A question on multiplicative linear functional on Banach algebra.

I am reading a book about C*-algebra. But i am confused with some of its content. It says

Assume $A$ is a non-unital C*-algebra and $\tilde{A}$ is its unitization (the elements of the form $a+\lambda$). If $\phi: A\rightarrow \mathbb{C}$, is a nonzero homomorphism($\mathbb{C}$ denotes complex field), then $\tilde{\phi}(a+\lambda)=\phi(a)+\lambda~(a\in A$ and $\lambda\in \mathbb{C})$. Let $P(A)$ and $P(\tilde{A})$ denote the nonzero multiplicative linear functional on A and $\tilde{A}$, respectively. Thus, $P(A)$ is a subset of $P(\tilde{A})$. In fact, $P(A)=P(\tilde{A})\cup\{\pi\}$, where $\pi$ is determined by $\tilde{A}/A=\mathbb{C}$, i.e., $\ker\pi=A$. Since $P(\tilde{A})$ is compact Hausdorff space, we conclude that $P(A)$ is a locally compact Hausdorff space.

My question are

1. Does the "=" in the equation $P(A)=P(\tilde{A})\cup\{\pi\}$ is under the meaning of isomorphism? Is it topological isomorphism? Why?

2. Why does the $P(\tilde{A})$ is compact Hausdorff space implies that $P(A)$ is a locally compact Hausdorff space?

First of all, you have to consider $P(A)$ as the set of all non-zero multiplicative linear functionals on $A$ with the weak$^*$-topology inherited from $A^*$, the dual of $A$. When $A$ is unital $P(A)$ is compact. So $P(\tilde{A})$ is always compact.
(1) It is $P(\tilde{A})=P(A)\cup \pi$. In fact, this equality reveals a more important fact which is $P(\tilde{A})$ is the one point compactification of $P(A)$.
(2) Yes. The reason is easy, because every compact subset of $P(\tilde{A})$ which does not contain $\{\pi\}$ (which is exactly the point at infinity) is compact in $P(A)$ too.
• Actually, I have explained this topic in my lecture notes on $C^*$-algebra, which is available at arxiv:1211.3404. More specifically, see Section 3.1. Of course, I denote $P(A)$ by $\Omega(A)$. – user56706 Oct 10 '13 at 17:35
• (2) is not quite as obvious as you say; not every subspace of a compact space is locally compact. (Consider $\mathbb{Q} \cap [0,1]$ as a subspace of $[0,1]$.) – Nate Eldredge Oct 10 '13 at 18:18
• @NateEldredge: Thanks Nate, you are right. We have to use the fact that $P(\tilde{A})$ is the one point compactification of $P(A)$. Then every compact neighborhood of a point $x\in P(A)$ in $P(\tilde{A})$ which does not meet $\{\pi\}$ is actually compact in $P(A)$ too. – user56706 Oct 10 '13 at 18:25