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Let $\alpha:A \rightarrow B$ be a *-homomorphism of $C^*$-algebras. Then $\alpha$ ist a stable homotopy equivalence if there exists a $*$-homomorphism $\beta: B \otimes \mathcal{K} \rightarrow A \otimes \mathcal{K}$ s.t.

$$ 1. \, \beta \circ (\alpha \otimes 1_{\mathcal{K}})\simeq 1_{A \otimes \mathcal{K}} \qquad 2. \, (\alpha \otimes 1_{\mathcal{K}}) \circ \beta \simeq 1_{B \otimes \mathcal{K}} $$ where $\mathcal{K}$ denotes the compact operators on $\ell^2(\mathbb{N})$.

Now, I have come across this lemma:

Let $H\neq 0$ be a separable Hilbert space and $p$ a rank-one projection on $H$ and $\varphi: \mathbb{C} \rightarrow \mathcal{K}$ the $*$-homomorphism with $\varphi(1)=p$. A $*$-homomorphism $\alpha: A \rightarrow B$ is a stable homotopy equivalence if there is a $*$-homomorphism $\beta: B \rightarrow A \otimes \mathcal{K}(H)$ such that $$ 1. \, \beta \circ \alpha \simeq \varphi \otimes 1_A \qquad 2. \, (1_{\mathcal{K(H)}} \otimes \alpha) \circ \beta \simeq \varphi \otimes 1_B $$ where we identify $A=A \otimes \mathbb{C}$ and $B=B\otimes \mathbb{C}$.

I am having a hard time understanding the argument. I even talked to somebody about it, thought I had understood it and now I am totally confused again. Could somebody outline the argument for me?

Thank you

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  • $\begingroup$ Your lemma implies that $\phi$ is itself a stable homotopy equivalence (apply it to $\phi$). Inversely it is easy to see that if $\phi$ is a stable homotopy equivalence then the lemma is true. The homotopy inverse of $\phi$ should be a map $\psi:\mathcal{K}(H\otimes H)\to\mathcal{K}(H)$ of the form $\psi(T)=ATB$ where $A,B$ are well choose bounded maps between the Hilbert spaces. $\endgroup$ – Omar Jan 26 '17 at 9:14

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