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Proposition 3.6.6. Let $A\subset B$ (C*-algebras) be an inclusion. Then the following are equivalent:

(1). there exists a c.c.p.(contractive completely positive) map $\phi: B\rightarrow A^{**}$ such that $\phi(a)=a$ for all $a\in A$;

(2). for every *-homomorphism $\pi: A\rightarrow B(H)$ there exists a c.c.p. map $\phi: B\rightarrow \pi(A)''$ such that $\phi(a)=\pi(a)$ for all $a\in A$.

It is easy to see $(1)\Rightarrow (2)$, but in order to show $(2)\Rightarrow(1)$, the author use "every representation of $A$ extends to a normal representation of $A^{**}$". However,

(1). Why is "every representation of $A$ extends to a normal representation"? (2). And I do not know how to use it to prove $(2)\Rightarrow (1)$.

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You left out two important words in your quote. Every representation of $A$ extends to a normal representation . . . of $A^{**}$. This is because $A^{**}$ can be identified with the universal enveloping von Neumann algebra of $A$. For further details, see section III.2 of Takesaki's Theory of Operator Algebras, volume I. You can also find the desired result in the paragraph following Theorem 1.4.1 in Brown and Ozawa's book (which I think you already possess).

To prove that $(2) \implies (1)$, take the representation $\pi$ to be the universal representation of $A$. Then $\pi(A)''=A^{**}$. See the paragraph immediately preceding Theorem 1.4.1 in Brown & Ozawa for further details.

Adding extra information requested in the comments:

Further information about the universal enveloping von Neumann algebra can be found in

Definition 2.3 and Theorem 2.4, Takesaki, Volume I, Section III.2, page 122 in my copy.

Here, I use the notation of Definition 2.3, where $M(\pi)=\pi(A)''$ is the von Neumann algebra generated by a representation $\pi$ of the $C^*$-algebra $A$.

Let $\pi$ be the universal representation of $A$. Then $M(\pi)=\pi(A)''=A^{**}$ is the universal enveloping von Neumann algebra of $A$. By definition, given any representation $(\rho,K)$ of $A$, there exists a $\sigma$-weakly continuous representation $\tilde \rho$ of $M(\pi)=A^{**}$ that extends $\rho$. This map $\tilde \rho: A^{**} \simeq \pi(A)''=M(\pi) \to M(\rho) \subseteq B(K)$ is the normal representation of $A^{**}$ desired.

The second half of the proof of Theorem 2.4 proves the existence of this representation, using Lemma 2.2. (In my copy, there is a typo that mistakenly directs you to Lemma 1.2.)

Why should any of these results hold? Well, it is because $A^* = (A^{**})_*$; the dual is the predual of the second dual. For example, a linear functional on $A$ is an element of $A^*$, which means it is an element of $(A^{**})_*$, which means it is a normal (=$\sigma$-weakly continuous) linear functional on $A^{**}$. This simple fact tightly links the weak topology on $A$ with the weak$^*$ (=$\sigma$-weak) topology on $A^{**}$.

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  • $\begingroup$ But, where to find the answer of the first question in Takesaki's book? $\endgroup$ – Yan kai Jun 2 '14 at 17:28
  • $\begingroup$ Definition 2.3 and Theorem 2.4, Takesaki, Volume I, Section III.2, page 122 in my copy. $\endgroup$ – Tom Cooney Jun 2 '14 at 17:53
  • $\begingroup$ Yes, the Definition 2.3 and Theorem 2.4 in Takesaki's book only say that $\pi(A)''=A^{**}$ (homeomorphism), $\pi$ is the universal representation. However, why a representation $\pi_{1}: A\rightarrow B(H)$ can be extend to a normal representation of $A^{**}$ (or $\pi(A)''$)? $\endgroup$ – Yan kai Jun 3 '14 at 7:38
  • $\begingroup$ Thanks for your detailed answer and your patience. $\endgroup$ – Yan kai Jun 4 '14 at 2:16

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