What is the Stinespring representation of the adjoint of a channel?

Question

For any completely positive trace nonincreasing map $N_{A\rightarrow B}$, the adjoint map is the unique completely positive linear map $N^\dagger_{B\rightarrow A}$ that satisfies

$$\langle N^\dagger(\sigma), \rho\rangle = \langle \sigma, N(\rho)\rangle$$

for all linear operators $\sigma \in \mathcal{L}(\mathcal{H}_B)$ and $\rho \in \mathcal{L}(\mathcal{H}_A)$.

Let $V_{A\rightarrow BE}$ be any isometry such that $\text{Tr}_E(V\rho V^\dagger) = N(\rho)$. This is the Stinespring representation of any completely positive map. Since $N^\dagger$ is also a completely positive map, it also has a Stinespring representation.

Question: Given $V$, can one write down the Stinespring representation of $N^\dagger$? Naively taking the transpose conjugate of $V$ to write down something like

$$\text{Tr}_E(V^\dagger\sigma V) = N^\dagger(\sigma)$$

doesn't even make sense since $V^\dagger_{BE\rightarrow A}$ whereas the isometry we are after should go from $B$ to $AE'$.

Answer 1

An adjoint to partial trace is just tensoring by $I$, i.e. $\text{Tr}_2^\dagger(\sigma) = \sigma \otimes I$.

So in this case we can write $$ N^\dagger(\sigma) = V^\dagger (\sigma \otimes I_E)V. $$

If $N^\dagger$ is trace-preserving then it can be written as $\text{Tr}_{E'}(U\sigma U^\dagger)$ for some isometry $U:B \rightarrow AE'$. But in general $N^\dagger$ is not trace-preserving, but unital. So such form $\text{Tr}_{E'}(U\sigma U^\dagger)$ may not be possible.