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I am working through one of the exercises in Rordam's book Introduction to $K$-Theory for $C^{*}$-algebras. Exercise 1.8:

Let $A$ and $B$ be $C^{*}$-algebras, and let $\varphi\colon A\to B$ be a unital $^{*}$-homomorphism. I need to show that $\operatorname{sp}(\varphi(a))=\operatorname{sp}(a)$ for all $a\in A$ provided $\varphi$ is injective.

Using the standard properties of unital $^{*}$-homomorphisms, I was able to prove that if $a-\lambda 1_{A}$ is invertible, then so is $\varphi(a)-\lambda 1_{B}$. Hence, $\operatorname{sp}(\varphi(a))\subseteq\operatorname{sp}(a)$. However, I don't know how to use the injectivity of $\varphi$ to prove the reverse containment.

I tried thinking about how I could maybe use the Continuous Functional Calculus in some way, but I wasn't sure how to proceed. Furthermore, I need to prove the result for all $a\in A$, not just for normal elements in $a$, so I'm not even sure if this is the correct approach.

Any help would be much appreciated. Thank you.

2 Answers 2

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The inclusion you have proven is the key. Now you know that $$\tag{*} \|\phi(a)\|^2=\|\phi(a)^*\phi(a)\|=\|\phi(a^*a)\|=\text{spr}\,(\varphi(a^*a)) \leq\text{spr}\,(a^*a)=\|a^*a\|=\|a\|^2. $$ So $\phi$ is a contraction (in particular, bounded).

Take $a$ selfadjoint. Suppose that $\text{sp}\,(\phi(a))\subsetneq\text{sp}\,(a)$. Then one can construct a function $f$ with $f(t)=0$ for all $t\in\text{sp}\,(\phi(a))$, but $f(t)=1$ for some $t\in\text{sp}\,(a)$. This implies that $f(\phi(a))=0$, while $f(a)\ne0$. But this is a contradiction, since $f(\phi(a))=\phi(f(a))$ (use polynomials, and the continuity of $\phi$), and the latter cannot be zero since $\phi$ is injective. So $\text{sp}\,(\phi(a))=\text{sp}\,(a)$.

An immediate consequence is that now we can repeat the idea as in $(*)$, but with equality: for arbitrary $a$, since $a^*a$ is selfadjoint, $$ \|\phi(a)\|^2=\|\phi(a)^*\phi(a)\|=\|\phi(a^*a)\|=\text{spr}\,(\varphi(a^*a)) =\text{spr}\,(a^*a)=\|a^*a\|=\|a\|^2. $$ It follows that $\phi$ is isometric. Now you can use this to show that $\phi (A) $ is closed, which was the only thing missing to make it a C$^*$-algebra. And then your original argument applies to $\phi^{-1} $: if $\phi (a)-\lambda I $ is invertible, so is $a-\lambda I $, and $\text {sp}(a)\subset \text {sp}(\phi (a))$.

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    Thank you very much for your help! Quick question: to prove that $\operatorname{sp}(\varphi(a))=\operatorname{sp}(a)$, you used the fact that $a$ is normal to apply the continuous functional calculus. Does that mean that the two sets may not be equal if $a$ is not normal? – 2017-02-09
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    @ervx Martin shows that $\varphi$ is an isometry. Now $A$ is (isometrically isomorphic to) a $C^*$-subalgebra of $B$ containing the unit of $B$, so apply the second result given in my answer. – 2017-02-09
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    @ervx: I have added two sentences to my answer. – 2017-02-09
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The easiest way I can see to prove it relies on the following results:

If $A,B$ are $C^*$-algebras and $\varphi:A\to B$ an injective $*$-homomorphism, then $\varphi$ is isometric.

and

If $A$ is a $C^*$-subalgebra of a unital $C^*$-algebra $B$ that contains the unit, then $\sigma_B(a)=\sigma_A(a)$ for all $a\in A$.

If your are interested in proofs, these are theorems $3.1.5$ and $2.1.11$ of Murphy's $C^*$-Algebras and Operator Theory.

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    Whoever downvoted, can you please let me know why? Downvoting without posting a comment does not help me improve my answer. – 2017-02-09