Proof of The Optimal Maps of a Uniquely Mapped Pair and of Its Reverse are Mutually Inverse Almost Everywhere

Each result cited below is universally quantified over the data in its own statement.

Put $\pi=(\mathrm{id},T)_{\#}\mu$ and $\pi'=(\mathrm{id},T')_{\#}\nu$. By Optimal Transport Maps and Uniquely Mapped Pairs of Probability Measures (i) these are optimal couplings of $\mu$ and $\nu$, respectively of $\nu$ and $\mu$.

1. The swap of $\pi$ is $\pi'$. Let $\sigma$ be the swap of Couplings on Euclidean Space: Product Coupling, Swap, Finiteness of the Cost, Push-Forward Couplings, Modifying One Marginal, Quantisation, Gluing over a Finitely Supported Measure, and the Lipschitz Bound (ii). By that clause $\sigma_{\#}\pi\in\Pi(\nu,\mu)$ and $I(\sigma_{\#}\pi)=I(\pi)$. Since $\pi$ is optimal, $I(\pi)=W_{2}(\mu,\nu)^{2}$, and $W_{2}(\mu,\nu)=W_{2}(\nu,\mu)$ by The Quadratic Wasserstein Distance is a Metric on the Wasserstein Space (i); hence $I(\sigma_{\#}\pi)=W_{2}(\nu,\mu)^{2}$, so $\sigma_{\#}\pi$ is an optimal coupling of $\nu$ and $\mu$. Because $(\nu,\mu)$ is uniquely mapped, there is an optimal map $R$ from $\nu$ to $\mu$ such that every optimal coupling of $\nu$ and $\mu$ equals $(\mathrm{id},R)_{\#}\nu$; applying this to the two optimal couplings $\sigma_{\#}\pi$ and $\pi'$ gives

$$\sigma_{\#}\pi=(\mathrm{id},R)_{\#}\nu=\pi' .$$

2. The first identity. Let

$$M=\bigl\{z\in\mathbb{R}^{d+d}:\ \mathrm{pr}_{2}(z)=T'\bigl(\mathrm{pr}_{1}(z)\bigr)\bigr\},$$

the graph of $T'$ in the sense of A Coupling Concentrated on the Graph of a Borel Map is the Push-Forward by That Map, a member of $\mathcal{B}(\mathbb{R}^{d+d})$ as shown there. By Pairs of Euclidean Points: Coordinate Projections, Pairings, the Product Measure on a Euclidean Space, Borel Norm Functions and Finite Sets (i) and Pairs of Euclidean Points: Coordinate Projections, Pairings, the Product Measure on a Euclidean Space, Borel Norm Functions and Finite Sets (ii), $(\mathrm{id},T')(y)=\iota(y,T'(y))$ lies in $M$ for every $y\in\mathbb{R}^{d}$, so $(\mathrm{id},T')^{-1}(M)=\mathbb{R}^{d}$ and

$$\pi'(M)=\nu\bigl((\mathrm{id},T')^{-1}(M)\bigr)=\nu(\mathbb{R}^{d})=1 .$$

On the other hand $\sigma(\iota(x,y))=\iota(y,x)$ by Pairs of Euclidean Points: Coordinate Projections, Pairings, the Product Measure on a Euclidean Space, Borel Norm Functions and Finite Sets (ii), so, using Pairs of Euclidean Points: Coordinate Projections, Pairings, the Product Measure on a Euclidean Space, Borel Norm Functions and Finite Sets (i),

$$\sigma^{-1}(M)=\bigl\{z\in\mathbb{R}^{d+d}:\ \mathrm{pr}_{1}(z)=T'\bigl(\mathrm{pr}_{2}(z)\bigr)\bigr\} ,$$

and $(\mathrm{id},T)^{-1}(\sigma^{-1}(M))=\{x\in\mathbb{R}^{d}:x=T'(T(x))\}$, since $(\mathrm{id},T)(x)=\iota(x,T(x))$ has first coordinate $x$ and second coordinate $T(x)$. Step 1 therefore gives

$$1=\pi'(M)=\sigma_{\#}\pi(M)=\pi\bigl(\sigma^{-1}(M)\bigr)=\mu\bigl(\{x\in\mathbb{R}^{d}:\ T'(T(x))=x\}\bigr).$$

The set $\{x\in\mathbb{R}^{d}:T'(T(x))\ne x\}$ is its complement and belongs to $\mathcal{B}(\mathbb{R}^{d})$ as recorded in the statement, so claim 3 of Basic Properties of a Measure gives $\mu(\{x\in\mathbb{R}^{d}:T'(T(x))\ne x\})=1-1=0$.

3. The second identity. The hypotheses of the statement are symmetric in the two triples $(\mu,\nu,T)$ and $(\nu,\mu,T')$: both $(\mu,\nu)$ and $(\nu,\mu)$ are uniquely mapped, $T$ is an optimal map from $\mu$ to $\nu$, and $T'$ is an optimal map from $\nu$ to $\mu$. Exchanging the two triples in steps 1 and 2 yields $\nu(\{y\in\mathbb{R}^{d}:T(T'(y))\ne y\})=0$. This proves claim 1 of the statement.