Prove that $Apreceq B$ implies $A=Psi(B)$ for some channel $Psi$

Question

Define $newcommand{PP}{mathbb{P}}newcommand{ket}[1]{lvert #1rangle}newcommand{tr}{operatorname{tr}}newcommand{ketbra}[1]{lvert #1rangle!langle #1rvert}PP_psiequivketbrapsi$, and let $ketpsi,ketphi$ be two bipartite states such that $tr_2(PP_psi)prectr_2(PP_phi)$.
Here, $Aprec B$ with $A,B$ positive operators means that the vector of eigenvalues of $A$ is majorised by that of $B$: $Apreceq BLongleftrightarrowlambda(A)preceqlambda(B)$.
A step to prove Nielsen's theorem, used in the proof of the theorem given here (pdf alert) is that $tr_2(PP_psi)prectr_2(PP_phi)$ implies $tr_2(PP_psi)=Psi(tr_2(PP_phi))$ for some mixed unitary channel $Psi$.
More precisely, it implies that $tr_2(PP_psi)=Psi( Wtr_2(PP_phi)W^dagger)$ for some mixed unitary channel $Psi$ and isometry $W$ (though these two statements seem pretty much equivalent to me).
To show this, an important observation seems to be the fact that, introducing the operators $X,Y$ with components $X_{ij}=psi_{ij}, Y_{ij}=phi_{ij}$ (that is, $ketpsi= operatorname{vec}(X)$ and $ketphi= operatorname{vec}(Y)$), we have
$$tr_2(PP_psi) = XX^dagger,qquad tr_2(PP_phi) = YY^dagger.$$
Suitably defining the underlying vector spaces, we can always assume $XX^dagger ,YY^dagger >0$. Moreover, $XX^daggerprec YY^dagger$ implies $operatorname{rank}(XX^dagger)geoperatorname{rank}(YY^dagger)$.
Why does this imply that the existence of a mixed unitary channel $Phi$ and isometry $W$ such that $XX^dagger = Psi(WYY^dagger W^dagger)$? The reason is probably trivial but I'm not seeing it right now.

keisuke.akira · Answer

Let $rho_{d}, sigma_{d}$  be the (simultaneously diagonal) density matrices whose eigenvalues are ${ p_{j} }, { q_{j} }$, respectively (represented as probability vectors below). Then, if $vec{p} succ vec{q}$, the following sequence of arguments can be observed:

There exists a bistochastic matrix $M$ such that $M vec{p} = vec{q}$ (basic result of majorization theory, see Marshall and Olkin, for example.)
Using Birkhoff's theorem, the bistochastic can be written as a convex combination of permutations: $M = sumlimits_{j} r_{j} P_{j}$.
$M$ can be ``quantized'' into a (mixed unitary) CPTP map, $M mapsto mathcal{M} = sumlimits_{j} r_{j} mathcal{U}_{P_{j}}$, where $mathcal{U}_{P_{j}}$ is the unitary superoperator, defined as $mathcal{U}_{P_{j}}(cdot) = P_{j} (cdot) P_{j}^{dagger}$. Recall that permutations have a unitary representation.
The action of $mathcal{M}$ is to transform $rho_{d} mapsto sigma_{d}$.

Why can we start from simultaneously diagonal states $rho_{d}, sigma_{d}$? Hint: the partial trace.
In several quantum resource theories the state transformation reduces to ``classical majorization'', i.e., majorization of vectors (as opposed to say matrix majorization), for example, resource theory of coherence, non-uniformity, etc.

Prove that $Apreceq B$ implies $A=Psi(B)$ for some channel $Psi$

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