In Breuer and Petruccione's book 'The theory of open quantum systems', section 2.4.3, 'representation theorem of quantum operations', a quantum operation $\phi_m$ corresponding to a measurement outcome m is defined as a map from one density matrix to another with following properties-

1. $0 \leq tr [\phi_m(\rho)] \leq 1$
2. Convex linearity: $\phi _m \left ( \Sigma a_i \rho _i\right )=\Sigma a_i \phi_m(\rho_i)$
3. The operator $(\phi_m \otimes I)$ is Completely postivite, where this new operator is defined as-

\begin{equation}(\phi_m \otimes I) \left ( \Sigma A_i \otimes B_i\right )=\Sigma \phi_m(A_i) \otimes B_i \tag{1}\label{1}\end{equation}

My question is regarding the definition of the operator $(\phi_m \otimes I)$ (on this auxiliary larger Hilbert space). For example, how will this operator act on a maximally entangled bipartite qubit state given by

$$|\psi\rangle = \frac {1}{\sqrt{2}}\left (|00\rangle + |11\rangle \right )$$

The desnity matrix is given as,

$$|\psi\rangle \langle \psi |= \frac {1}{2}\begin{pmatrix} 1 & 0 & 0 & 1\\ 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0\\ 1 & 0 & 0 & 1 \end{pmatrix}$$

Now, in order for us to use the definition \eqref{1}, we need a decomposition of $|\psi\rangle \langle \psi |$ given as

$$|\psi\rangle \langle \psi |= \Sigma A_i \otimes B_i$$

But I don't think it is possible to express $|\psi\rangle \langle \psi |$ in this form in this basis because the two qubits are entangled. That is, I think a maximally entangled qubit cannot be written as classical mixture of unentangled qubits.

Then how do we find $ (\phi_m \otimes I)(|\psi\rangle \langle \psi |)$?