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Let $A \to B$ be a proper morphism of $C^*$-algebras. A nondegenerate representation of $B$ induces a nondegenerate representation of $A$. Does the converse hold?

I.e.: let $A \to B$ be a morphism of $C^*$-algebras such that every nondegenerate representation of $B$ induces a nondegenerate representation of $A$. Does the morphism result proper? I guess not.

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What is a "proper" morphism of C*-algebras in your view? –  Paul Siegel Nov 21 '12 at 3:26
    
It is a morphism that maps an approximate unit to an approximate unit (equivalently: it maps every approximate unit to an approximate unit). –  Sergio A. Yuhjtman Nov 21 '12 at 18:16

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up vote 5 down vote accepted

This is true. Factoring by the kernel of the homomorphism, we may assume that $A$ is a C*-sub-algebra of $B$ and the homomorphism is just the inclusion. So assume that $A\subseteq B$ and

(a) Every non-degenerate representation of $B$ restricted to $A$ is non-degenerate.

Then $A$ cannot be contained in the kernel of a state $\lambda$ of $B$. Otherwise, the GNS construction would give us a representation $\pi_\lambda\colon B\to B(H_\lambda)$ that is non-degenerate but restricted to $A$ is degenerate (since there exists $h\in H$ such that $\lambda(a)=\langle\pi_\lambda(a)h,h\rangle$ for all $a\in B$). So, (a) implies

(b) $A$ is not contained in the kernel of a state of $B$.

This implies that $A$ must generate $B$ as a closed left ideal, by Theorem 3.10.7 of Pedersen's "C*-algebras and their automorphims". Now let $(a_i)$ be an approximate unit for $A$. Then for every element of the form $ab$, with $a\in A$ and $b\in B$, we have $a_iab\to ab$. But the linear span of these elements elements is dense in $B$. So, (b) implies

(c) Any approximate unit of $A$ is also an approximate unit of $B$.

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I've updated the answer to account for the general case. –  Leonel Robert Nov 29 '12 at 5:11
    
Thanks a lot! . –  Sergio A. Yuhjtman Nov 29 '12 at 18:06

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