Do channels which do not affect a subsystem always have the form $\mathcal E=\mathcal T_{\mathcal A}\otimes\text{id}_B$? consider a bipartite state on two Hilbert spaces $\rho_{AB}$. Usually, I think of a channel $\mathcal E$ as acting only on $A$ if
$$\mathrm{Tr}_A(\mathcal E(\rho_{AB}))=\mathrm{Tr}_A(\rho_{AB}) \tag A$$
i.e., the action of the channel on a system does not influence the reduced state on the other system. My question is: is this equivalent to
$$\mathcal E=\mathcal{T_A}\otimes\mathrm{id}_B$$
where $\mathcal T_A:A\rightarrow A$?
That the second line imples the first is obvious, the other direction doesn't look obvious to me, there could be some action on $B$ that somehow gets canceled by taking the trace somehow. Is it true?
 A: The idea is to prove that (A) implies that the Kraus operators $A_a$ of $\mathcal E$ are local, that is, they have the form $A_a=\tilde A_a\otimes I$. For the purpose, I will leverage here the representation of the channel in terms of an enlarged unitary, and use it to prove the result on the Kraus operators.
Enlarged unitary representation
Every map CPTP $\mathcal E$ can be written as $\newcommand{\ketbra}[1]{|#1\rangle\!\langle #1|}\newcommand{\calU}{{\mathcal{U}}}\newcommand{\calE}{{\mathcal{E}}}\newcommand{\calH}{{\mathcal{H}}}\newcommand{\tr}{\operatorname{tr}}\calE(\rho)=\tr_E[\calU(\rho\otimes\ketbra{0_E}) \calU^\dagger]$ for some (unique) unitary $\calU$.
Explicitly, denoting with $K(\calE)$ the natural representation of $\calE$, the relation between $\calE$ and $\calU$ can be written as
$$K(\calE)_{ij}^{k\ell} = \sum_a \calU_{ia}^{k0} (\calU^*)_{ja}^{\ell0}.$$
Feel free to completely ignore the placement of up/down indices, I only do it to make the formula a bit easier to read.
If this looks similar to the Kraus decomposition of the channel, that's not a coincidence. The relation between the unitary $\calU$ and the Kraus operators $A_a$ is $(A_a)_i^k = \calU_{ia}^{k0}$. 
While the above notation is for maps acting on a single system, the formulae translate seamlessly to a bipartite scenario by "duplicating" each index: $i\to(i_1,i_2)$, $j\to (j_1,j_2)$ etc.
What (A) implies for $\mathcal U$
Note that if (A) holds for all states $\rho_{AB}$, it holds in particular for all pure product states $|\psi\rangle|\phi\rangle$. On these states, $\calU$ must act in such a way that tracing out the subsystem $A$ still gives a pure state on $B$ (because $\tr_A(\ketbra\psi\otimes\ketbra\phi)=\ketbra\phi$ is pure). We must thus have
$$
  \calU\, |\psi\rangle_A|\phi\rangle_B|0\rangle_E = |\Psi_{AE}\rangle\otimes|\phi\rangle_B,
$$
where $|\Psi_{AE}\rangle$ is some state in $\calH_A\otimes\calH_E$ (the joint space of $A$ and the auxiliary space $E$).
More explicitly, this expression reads
$$
  \calU_{ima}^{kn0} \psi_{k}\phi_n = \Psi_{ia} \phi_m.
$$
If we choose $|\psi\rangle,|\phi\rangle$ to be elements of the computational basis, we conclude that
$\calU_{ima}^{kn0} = \delta_{nm} \Psi^k_{ia}$, where $\sum_{ia}\Psi^k_{ia}|i,a\rangle$ is some pure state in $\calH_A\otimes\calH_E$ for each $k$. Now all that remains is to see why this implies the original statement  about the form of $\calE$.
Conclude that $\mathcal E$ is local
We proved that the Kraus operators of $\calE$ have the form
$$(A_a)_{im}^{kn} = \delta_{mn} \Psi^{k}_{ia} \Longleftrightarrow
A_a = \tilde A_a\otimes I$$
for some set of operators $\tilde A_a$ acting on $\calH_A$.
This is enough to prove the statement: a map is local if and only if its Kraus operators are. If we want to be more explicit, we can write the expressions componentwise as:
$$
\calE(\rho) =
\sum_a A_a \rho A_a^\dagger
\Longleftrightarrow (\calE(\rho))_{im,jn} = (A_a)_{im}^{k\ell} (A_a^*)_{jn}^{pq} \rho_{k\ell,pq}
= \Psi^k_{ia}(\Psi^*)^p_{ja}\rho_{km,pn}.
$$
We can then define the operator $\mathcal T$ with components
$$(\mathcal T)_{ij}^{kp} = \sum_a \Psi^k_{ia}(\Psi^*)^p_{ja},$$
and observe that $\calE = \mathcal T\otimes I$.
TL;DR: Consider the action of the map on pure product states, and observe that (A) implies that its Kraus operators are local.
A: An simple argument goes via the Choi-Jamiolkowski isomorphism:
The Choi state is obtained by applying the channel to the left half of a maximally entangled state,
$$
\vert\Omega\rangle = \sum_{ij} \vert i,j\rangle_{AB}\vert i',j'\rangle_{A'B'}\ .
$$
The Choi state is then
$$
\sigma_{\mathcal E}= (\mathcal E_{AB}\otimes I_{A'B'})\,(\vert\Omega\rangle\langle\Omega\vert)\ .
$$
Now consider the channel you describe above, namely $\mathrm{tr}_A\circ \mathcal E$.  Its Choi state is $\chi=\mathrm{tr}_A\,\sigma$.  On the other hand, your equation (A) tells us that
$$
\mathrm{tr}_A\,\sigma = \chi = \rho_{A'}\otimes \vert\omega\rangle\langle\omega\vert_{BB'}\ ,
$$
where $\vert\omega\rangle = \sum_i \vert i\rangle_B\vert i\rangle_{B'}$ (since (A) says that $\mathrm{tr}_A\circ \mathcal E$ acts as the identity channel on $B$. This implies that $\sigma$ must be of the form
$$
\sigma = \rho_{AA'}\otimes \vert\omega\rangle\langle\omega\vert_{BB'}\ ,
$$
which is the Choi state of a channel of the form
$$
\mathcal T_A\otimes I_B\ .
$$
