Are products of quantum channels (CPTPMs) also quantum channels? A quantum channel $\mathcal E$ is a completely positive trace preserving map.
Is $\mathcal E \otimes \mathcal F$ a quantum channel if $\mathcal E,\mathcal F$ are quantum channels?
What I have:


*

*It is sufficient to answer this for $\mathcal E \otimes \mathit{id}$ since $\mathcal E \otimes \mathcal F = (\mathcal E \otimes \mathit{id}) \circ (\mathit{id} \otimes \mathcal F)$.

*Complete positivity of $\mathcal E \otimes \mathit{id}$ follows by definton from complete positivity of $\mathcal E$.

*The fact that $\mathcal E\otimes\mathit{id}$ is trace preserving is equivalent to showing that the diamond norm (aka completely bounded trace norm) of $\mathcal E$ is 1. This is stated in Watrous, Prop. 3.44.2. However, the proof of that property already assumes (as far as I could tell), that $\mathcal E \otimes \mathit{id}$ is a channel.


Bonus questions:
It would be great if the answer also covers completely positive trace reducing maps as well as infinite Hilbert spaces.
 A: Your first two points are correct, and the fact that $\mathcal E\otimes \mathrm{id}$ is trace preserving is simply a consequence of the fact that, for any state $\rho_{AB}$, 
$$\mathrm{Tr}_B(\mathcal{E}_A\otimes \mathrm{id}_B(\rho_{AB}))=\mathcal{E}(\mathrm{Tr}_B(\rho_{AB})) $$
which is very easy to show either by writing down the partial trace explicitly or by realizing that $\mathrm{Tr}_B$ commutes with $\mathcal E_A$ as they act on different spaces. This works for both trace preservingness or trace non increasing-ness or any trace property you might think of.
Another way to show this is that if a channel is trace preserving on a basis of matrices then it is trace preserving on any state. You can write any $\rho_{AB}$ as
$$ \rho_{AB}=\sum_{ij}\lambda_{ij}\sigma_A^i\otimes \tau_B^j$$
for some matrices $\sigma_A^i$, $\tau_B^j$, hence
$$ \mathrm{Tr}(\mathcal{E}\otimes \mathcal F(\rho_{AB}))=\sum_{ij}\lambda_{ij}\mathrm{Tr}(\mathcal{E}(\sigma^i))\mathrm{Tr}(\mathcal{F(\tau^j)})$$
which clearly implies everything you're looking for
A: If $\mathcal E$ is a channel, then it can be expressed in terms of Kraus operators,
$$
\mathcal E(\rho) = \sum K_i\rho K_i^\dagger\ .
$$
Then, the map $\mathcal E\otimes I$ is of the form
$$
(\mathcal E\otimes I)(\rho) = \sum (K_i\otimes I)\rho(K_i\otimes I)^\dagger\ .
$$
It is thus also a channel, with Kraus operators $K_i\otimes I$. In particular, if $\mathcal E$ is trace-preserving, it is immediate to see that $\mathcal E\otimes I$ is trace-preserving.
