Definition 2.3.1. A C*-algebra $A$ is nuclear if the identity map $id_{A}: A\rightarrow A$ is nuclear.
Definition 2.3.2. A C*-algebra $A$ is exact if there exists a faithful representation $\pi:A \rightarrow B(H)$ such that $\pi$ is nuclear.
There is quotation below:
Let $\pi: A\rightarrow B(H)$ be a faithful representation, then ,
(1). $A$ is nuclear if and only if $\pi$ is nuclear when regarded as taking values in $\pi(A)$.
(2). While $A$ is exact if and only if $\pi$ is nuclear when regarded as taking values in $B(H)$.
My question is how to explain regarding as taking values in $\pi(A)$ and $B(H)$ above, do they make any difference in estimating nuclearness and exactness?
Proof. (1). "only if" Since $A$ is nuclear, then we have c.c.p. $\bar{\phi}_{n}: A \rightarrow M_{k(n)(C)}$ and $\bar{\psi}_{n}: M_{k(n)}(C)\rightarrow A$ such that $\bar{\psi}_{n} \circ \bar{\phi}_{n} \rightarrow I_{A}$ in point-norm topology. Then, we consider $\phi_{n}=\bar{\phi}_{n}$ and $\psi_{n}=\pi \circ \bar{\psi}_{n}$, we can check $\psi \circ \phi_{n} \rightarrow \pi$ in point-norm topology.
"if " Since $\pi$ is nuclear (and we take values in $\pi(A)$), we have c.c.p. $\bar{\phi}_{n}: A \rightarrow M_{k(n)(C)}$ and $\bar{\psi}_{n}: M_{k(n)}(C)\rightarrow \pi(A)$ such that $\bar{\psi}_{n} \circ \bar{\phi}_{n} \rightarrow \pi$ in point-norm topology. Then we consider $\phi_{n}=\bar{\phi}_{n}$ and $\psi_{n}=\pi^{-1} \circ \bar{\psi}_{n}$, we can check $\psi \circ \phi_{n} \rightarrow I_{A}$ in point-norm topology.
(Because of utilizing the invertibility of $\pi$, we need to consider the value in $\pi(A)$, right?)
(2). "if" is clear from the definition. But how to verify the "only if"? Could you give me some hints?
