Does nuclearity pass to un-tensoring? Let $A$ be a C*-algebra such that $A \otimes_{\min} A$ is nuclear.
Does it follow that $A$ is nuclear?
 A: I have in mind the following argument, which I am a little suspicious of since it seems dangerously simple, but here goes: since $A\otimes_{min}A$ is nuclear, there is a net of cp maps $\pi_\lambda:A\otimes_{min}A\to A\otimes_{min}A$ factoring through matrix algebras $M_{n_\lambda}(\mathbb C)$ and converging point-norm to the identity map. We can view $A$ as a subalgebra of $A\otimes_{min}A$ by inclusion in the first factor: $\iota:A\to A\otimes_{min}A$ by $\iota(a) = a\otimes 1$.  On the other hand, if we fix a state $\rho$ on $A$ then we get a ucp map $\text{id}_A\otimes \rho:A\otimes_{min}A\to A$ defined by $a\otimes b\to \rho(b)a$.  If we then form the composition
$$
\psi_\lambda = (\text{id}_{A}\otimes \rho) \circ \pi_\lambda \circ \iota
$$
we get a net of factorable cp maps $\psi_\lambda:A\to A$ converging point-norm to $\text{id}_A$.
In fact it seems to me this argument works whenever $A\otimes_{min}B$ is nuclear, and would work for any nuclear tensor product $A\otimes B$ as long as the natural maps $a\to a\otimes 1_B$ and $a\otimes b\to \rho(b)a$ are completely positive.
