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This question relates to one on topology and C^*-algebras that was asked two days ago, namely at the link: C*-algebras with no nontrivial endomorphisms link text

Let D be the closed unit disk in the plane. Let C(D) be the unital ring of continuous complex-valued functions on D. Then, C(D) is naturally a Banach algebra with pointwise addition and multiplication as the ring operations. Furthermore, the "star-operation" on elements of C(D) can be defined by: $g*(x)$ to be the complex conjugate (pointwise) of g(x), any x in D, and for any function g in C(D).

The question in reference 1 above was related to injective star-endomorphisms of some $C*$ -algebras. Without saying so, I think the famous commutative Gelfand-Naimark theorem and the Gelfand representation figured "in the background", because of the interplay of commutative $C*$-algebras and topology on compact spaces ... If $\alpha$ is an injective star-morphism of C(D) to itself, is it possible for $\alpha$(C(D)) to be a proper (i.e. a `strict') star-sub-algebra of the $C*$-algebra C(D) ?

If so, I cannot find such a special *-morphism $\alpha$, hence my question.

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    $\begingroup$ Am I missing something? Can't you take $\alpha(f)= f(g)$ where $g$ is a continuous surjection from $D$ onto $D$ that is not injective? $\endgroup$
    – Bill Johnson
    Oct 18, 2012 at 15:18

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Assume $D$ lies in $\mathbb{R}^2$ and define $f:D\rightarrow D$ by $f(x):=2x$ if $\|x\|\leq 1/2$ and $f(x):=\frac{x}{\|x\|}$ if $\|x\|\geq 1/2$. Then, $f$ is onto and continuous, but it is not injective. It is easy to see that $f^\ast :C(D)\rightarrow C(D)$ is an injective $^\ast$-homomorphism, but it is not onto.

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  • $\begingroup$ Thanks, Vahid Shirbisheh. I'm beginning to understand your answer. I'm not familiar with the definition or naming of $f*$ , given $f$ . I have heard of pull-back maps, but I don't know what they are. The same goes for so-called "push-forward" maps. David Bernier $\endgroup$
    – David Bernier
    Oct 18, 2012 at 21:24
  • $\begingroup$ $f^\ast$ is defined by $f^\ast (g) (x):= g(f(x))$ for every $g\in C(D)$ and $x\in D$. $\endgroup$
    – user23860
    Oct 19, 2012 at 6:24
  • $\begingroup$ Thanks. $f*$ is obtained by "pre-composing" through the map $f: D -> D$. It's clear to me now. $\endgroup$
    – David Bernier
    Oct 19, 2012 at 7:40

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