= Solution
The channel acts only on the input system $Q$; the reference $R$ is untouched. Write
$$
\omega_{RB}=(\operatorname{id}_R\otimes\Phi)(|\Psi^\rho\rangle\langle\Psi^\rho|_{RQ}).
$$
The <coherent information> is
$$
\boxed{I_c(\Phi,\rho)=S(B)_\omega-S(RB)_\omega=-S(R\mid B)_\omega}.
$$
The reference marginal remains $\rho_R$, whose nonzero eigenvalues agree with those of $\rho$. Unlike a classical information quantity, coherent information can be negative.
\Image[/past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2010/iii/paper-48-channel.png]
{title=Quantum channel acting on one half of a purification, with untouched reference and discarded environment}
The lower diagram uses a <Stinespring dilation>: append an environment in a fixed <pure state>, apply a joint <unitary operator>, and discard the environment by <partial trace>. Before discarding it, the global $RBE$ state is pure.
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