An exercise on nuclear maps in C*-algebra Definition 2.1.1 Let $A, B$ be the C*-algebra, a map $\theta: A\rightarrow B$ is called nuclear if there exist contractive completely positive maps $\phi_{n}: A\rightarrow M_{k(n)}(\mathbb{C})$ and $\psi_{n}:M_{k(n)}(\mathbb{C}) \rightarrow B$ such that 
$$||\psi_{n}\circ\phi_{n}(a)-\theta(a)||\rightarrow0$$
for all $a\in A$.
Then, there is an exercise below( I do not know how to do the exercise):
Exercise 2.1.1 Show that $\theta:A \rightarrow B$ is nuclear if and only if for each finite set $F\subset A$ and $\varepsilon>0$ there exsit $n\in \mathbb{N}$ and contractive completely positive maps $\phi: A\rightarrow M_{n}(\mathbb{C})$, $\psi:M_{n}(\mathbb{C})\rightarrow B$ such that $||\theta(a)-\psi\circ\phi(a)||<\varepsilon$ for all $a\in F$.
I suppose the "if" is clear from the definition. But how to prove the "only if"? I guess we need to use the finite set to construct a sequence of contractive completely positive map (just as in the definition.)  
 A: Consider the net $$\mathcal F=\{(F,t):\ F\subset A\,\text{ finite},\ t>0\},$$
ordered by $(F_1,t_1)\leq (F_2,t_2)$ if $F_1\subset F_2$, $t_1\geq t_2$. For each $\alpha=(F,t)\in\mathcal F$, there exist $n_\alpha\in\mathbb N$ and $\phi_\alpha:A\to M_{n_\alpha}(\mathbb C)$, $\psi_\alpha:M_{n_\alpha}(\mathbb C)\to A$ such that 
$$\|\psi_\alpha\circ\phi_\alpha(a)-\theta(a)\|<t,\ \ a\in F.$$
This net satisfies the requirement, then. For if $\varepsilon>0$ and $a\in A$ are given, we can choose $(F,\varepsilon)\in\mathcal F$ with $a\in F$, and then $\|\psi_\alpha\circ\phi_\alpha(a)-\theta(a)\|<\varepsilon$. And if $\beta\geq\alpha$, then $\beta=(F',s)$ with $F\subset F'$, $s\leq \varepsilon$, so 
$$
\|\psi_\beta\circ\phi_\beta(a)-\theta(a)\|<s<\varepsilon.
$$ 
We have shown that for any $\varepsilon>0$, there exists $\alpha_0\in\mathcal F$ such that, for any $\alpha\geq\alpha_0$, $\|\psi_\alpha\circ\phi_\alpha(a)-\theta(a)\|<\varepsilon$. That is,
$$
\lim_\alpha\|\psi_\alpha\circ\phi_\alpha(a)-\theta(a)\|=0.
$$
