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2

Define $\phi:(B+I)/I\to B/B\cap I$ by $$\phi(b+j+I)=b+B\cap I,\ \ \ \ \ b\in B,\ j\in I.$$ Of course we need to check that this is well-defined. If $b_1+j_1=b_2+j_2$, then $$b_1-b_2=j_2-j_1\in B\cap I,$$ so $b_1+B\cap I=b_2+B\cap I$. The map is obviously linear, multiplicative, $*$-preserving, and onto. As for injectivity, if $b_1+B\cap I=b_2+B\cap I$, ...

3

You can write the similarity as $NS=SM$. As $N$ and $M$ are normal, the Fuglede-Putnam theorem guarantees that $N^*S=SM^*$. Taking adjoints, $S^*N=MS^*$. Then $$S^*SM=S^*NS=MS^*S.$$ Using this identity repeteadly, $p (S^*S)M=Mp (S^*S )$ for all polynomials; taking limits, $f (S^*S)M=Mf (S^*S)$ for all continuous functions $f$. In particular, if ...

2

I don't know why you say that $f(\sigma(x))=F_1$. A point in $\sigma(x)$ is either in $F_1$ or in $F_2$, and so $f(x)$ is either $0$ or $1$; and then $f(\sigma(x))=\{0,1\}$.

1

This is a corollary of the fact that images of $\ast$-homomorphism of $C^\ast$-algebras are closed. More precisely, denote by $\pi$ the canonical projection $A\to A/I$. Then $\pi(B)\subset A/I$ is closed and so is $B+I=\pi^{-1}(\pi(B))\subset A$ as the preimage of a closed set under a continuous map.

2

Let's define for $a\in A$ the left multiplication operator and the right multiplication operator $$L_a : A \rightarrow A, L_a(x)=ax \quad \text{and} \quad R_a: A \rightarrow A, R_a(x)=xa.$$ As multiplication is assumed to be continuous we get that $L_a$ and $R_a$ are both continuous for all $a\in A$. Let $B=B_1(0,A)$ denote the unit ball in $A$. Then ...

3

For an example of a non-proper $*$-morphism that isn't trivial, consider $A=B=\mathbb C\oplus \mathbb C$, and $\phi(a,b)=(a,0)$. For the existence of $\varphi_*$, the key property is that for any $*$-morphism $\varphi:C_0(X)\to C_0(Y)$, there exists $\varphi_*:Y\to X$, continuous, such that \tag{1}\varphi(f)(y)=f(\varphi_*(y))\ \ \ \ \text{ for all }y\in ...

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