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Let $T: \mathcal{H}_A \rightarrow \mathcal{H}_B$ be a CPTP map with Stinespring extension $U: \mathcal{H}_{A} \rightarrow \mathcal{H}_{B} \otimes \mathcal{H}_E$. That is $U$ is an isometry such that for all states $\rho_A$, we have $Tr_{E} \left( U \rho_A U^\dagger\right)= T(\rho_A)$.

I am interested in Stinespring dilation of $T \otimes \mathbb{I}_C$ where $C$ is some additional register. My guess is $U \otimes \mathbb{I}_C$ should work but I am unable to prove.

That is, consider a state $\rho_{AC}$ and $\sigma_{BC}= T_A \otimes \mathbb{I}_C (\rho_{AC})$. Will the following equality hold:

$Tr_{E} \left( (U\otimes \mathbb{I}_C) \rho_{AC} (U\otimes \mathbb{I}_C)^\dagger\right)= \sigma_{BC}$.

Ultimately, I want to extend this to $U \otimes V$ but I suppose I can break it up to ($U \otimes \mathbb{I}) (\mathbb{I} \otimes V)$.

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Yes. Note that in general ${\rm Tr}_1(\rho_{12} \otimes \rho_3) = {\rm Tr}_1(\rho_{12}) \otimes \rho_3$.

It's easy to verify your equality for $\rho_{AC} = \rho_A \otimes \rho_C$ :

$$ {\rm Tr}_{E} \left( (U\otimes \mathbb{I}_C) \rho_{AC} (U\otimes \mathbb{I}_C)^\dagger\right) = {\rm Tr}_{E} \left( U \rho_{A} U^\dagger \otimes \rho_C \right) = {\rm Tr}_{E} \left( U \rho_{A} U^\dagger \right) \otimes \rho_C = (T \otimes \mathbb{I}_C)(\rho_{AC}). $$

But since both parts are linear over $\rho_{AC}$ (it's common in QM) the equality is true for any $\rho_{AC}$.

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