= Solution
First prove monotonicity for a <partial trace>. Let $r=\dim\mathcal H_B$ and define the <Heisenberg-Weyl twirling channel>
$$
\mathcal T_B(X)=\frac1{r^2}\sum_{k,m}(I_A\otimes W_{k,m})X(I_A\otimes W_{k,m})^\dagger
=(\operatorname{Tr}_BX)\otimes\frac{I_B}{r}.
$$
The supplied identity on individual operators extends to bipartite operators by expanding them in a <tensor-product basis>. The <joint convexity of quantum relative entropy> and its unitary invariance imply
$$
D\left(\rho_A\otimes\frac{I_B}r\middle\|\sigma_A\otimes\frac{I_B}r\right)
\leq\frac1{r^2}\sum_{k,m}D(U_{k,m}\rho_{AB}U_{k,m}^\dagger\|U_{k,m}\sigma_{AB}U_{k,m}^\dagger)
=D(\rho_{AB}\|\sigma_{AB}).
$$
The <additivity of quantum relative entropy> makes the left side $D(\rho_A\|\sigma_A)$, proving the partial-trace case.
For a general deterministic quantum operation, meaning a <quantum channel>, use a <Stinespring dilation> $\Lambda(X)=\operatorname{Tr}_E(VXV^\dagger)$ with $V^\dagger V=I$. An <isometric embedding> preserves <quantum relative entropy>, because restricting the output to the common image of $V$ preserves the eigenvalues and trace formula. Applying the partial-trace result gives
$$
\boxed{D(\Lambda(\rho)\|\Lambda(\sigma))
\leq D(V\rho V^\dagger\|V\sigma V^\dagger)=D(\rho\|\sigma)}.
$$
The other properties used are unitary invariance and invariance under adjoining an identical ancillary state; both follow directly from the relative-entropy trace formula. This is the <Lindblad-Uhlmann monotonicity theorem>. The proof applies to deterministic channels; normalized postselection is not such a linear channel.
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