Every finite extension is obtained from a purification by a channel

Choose a finite purification $\mathrm{\mid }\mathrm{\Omega }{⟩}_{ZAF}$ of ${\omega }_{ZA}$ by [F1] and reorder its factors as $A(ZF)$. Let ${\rho }_A={\sum }_{j=1}^r{\lambda }_j\mathrm{\mid }{a}_j⟩⟨a_j\mathrm{\mid }$ be a spectral decomposition on its positive support, obtained from [F7]. Expand each purification in the product basis using this same eigenbasis on $A$: $\mathrm{\mid }\psi {⟩}_{RA}={\sum }_{j=1}^r\mathrm{\mid }{r}_j^0{⟩}_R\mathrm{\mid }{a}_j{⟩}_A,\mathrm{\mid }\mathrm{\Omega }{⟩}_{A(ZF)}={\sum }_{j=1}^r\mathrm{\mid }{a}_j{⟩}_A\mathrm{\mid }{z}_j^0{⟩}_{ZF}\mathrm{.}$ The partial-trace coordinate formula [F2] and the fact that both $A$ marginals equal ${\rho }_A$ show that the coefficient vectors have Gram matrices $⟨r_j^0,r_k^0⟩={\lambda }_j{\delta }_{jk}$ and $⟨z_j^0,z_k^0⟩={\lambda }_j{\delta }_{jk}$. Thus $\mathrm{\mid }{r}_j⟩:=\mathrm{\mid }{r}_j^0⟩\mathrm{/}\sqrt{{\lambda }_j}$ and $\mathrm{\mid }{z}_j⟩:=\mathrm{\mid }{z}_j^0⟩\mathrm{/}\sqrt{{\lambda }_j}$ are orthonormal and $\mathrm{\mid }\psi {⟩}_{RA}={\sum }_{j=1}^r\sqrt{{\lambda }_j}\mathrm{\mid }{r}_j{⟩}_R\mathrm{\mid }{a}_j{⟩}_A,\mathrm{\mid }\mathrm{\Omega }{⟩}_{A(ZF)}={\sum }_{j=1}^r\sqrt{{\lambda }_j}\mathrm{\mid }{a}_j{⟩}_A\mathrm{\mid }{z}_j{⟩}_{ZF}\mathrm{.}$ This coefficient argument also handles repeated positive eigenvalues because the chosen basis is fixed before either expansion.

Let $d=\dim R=\dim A$. In every case set $G={\mathbb{C}}^d$, fix its first coordinate unit vector $\mathrm{\mid }0{⟩}_G$, and append it to $\mathrm{\mid }\mathrm{\Omega }⟩$. Since $Z$ and the purifying space $F$ are nonzero, $\dim (Z\otimes F\otimes G)\ge d$. Denote the resulting orthonormal Schmidt partners by $\mathrm{\mid }{z}_j^{\mathrm{\prime }}⟩=\mathrm{\mid }{z}_j⟩\mathrm{\mid }0{⟩}_G$. The map $U_0:\mathrm{span}\{\mathrm{\mid }{r}_j⟩\}\to \mathrm{span}\{\mathrm{\mid }{z}_j^{\mathrm{\prime }}⟩\}$ given by $U_0\mathrm{\mid }{r}_j⟩=\mathrm{\mid }{z}_j^{\mathrm{\prime }}⟩$ is an isometry. Extend $\{\mathrm{\mid }{r}_j⟩\}$ to an orthonormal basis of $R$ and choose the same number of additional orthonormal vectors in the orthogonal complement of the $z_j^{\mathrm{\prime }}$ in $Z\otimes F\otimes G$; [F5] and the target dimension bound ensure this is possible. Sending the added basis vectors to those target vectors extends $U_0$ to a linear isometry $U:R\to Z\otimes F\otimes G$ on all of $R$. By construction, $(U\otimes I_A)\mathrm{\mid }\psi ⟩$ is exactly the padded purification $\mathrm{\mid }\mathrm{\Omega }⟩\otimes \mathrm{\mid }0{⟩}_G$, after reordering to $Z(F\otimes G)A$. This construction includes the singular case $r<d$, since it extends $U$ on all zero-Schmidt directions of $R$.

Define $\mathrm{\Lambda }(X)={\mathrm{Tr}}_{F\otimes G}(UX{U}^{\ast })$ from $\mathcal{L}(R)$ to $\mathcal{L}(Z)$. Since $U$ is an isometry, [F4] makes $\mathrm{\Lambda }$ completely positive and trace-preserving on the full operator space. Applying it to the reference factor of $\mathrm{\mid }\psi ⟩⟨\psi \mathrm{\mid }$ traces $F\otimes G$ from the pure state $\mathrm{\mid }(U\otimes I_A)\psi ⟩⟨(U\otimes I_A)\psi \mathrm{\mid }$, leaving its $ZA$ marginal, which is ${\omega }_{ZA}$ by construction and is unchanged by the appended pure $G$ register [F6]. Thus $(\mathrm{\Lambda }\otimes {\mathrm{id}}_A)(\mathrm{\mid }\psi ⟩⟨\psi \mathrm{\mid })={\omega }_{ZA}$, as required.