MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Hi everyone On page 147 of the note "Group C*-Algebras and K-theory" by N.Higson and E.Guentner there are something about the stabilized homotopy category of graded C* algebra, which is a category whose objects are the graded C* -algebras and morphisms from A to B are the homotopy classes of graded $\ast$-homomorphisms from A to $B\otimes K(H)$. But the exact definition of composition of morphisms and the identity morphisms are not mentioned. I think the definition is dual to the defition of amplified category of graded C*-algebras, is there someone knows some references about this?\newline Thinks

share|cite|improve this question
up vote 2 down vote accepted

In what follows, all tensor products are graded.

The comments about the existence of canonical (up to homotopy) $\ast$-homomorphisms $\mathbb{C} \to K(H)$ and $K(H) \otimes K(H) \to K(H)$ right before the definition of the category in question are key. If you have $\ast$-homomorphisms $A \to B \otimes K(H)$ and $B \to C \otimes K(H)$ then the composition is just $A \to B \otimes K(H) \to C \otimes K(H) \otimes K(H) \to C \otimes K(H)$. And the identity morphism is just the map $A \to A \otimes K(H)$ given by $a \mapsto a \otimes e$ where $e$ is the projection onto a one dimensional grading-degree zero subspace of $H$ (well-defined up to homotopy).

I'm not totally sure where you can find further discussion of this category, but you might try the AMS Memoire "Equivariant E-Theory for C* algebras" by Higson, Guentner, and Trout.

share|cite|improve this answer
I can not find Stabilized homotopy category in "Equivariant E-Theory for C* algebras". ..but thanks – m07kl May 28 '11 at 10:11
By the way do you know a solution of Exercise 1.9 without showing two definitions are natrurally isomorphic? – m07kl Jun 2 '11 at 9:24

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.