The finite-dimensional erasure map is a quantum channel

Let $E=A\oplus \mathbb{C}{e}_E$ with $e_E$ a unit vector orthogonal to the data summand, and write $J_B,J_E$ for the data inclusions. Define $V_{ϵ}x:=\sqrt{1-ϵ}{J}_{B}x\otimes e_E+\sqrt{ϵ}e\otimes J_{E}x$. The summands lie in orthogonal subspaces of $B\otimes E$, so for all $x,y\in A$, $⟨V_{ϵ}x,V_{ϵ}y⟩=(1-ϵ)⟨x,y⟩+ϵ⟨x,y⟩=⟨x,y⟩$. Thus $V_{ϵ}$ is a linear isometry, including at $ϵ=0$ and $ϵ=1$.

Expanding $V_{ϵ}X{V}_{ϵ}^{\ast }$ gives two diagonal terms and two cross terms. The first diagonal term is $(1-ϵ)J_{B}X{J}_B^{\ast }\otimes \mathrm{\mid }{e}_E⟩⟨e_E\mathrm{\mid }$, whose partial trace over $E$ is $(1-ϵ)J_{B}X{J}_B^{\ast }$. The second is $ϵ\mathrm{\mid }e⟩⟨e\mathrm{\mid }\otimes J_{E}X{J}_E^{\ast }$, whose partial trace is $ϵ\mathrm{tr}(X)\mathrm{\mid }e⟩⟨e\mathrm{\mid }$. Each cross term has a partial-trace coefficient equal to the inner product of the orthogonal erasure flag with the data subspace, hence vanishes. Therefore ${\mathrm{Tr}}_E(V_{ϵ}X{V}_{ϵ}^{\ast })={\mathcal{E}}_{d,ϵ}(X)$ for every $X$.

By step 1.1, $V_{ϵ}$ is an isometry; [F2] says its partial-trace map is completely positive and trace-preserving on the full operator space. Step 2.1 identifies that map with ${\mathcal{E}}_{d,ϵ}$, proving the Statement.