Suppose a composite system $AB$ initially in an unknown quantum state $\rho$ is brought into contact with a composite system $CD$ initially in some standard state $|0\rangle$, and the two systems interact according to a unitary interaction $U$. After the interaction we discard systems $A$ and $D$, leaving a state $\rho\:'$ of the system $BC$. Show that the map $\mathcal{E}(\rho)=\rho\:'$ satisfies $$ \mathcal{E}(\rho)=\sum_k E_k\rho E_k^\dagger $$ for some set of linear operators $E_k$ from the state space of system $AB$ to the state space of system $BC$, and such that $\sum_kE_k^\dagger E_k=I$.
This is given, in Exercise 8.3, Page 361, Operator-sum representation, Chapter 8, Quantum Computation and Quantum Information by Nielsen and Chuang.
Overall state of the system after the interaction is $U(\rho\otimes|0\rangle\langle 0|)U^\dagger$
Let $|a_m\rangle,|b_n\rangle,|c_p\rangle,|d_q\rangle$ be orthonormal basis for the state space of the systems $A,B,C,D$, respectively.
Discarding systems $A$ and $D$, i.e., tracing out system $A$ first then tracing out the system $D$ obtains
\begin{align} \mathcal{E}(&\rho)=tr_D\Big[tr_A\big[U(\rho\otimes|0\rangle\langle 0|)U^\dagger\big]\Big]\\ &=tr_D\Big[\sum_m(\langle a_m|\otimes I\otimes I\otimes I) \big[U(\rho\otimes|0\rangle\langle 0|)U^\dagger\big](|a_m\rangle\otimes I\otimes I\otimes I)\Big]\\ &=\sum_q(I\otimes I\otimes I\otimes \langle d_q|)\\ &\times\Big[\sum_m(\langle a_m|\otimes I\otimes I\otimes I) \big[U(\rho\otimes|0\rangle\langle 0|)U^\dagger\big](|a_m\rangle\otimes I\otimes I\otimes I)\Big](I\otimes I\otimes I\otimes |d_q\rangle)\\ &=\sum_{m,q}(\langle a_m|\otimes I\otimes I\otimes \langle d_q|) \big[U(\rho\otimes|0\rangle\langle 0|)U^\dagger\big](|a_m\rangle\otimes I\otimes I\otimes |d_q\rangle) \end{align}
Expanding $\rho\otimes|0\rangle\langle 0|$ into the products
\begin{align} \rho\otimes|0\rangle\langle 0|&=\rho_{AB}\otimes|0_{CD}\rangle\langle 0_{CD}|=\color{blue}{(\rho_{AB}\otimes I_{CD})(I_{AB}\otimes|0_{CD}\rangle)}(I_{AB}\otimes\langle 0_{CD}|)\\ =&\color{blue}{(\rho_{AB}I_{AB}\otimes I_{CD}|0_{CD}\rangle)}(I_{AB}\otimes\langle 0_{CD}|)=\color{blue}{(I_{AB}\rho_{AB}\otimes |0_{CD}\rangle.1)}(I_{AB}\otimes\langle 0_{CD}|)\\ =&\color{blue}{(I_{AB}\otimes |0_{CD}\rangle)(\rho_{AB}\otimes 1)}(I_{AB}\otimes\langle 0_{CD}|)=\color{blue}{(I_{AB}\otimes |0_{CD}\rangle)\rho_{AB}}(I_{AB}\otimes\langle 0_{CD}|)\\ =&\color{blue}{(I_{AB}\otimes |0_{CD}\rangle)\rho}(I_{AB}\otimes\langle 0_{CD}|) \end{align} Substituting back into the expression, \begin{align} \mathcal{E}(&\rho)=\\\\ =&\sum_{m,q}(\langle a_m|\otimes I\otimes I\otimes \langle d_q|) U(I_{AB}\otimes |0_{CD}\rangle)\rho(I_{AB}\otimes\langle 0_{CD}|)U^\dagger(|a_m\rangle\otimes I\otimes I\otimes |d_q\rangle)\\\\ =&\sum_{m,q} E_{m,q}\rho E_{m,q}^\dagger \end{align} where $E_{m,q}=\langle a_m|\otimes I\otimes I\otimes \langle d_q|) U(I_{AB}\otimes |0_{CD}\rangle)$
\begin{align} &\color{red} {\text{How do I show that this is a linear operator from the state space of $AB$ to the state}}\\ &\color{red} {\text{ space of $BC$ ?}} \end{align} My Attempt
If $|a_m\rangle,|b_n\rangle,|c_p\rangle,|d_q\rangle$ be orthonormal basis for the state space of the systems $\mathbb{C}^{i},\mathbb{C}^{j},\mathbb{C}^{k},\mathbb{C}^{l}$ respectively, then $|a_m\rangle\otimes|b_n\rangle\otimes|c_p\rangle\otimes|d_q\rangle$ is an orthonormal basis for the combined system $\mathbb{C}^{i}\otimes\mathbb{C}^{j}\otimes\mathbb{C}^{k}\otimes\mathbb{C}^{l}=\mathbb{C}^{ijkl}$. \begin{align} &\color{gray}{\text{Lemma: If $V$ and $W$ are vector spaces with basis $|v_i\rangle$ and $|w_j\rangle$ respectively. Then }}\\ &\color{gray}{|v_i\rangle\otimes_{outer}|w_j\rangle=|v_i\rangle\langle w_j| \text{form a basis for the vector space $V\otimes_{outer}W$.}} \end{align} Therefore, $(|a_m\rangle\otimes|b_n\rangle\otimes|c_p\rangle\otimes|d_q\rangle)(\langle a_{m'}|\otimes\langle b_{n'}|\otimes\langle c_{p'}|\otimes\langle d_{q'}|)=|a_m\rangle\langle a_{m'}|\otimes|b_n\rangle\langle b_{n'}|\otimes|c_p\rangle\langle c_{p'}|\otimes|d_q\rangle\langle d_{q'}|$ for a basis for the space $\mathbb{C}^{ijkl}\otimes_{outer}\mathbb{C}^{ijkl}$.
$\therefore U\in\mathbb{C}^{ijkl}\otimes_{outer}\mathbb{C}^{ijkl}$ can be written as a linear combination of the basis $(|a_m\rangle\otimes|b_n\rangle\otimes|c_p\rangle\otimes|d_q\rangle)(\langle a_{m'}|\otimes\langle b_{n'}|\otimes\langle c_{p'}|\otimes\langle d_{q'}|)=|a_m\rangle\langle a_{m'}|\otimes|b_n\rangle\langle b_{n'}|\otimes|c_p\rangle\langle c_{p'}|\otimes|d_q\rangle\langle d_{q'}|$.
i.e., $$ U=\sum_{m',n',p',q'}\nu_{m',n',p',q'} (|a_{m'}\rangle\otimes|b_{n'}\rangle\otimes|c_{p'}\rangle\otimes|d_{q'}\rangle)(\langle a_{m''}|\otimes\langle b_{n''}|\otimes\langle c_{p''}|\otimes\langle d_{q''}|)\\ =\sum_{m',n',p',q'}\nu_{m',n',p',q'} |a_{m'}\rangle\langle a_{m''}|\otimes|b_{n'}\rangle\langle b_{n''}|\otimes|c_{p'}\rangle\langle c_{p''}|\otimes|d_{q'}\rangle\langle d_{q''}| $$
And $|0_{CD}\rangle$ being a standard state of the system $CD$ can be represented in terms of the orthonormal basis vectors as, $$ |0_{CD}\rangle=\sum_{p,q}\eta_{p,q} |c_p\rangle\otimes|d_q\rangle $$
Substituting back into the expression for $E_{m,q}$ obtains \begin{align} &E_{m,q}=\\ &=\langle a_m|\otimes I\otimes I\otimes \langle d_q|) U(I_{AB}\otimes |0_{CD}\rangle)\\ &=\langle a_m|\otimes I_B\otimes I_C\otimes \langle d_q|) U\sum_{p''',q'''}\eta_{p''',q'''}(I_{A}\otimes I_{B}\otimes|c_{p'''}\rangle\otimes|d_{q'''}\rangle)\\ &=\langle a_m|\otimes I_B\otimes I_C\otimes \langle d_q|)\Big[ \sum_{m',n',p',q'\\m'',n'',p'',q''}\nu_{m',n',p',q'\\m'',n'',p'',q''} |a_{m'}\rangle\langle a_{m''}|\otimes|b_{n'}\rangle\langle b_{n''}|\otimes|c_{p'}\rangle\langle c_{p''}|\otimes|d_{q'}\rangle\langle d_{q''}|\Big]\\ &\times\sum_{p''',q'''}\eta_{p''',q'''}(I_{A}\otimes I_{B}\otimes|c_{p'''}\rangle\otimes|d_{q'''}\rangle)\\ &=\langle a_m|\otimes I_B\otimes I_C\otimes \langle d_q|)\Big[ \sum_{m',n',p',q'\\m'',n'',p'',q''}\sum_{p''',q'''}\eta_{p''',q'''}\nu_{m',n',p',q'\\m'',n'',p'',q''} |a_{m'}\rangle\langle a_{m''}|\otimes|b_{n'}\rangle\langle b_{n''}|\otimes|c_{p'}\rangle\langle c_{p''}|\otimes|d_{q'}\rangle\langle d_{q''}|\Big]\\ &\times(I_{A}\otimes I_{B}\otimes|c_{p'''}\rangle\otimes|d_{q'''}\rangle)\\ \\ &=\sum_{m',n',p',q'\\m'',n'',p'',q''}\sum_{p''',q'''}\eta_{p''',q'''}\nu_{m',n',p',q'\\m'',n'',p'',q''} \langle a_{m}|a_{m'}\rangle\langle a_{m''}|\otimes|b_{n'}\rangle\langle b_{n''}|\otimes|c_{p'}\rangle\langle c_{p''}|c_{p'''}\rangle\otimes\langle d_{q}|d_{q'}\rangle\langle d_{q''}|d_{q'''}\rangle\\ \\ &=\sum_{m',n',p',q'\\m'',n'',p'',q''}\sum_{p''',q'''}\eta_{p''',q'''}\nu_{m',n',p',q'\\m'',n'',p'',q''} \delta_{m,m'}\langle a_{m''}|\otimes|b_{n'}\rangle\langle b_{n''}|\otimes|c_{p'}\rangle\delta_{p'',p'''}\otimes\delta_{q,q'}\delta_{q'',q'''}\\ \\ &=\sum_{n',p',q,m'',n'',p'',q''}\eta_{p'',q''}\nu_{m,n',p',q\\m'',n'',p''} \langle a_{m''}|\otimes|b_{n'}\rangle\langle b_{n''}|\otimes|c_{p'}\rangle\otimes 1\\ &=\sum_{n',p',q,m'',n'',p'',q''}\eta_{p'',q''}\nu_{m,n',p',q\\m'',n'',p''} (1.\langle a_{m''}|\otimes|b_{n'}\rangle\langle b_{n''}|\otimes|c_{p'}\rangle.1\otimes 1.1)\\ &=\sum_{n',p',q,m'',n'',p'',q''}\eta_{p'',q''}\nu_{m,n',p',q\\m'',n'',p''} (1\otimes|b_{n'}\rangle\otimes|c_{p'}\rangle\otimes 1)(\langle a_{m''}|\otimes\langle b_{n''}|\otimes 1\otimes 1)\\\end{align}
$\implies E_{m,q}$ is a sum of operators from $AB$ to $BC$.
