# What is the Stinespring dilation of $T\otimes I$ for some CPTP map $T$?

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 ($$U \otimes \mathbb{I}) (\mathbb{I} \otimes V)$$.

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 $$\rho_{AC}$$.