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This question relates to one on topology and C^*-algebras that was asked two days ago, namely at the link: http://mathoverflow.net/questions/109772/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.

asked Oct 18 at 14:45

David Bernier

community wiki

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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? – Bill Johnson Oct 18 at 15:18

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Why is this community wiki? – Bill Johnson Oct 18 at 15:19

Regarding having the question in community wiki, I now think it probably wasn't a wise decision. In the usual, non-community wiki mode, the question could have up-votes and down-votes; also, the answer(s) could have up-votes and down-votes. In my opinion, non-community wiki mode would have been preferable. It's an unfortunate decision for which I must take the responsibility. I hope to improve over time :) . – David Bernier Oct 18 at 21:09

Vahid Shirbisheh · Accepted Answer · 2012-10-18 15:34:40Z UTC

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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.

answered Oct 18 at 15:34

Vahid Shirbisheh

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	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 – David Bernier Oct 18 at 21:24
	$f^\ast$ is defined by $f^\ast (g) (x):= g(f(x))$ for every $g\in C(D)$ and $x\in D$. – Vahid Shirbisheh Oct 19 at 6:24
	Thanks. $f*$ is obtained by "pre-composing" through the map $f: D -> D$. It's clear to me now. – David Bernier Oct 19 at 7:40

injective morphisms of C(D), the C^ algebra of continuous functions on the closed unit disk D

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