I want to understand the relation between the $KK$ groups (in the sense of Kasparov) and ordinary operator $K$-theory, i. e. I would like to prove something like this: For $B$ trivially graded $\sigma$-unital $C^*$ algebra it holds that $KK(\mathbb{C}, B) \cong K_0(B)$. The proof given in "K-Theory for Operator Algebras" by Bruce Blackadar (2nd edition) is somewhat sketchy (see section 17.5). The following things are unclear:
- In Proposition 17.4.3 certain normalizations of Kasparov modules are discussed. One can prove that in the definition of $KK_{oh}(A, B)$ and $KK_h(A, B)$ it suffices to consider triples $(E, \phi, F)$ with $F = F^*$ and $\lVert F \rVert \leq 1$. This is clear. However, it is stated that for $A$ unital one can in addition assume that $F^2 - 1 \in \mathbb{K}(E)$. This I don't understand: In the proof it says that one considers $\tilde{F} = PFP + (1-P)$ instead of $F$ where $P = \phi(1)$. I don't see why $\tilde{F}^2 - 1 \in \mathbb{K}(E)$ and moreover I don't see why we still have $\lVert \tilde{F} \rVert \leq 1$.
- In section 17.5 Blackadar introduces the "Fredholm picture" of the KK groups which ultimately leads to the statement I'm trying to prove. If $A$ is unital, one can apply Proposition 17.4.3 and replace $E$ by $PE$ and $F$ by $FPF$ so that one can assume that $\phi$ is unital. In the next sentence, Blackadar says "If $B$ is $\sigma$-unital and there is a unital (graded) $*$-homomorphism from $A$ to $\mathbb{B}(\hat{\mathbb{H}}_B)$ [...] we may in addition restrict to the case $E = \hat{\mathbb{H}}_B$ (with $\phi$ unital)". My question is: Why can't we apply Proposition 17.4.1 which already states that we may restrict to $E = \hat{\mathbb{H}}_B$? Also: How does existence of a unital (graded) $*$-homomorphism from $A$ to $\mathbb{B}(\hat{\mathbb{H}}_B)$ help us to get $E = \hat{\mathbb{H}}_B$?