps among $\nu_0,\nu_{1/2},\nu_1$ satisfy $t_{\nu_0}^{\nu_1} = t_{\nu_{1/2}}^{\nu_1}\circ t_{\nu_0}^{\nu_1{/2}}$. For example, Gaussian measures are compatible if their covariance matrices are simultaneously diagonalizable. You may find some more examples in section 2.3.2 in An Invitation to Statistics in Wasserstein Space. $\endgroup$ Commented Dec 31, 2023 at 0:27
  • $\begingroup$ The answer is definite if we are considering the quadratic cost and $X=Y=Z = \mathbb{R}^d$. Using Brenier's theorem says that a transport map is optimal between $\mu \in \mathcal{P}_{ac}(\mathbb{R}^d)$ and $\nu := T\#\mu$ if and only if it is of the form $T:=\nabla \varphi$, where $\varphi :\mathbb{R}^d \to \mathbb{R}$ is convex. Thus, if $T\circ S$ is still the gradient of some convex function, we may give a definite answer. $\endgroup$ Commented Feb 13 at 2:52

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