Degradability of unitary channels
A quantum channel $\Phi$ is said to be degradable if there is a (generally noisy) map $\Lambda$ that connects $\Phi$ and its complementary map $\Phi_E$ as $$\Phi_E=\Lambda\circ \Phi. $$ The most basic example I can think of for this concept is the unitary mapping $\Phi_{U_S}(\rho_S)=U_S\rho_S U_S^\dagger$. Since the time evolution operator simplifies as $U_{SE}=U_S\otimes 1_E$, in Stinespring representation the action of the complementary map $\Phi_E$ is trivial and gives $\Phi_E(\rho_S)=|0\rangle_E\langle 0|$ for any input $\rho_S$. All the transmitted information is thus found in $\Phi(\rho_S)=\rho_S'$.
My question is actually more of a notational doubt. 680
I have seen the map $\Lambda$ for this case written simply as $\Lambda=|0\rangle_E\langle 0|$. Is this just a shorthand for the more precise $$\Lambda=0_S\otimes |0\rangle_E\langle0|, $$ where (the non-linear operator) $0_S$ acts as $0_S \rho_S=1_S$ for every state $\rho_S$? We would then obtain $$\Lambda(\Phi(\rho_S))=(0_S\otimes |0\rangle_E\langle 0|)(\rho_S'\otimes 1_E)=1_S\otimes |0\rangle_E\langle 0| $$ and then we can freely ignore the identity $1_S$ to write $\Phi_E(\rho_S)=|0\rangle_E\langle 0|$ as a map $\mathscr S(\mathcal H_S)\to \mathscr S(\mathcal H_E)$.
Am I overcomplicating things? In general, we omit to write the trivial factors of a separable state living in the tensor product of multiple spaces (i.e. $1_S\otimes |0\rangle_E\langle 0|\sim |0\rangle_E\langle 0|$ as above) but we should restore them to avoid mistakes such as writing $$\Lambda(\Phi(\rho_S))=|0\rangle_E\langle 0|\psi'\rangle_S\langle \psi'|$$ which clearly doesn't make sense, as we are trying to multiply things living in different spaces. But my way of doing things also implies thinking (at least temporarily) about the image of the map $\Phi(\rho_S)$ as something that lives in $\mathcal H_S\otimes \mathcal H_E$ rather than simply $\mathcal H_S$ (even though the second factor is trivial and can be forgotten) which also strikes me as imprecise.
So it appears I've overthought myself into a corner. How do I get out? 680