A simple question about Lance's weak expectation property.

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Here is a quotation of a book:

Definition 3.6.7. A C*-algebra $A\subset B(H)$ is said to have Lance's weak expectation property (WEP) if there exists a u.c.p map $\Phi: B(H)\rightarrow A^{**}$ such that $\Phi(a)=a$ for all $a\in A$.

A simple application of Arveson's Extension Theorem shows that the WEP is independent of the choice of faithful representation.

However, how to use Arveson's Extension Theorem to show the independence? I suppose it may be a very simple question, but could someone give me some hints?

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Suppose that we are given two faithful representations $\pi:A \to B(H)$ and $\rho: A \to B(K)$. Suppose that we also know that $A$ has the weak expectation property (with respect to $B(H)$). So we know that there exists a unital completely positive map $\Phi: B(H) \to A^{**}$ such that $\Phi(a) =a$ for all $a \in A$. We seek to show that there exists a corresponding map $\tilde \Phi:B(K) \to A^{**}$.

The map $\pi :A \to B(H)$ extends to a map $\tilde \pi :B(K) \to B(H)$ by Arveson's extension theorem. The desired map is then given by $$\tilde \Phi= \Phi \circ \tilde \pi: B(K) \to A^{**}.$$