# Subspace Detours Meet Gromov-Wasserstein
## Appendix B Knothe-Rosenblatt
### Proof of Theorem 3
Now, we still have to show that the marginals of \(\gamma^{2|1}((x_{1},y_{1}),\cdot)\) and \(\gamma^{2,1}_{K}((x_{1},y_{1}),\cdot)\) are well the same, i.e., \(\mu^{2|1}(x_{1},\cdot)\) and \(\
u^{2|1}(y_{1},\cdot)\). Let \(\phi\) and \(\psi\) be continuous functions, then we have to show that for \(\gamma^{1}\)-a.e. \((x_{1},y_{1})\), we have  
\[ \begin{cases}\int\phi(x_{2})\gamma^{2|1}((x_{1},y_{1}),(\mathrm{d}x_{2}, \mathrm{d}y_{2}))=\int\phi(x_{2})\mu^{2|1}(x_{1},\mathrm{d}x_{2})\\ \int\psi(y_{2})\gamma^{2|1}((x_{1},y_{1}),(\mathrm{d}x_{2},\mathrm{d}y_{2}))= \int\psi(y_{2})\
u^{2|1}(y_{1},\mathrm{d}y_{2}).\end{cases} \]  
As we want to prove it for \(\gamma^{1}\)-a.e. \((x_{1},y_{1})\), it is sufficient to prove that for all continuous function \(\xi\),  
\[ \begin{cases}\iint\xi(x_{1},y_{1})\phi(x_{2})\gamma^{2|1}((x_{1},y_{1}),( \mathrm{d}x_{2},\mathrm{d}y_{2}))\mathrm{d}\gamma^{1}(x_{1},y_{1})\\ =\iint\xi(x_{1},y_{1})\phi(x_{2})\mu^{2|1}(x_{1},\mathrm{d}x_{2})\mathrm{d} \gamma^{1}(x_{1},y_{1})\\ \iint\xi(x_{1},y_{1})\psi(y_{2})\gamma^{2|1}((x_{1},y_{1}),(\mathrm{d}x_{2}, \mathrm{d}y_{2}))\mathrm{d}\gamma^{1}(x_{1},y_{1})\\ =\iint\xi(x_{1},y_{1})\psi(y_{2})\
u^{2|1}(y_{1},\mathrm{d}y_{2})\mathrm{d} \gamma^{1}(x_{1},y_{1}).\end{cases} \]  
First, we can use the projections \(\pi_{x}(x,y)=x\) and \(\pi_{y}(x,y)=y\). Moreover, we know that \(\gamma^{1}=(Id\times T^{1}_{K})_{\sharp}\mu^{1}\). The alternate Knothe-Rosenblatt rearrangement is, as the usual one, bijective (because \(\mu\) and \(\
u\) are absolutely continuous), and thus, as we suppose that \(\
u\) satisfies the same hypothesis than \(\mu\), we also have \(\gamma^{1}=((T^{1}_{K})^{-1},Id)_{\sharp}\
u^{1}\). Let us note \(\hat{T}^{1}_{K}=(T^{1}_{K})^{-1}\). Then, the equalities that we want to show are:  
\[ \begin{cases}\iint\xi(x_{1},T^{1}_{K}(x_{1}))\phi(x_{2})\gamma^{2|1}_{\tau}((x_ {1},T^{1}_{K}(x_{1})),\mathrm{d}x_{2})\mathrm{d}\mu^{1}(x_{1})\\ =\iint\xi(x_{1},T^{1}_{K}(x_{1}))\phi(x_{2})\mu^{2|1}(x_{1},\mathrm{d}x_{2}) \mathrm{d}\mu^{1}(x_{1})\\ \iint\xi(\hat{T}^{1}_{K}(y_{1}),y_{1})\psi(y_{2})\gamma^{2|1}_{\tau}((T^{1}_{K}(y _{1}),y_{1}),\mathrm{d}y_{2})\mathrm{d}\
u^{1}(y_{1})\\ =\iint\xi(\hat{T}^{1}_{K}(y_{1}),y_{1})\psi(y_{2})\
u^{2|1}((y_{1},\mathrm{d}y_{2 })\mathrm{d}\
u^{1}(y_{1}).\end{cases} \]In addition, we have indeed  
\[\iint\xi(x_{1},T_{K}^{1}(x_{1}))\phi(x_{2})\gamma_{x}^{2|1}((x_{1},T _{K}^{1}(x_{1})),\mathrm{d}x_{2})\mathrm{d}\mu^{1}(x_{1})\] \[=\iint\xi(x_{1},T_{K}^{1}(x_{1}))\phi(x_{2})\mathrm{d}\gamma^{1,2} ((x_{1},x_{2}),(y_{1},y_{2}))\] \[=\iint\xi(x_{1},T_{K}^{1}(x_{1}))\phi(x_{2})\mathrm{d}\gamma_{x}^{ 1,2}(x_{1},x_{2})\] \[=\iint\xi(x_{1},T_{K}^{1}(x_{1}))\phi(x_{2})\eta^{2|1}(x_{1}, \mathrm{d}x_{2})\mathrm{d}\mu^{1}(x_{1}).\]  
We can do the same for the \(\
u\) part by symmetry. _Part 3:_  
Now, we can proceed the same way by induction. Let \(\ell\in\{2,\ldots,d\}\) and suppose that the result is true in dimension \(\ell-1\) (i.e., \(\gamma^{1:\ell-1}=\pi_{\#}^{1:\ell-1}\gamma=\gamma_{K}^{1:\ell-1}\)). For this part of the proof, we rely on [19] (Theorem 2.23). We can build a measure \(\gamma_{K}^{t}\in\mathcal{P}(\mathbb{R}^{d}\times\mathbb{R}^{d})\) such that:  
\[ \begin{cases}\pi_{\#}^{x}\gamma_{K}^{t}=\mu\\ \pi_{\#}^{y}\gamma_{K}^{t}=\
u\\ \pi_{\#}^{1:\ell-1}\gamma_{K}^{t}=\eta_{t,\ell}\end{cases} \] (A8)  
where \(\eta_{t,\ell}\) is the optimal transport plan between \(\mu^{\ell}=\pi_{\#}^{1:\ell-1}\mu\) and \(\
u^{\ell}=\pi_{\#}^{1:\ell-1}\
u\) for the objective:  
\[\iint\sum_{k=1}^{\ell-1}\Big{(}\prod_{i=1}^{k-1}\lambda_{t}^{(i)}\Big{)}(x_{k }x_{k}^{\prime}-y_{k}y_{k}^{\prime})^{2}\;\mathrm{d}\gamma(x,y)\mathrm{d} \gamma(x^{\prime},y^{\prime}).\]  
By induction hypothesis, we have \(\eta_{t,\ell}\xrightarrow[t\to 0]{\mathcal{D}}\pi_{\#}^{1:\ell-1}\gamma_{K}\). To build such a measure, we can first disintegrate \(\mu\) and \(\
u\):  
\[ \begin{cases}\mu=\mu^{1:\ell-1}\otimes\mu^{\ell:d|1:\ell-1}\\ \
u=\
u^{1:\ell-1}\otimes\
u^{\ell:d|1:\ell-1},\end{cases}