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Is there any $*$ homomorphism $T$ from $A$ to $B$, wherein $B$ is a $*$ closed subalgebra of $C^*$ algebra $A$, containing the unit of $A$, such that $T(b)=b$ for all $b\in B$ and $\|T(a)\|=\|a\|$ for all $a\geq 0$ ?

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I'm not quite sure what you're asking: Are you asking whether this situation can occur at all or are you asking whether such a $T$ exists for every $\ast$-closed subalgebra $B$ of $A$ containing the identity of $A$? If the latter is intended: why would you think this is the case? (it's not true: there need not even be an idempotent continuous linear map with range $B$) – t.b. Jun 25 '12 at 9:48

Note that if $\|T(a)\|=\|a\|$ for all $a\geq 0$, then for all $a\in A$, $\|T(a)\|^2=\|T(a)T(a)^*\|=\|T(aa^*)\|=\|aa^*\|=\|a\|^2$, so $T$ is isometric. Suppose that $B\neq A$, and let $a$ be in $A\setminus B$. Then $a-T(a)\neq 0$, but $T(a-T(a))=0$, a contradiction. It follows that the only way this is possible is when $A=B$ and $T$ is the identity operator on $A$.

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