# Injective $*$-homomorphism is isometric

I am aware there are other proofs of line of this statement. But I am interested in the argument outlined here on page 62-63

Corollary II.2.2.9 Let $$A$$ and $$B$$ be $$C^*$$ algebras, $$\phi:A \rightarrow B$$ be injective $$*$$-homomoprhism. Then $$\phi$$ is isometric, i.e. $$||\phi(x) || = ||x||$$ for all $$x \in A$$.

The proof goes as follows: Wlog we may assume $$A,B$$ are commutative (I got this ), and it is obvious from II.2.2.4. (as below).

Theorem II.2.2.4 If $$A$$ is a commutative $$C^*$$ algebra, then the Gelfand trasform is an isometric $$*$$-isomorphism from $$A$$ onto $$C_0(\hat{A})$$.

How does II.2.2.4 imply 2.2.9?

That would be II.2.2.4, the Gelfand Transform, a few theorems back. If you are new to C$$^*$$-algebras, reading Blackadar (with almost no proofs and a general point of view) is probably not the best idea.
Answer to the edit: as Blackadar says, the C$$^*$$-identity lets you restrict the problem to $$A,B$$ abelian. So you have maps $$C_0(\hat A)\xrightarrow{\ \ \ \ \ } A\xrightarrow{\ \ \phi\ \ } B\xrightarrow{\ \ \ \ \ } C_0(\hat B),$$ where the three maps are injective ($$\phi$$ by hypothesis and the other two by II.2.2.4). Now when restricted to algebras of continuous functions you can use II.2.2.5/II.2.2.7: say $$\psi:C(X)\to C(Y)$$ is an injective homomorphism. Then $$\breve\psi:Y\to X$$ is surjective. Now, for $$f\in C(X)$$, you have (I do the compact case to avoid a few epsilons) $$\|f\|=|f(x_1)|=|f(\breve\psi (y_1))|=|(\psi f)(y_1)|\leq \|\psi f\|.$$ Also, $$\|\psi f\|=|(\psi f)(y_0)|=|f(\breve\psi(y_0))|\leq \|f\|.$$ Thus $$\|f\|=\|\psi f\|$$.
• No, he says "For the first part, it remains only to show that $\Gamma$ is isometric (then the range will be closed, hence all of $C_0(\tilde A)$ by the Stone-Weierstrass Theorem). By the C$^*$-axiom and II.1.6.6., it suffices to show that if $x = x^∗$, then $\|\hat x\|\geq\|x\|$. By II.2.1.2. there is $λ \in \sigma_A(x)$ with $|λ| = \|x\|$, so there is a φ with $|\tilde x(φ)| = \|x\|$." – Martin Argerami Apr 12 at 18:05
• I mean - how did you get the last $\le$ for both of your lines? – Cy L Shih Apr 12 at 21:05
• They are points. The norm of a function $f$ is $\|f\|=\max\{|f(t)|: t\}$. So you always have $|f(t)|\leq \|f\|$. – Martin Argerami Apr 12 at 21:07