In the section 11.1 of Rordam's book, there is a remark : If $A$ is a unital $C^*$-algebra, denote $SA$ by the suspension of $A$, $\tilde{SA} $ is the unitalization of $SA$. We can identify $M_n(\tilde{SA})$ with the set $E$ of functions $f$ in $C(\Bbb T,M_n (A))$ such that $f(1)$ belongs to $M_n(\Bbb C 1_A)$.
I want to construct the $*$-isomorphism between $M_n(\tilde{SA})$ and $E$.
My thought:
Define $\phi :M_n(\tilde{SA})\to E $ as following
$\phi((f_{ij}))(z)=(f_{ij}(z))$,where each $f_{ij}\in \tilde{SA},z\in \Bbb T$, but how to construct a $*$-isomorphism between $\tilde{SA}$ and $F=\{f\in C(\Bbb T,A):f(1)\in \Bbb C 1_A\}$ when $A$ is unital ?
I tried to define a map $\psi: \tilde{SA} \to F$ such that $\psi((f,\alpha))(z)=f(z)+\alpha 1_A$,but it is not an isomorphism.Can anyone give me some hints?Thanks！
To answer the question in the simple case of $n = 1$, we have $F = \{f \in C(\mathbb{T},A) \mid f(1) \in \mathbb{C} 1_A\}$. I think that your $\psi$ actually does work. $F$ is clearly a C*-algebra and you can check that $\psi$ really is a *-homomorphism, so lets see why its surjective and injective. Recall that we identify $SA = \{f \in C(\mathbb{T},A) \mid f(1) = 0\}$.
If $g \in F$, then $g(1) \in \mathbb{C}1_A$, so we can write $g(1) = \alpha1_A$ for some $\alpha \in \mathbb{C}$. Now define $f(z) = g(z) - \alpha1_A$, which is in $SA$ since $f(1) = 0$. Then $$ \psi(f + \alpha1_{\widetilde{CA}})(z) = f(z) + \alpha1_A = g(z) - \alpha 1_A + \alpha1_A = g(z). $$ So $\psi$ is surjective. To see injectivity, suppose that $\psi(f + \alpha1_{\widetilde{CA}})(z) = 0$ for all $z \in \mathbb{T}$. Thus $f(z) + \alpha1_A = 0$ for all $z \in \mathbb{T}$, so that $f(z) = -\alpha1_A$ for all $z \in \mathbb{T}$. But since $f(1) = 0$ ($f \in SA$), we have that $0 = f(1) = -\alpha1_A$, so that $\alpha = 0$. So both $\alpha$ and $f$ are zero. Thus $\psi$ is injective.
You can work out the details when moving up to the matrix algebras.