most recent 30 from http://mathoverflow.net 2013-06-19T10:32:07Z http://mathoverflow.net/feeds/question/110017 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/110017/injective-morphisms-of-cd-the-c-algebra-of-continuous-functions-on-the-clos David Bernier 2012-10-18T14:45:08Z 2012-10-18T15:34:40Z

<p>This question relates to one on topology and C^*-algebras that was asked two days ago, namely at the link: <a href="http://mathoverflow.net/questions/109772/c-algebras-with-no-nontrivial-endomorphisms" rel="nofollow">http://mathoverflow.net/questions/109772/c-algebras-with-no-nontrivial-endomorphisms</a> <a href="http://mathoverflow.net/questions/109772/c-algebras-with-no-nontrivial-endomorphisms" rel="nofollow">link text</a></p> <p>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).</p> <p>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) ?</p> <p>If so, I cannot find such a special *-morphism $\alpha$, hence my question.</p>

http://mathoverflow.net/questions/110017/injective-morphisms-of-cd-the-c-algebra-of-continuous-functions-on-the-clos/110021#110021 Vahid Shirbisheh 2012-10-18T15:34:40Z 2012-10-18T15:34:40Z

<p>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.</p>