commutativity of a diagram in cohomology of $C^*$-algebras

Question

The setting is the same as in my last question commutative diagram with $K_{i+1}(A)\to K_i(A\rtimes_{\rho} \mathbb{R})$ (for $C^*$-algebras) :

Let $A$ be in the bootstrap category (=N in the other thread) and let $B$ be a $\sigma$-unital $C^*$-algebra such that $K_*(B)$ is an injective $\mathbb{Z}$-module. Let $$\gamma(A,B):KK_*(A,B)\to Hom(K_*(A),K_*(B))$$ $$\epsilon_2 \mapsto (\epsilon_1 \mapsto \epsilon_1 \otimes \epsilon_2)$$ (The map $\gamma$ comes from the Kasparov product).

Given a short exact sequence $$0\to J \xrightarrow{\text{i}} A\xrightarrow{\text{j}} A/J\to 0$$ of $C^*$-algebras in the bootstrap category. Then there are maps $\delta_1^*:KK_*(J,B)\to KK_{*+1}(A/J,B)$ and $\delta_2^*:Hom(K_*(J),K_*(B))\to Hom(K_{*+1}(A/J),K_*(B))$ and a diagram with long exact sequences rows

$$\require{AMScd} \begin{CD} ... KK_n(A/J,B) @>j^n >> KK_n(A,B) @>i^n>> KK_n(J,B) @>\delta_1^n>> KK_{n+1}(A/J,B) @> >>.. \\ @VV \gamma_n(A/J,B) V @VV \gamma_n(A,B) V @VV \gamma_n(J,B) V @VV \gamma_{n+1}(A/J,B) V \\ ... Hom(K_n(A/J),K_*(B)) @>(j')^n >> Hom(K_n(A),K_*(B)) @>(i')^n>> Hom(K_n(J),K_*(B)) @>\delta_2^n>> Hom(K_{n+1}(A/J),K_*(B)) @> >>.. \\ \end{CD} $$

I want to check $\gamma_{n+1}(A/J,B)\circ \delta_1^n =\delta_2^n\circ \gamma_n(J,B)$ such that the diagram with the long exact rows is commutative (since $\gamma$ is a natural transormation, the other squares commute). But again I don't how to write down the maps $\delta_1^n$ and $\delta_1^n$ for $n\in\mathbb{N}$ explicitely. Thus my question is

Why $\gamma_{n+1}(A/J,B)\circ \delta_1^n =\delta_2^n\circ \gamma_n(J,B)$?

Best

Edit: Meanwhile I found a description of $\delta_1$ (theorem 19.5.7- Six-Term exact sequence for KK) in Blackadars book "K-theory for operator algebras": There exists an element $\sigma\in KK_1(A/J,J)$ such that $\delta_1$ is just multiplication (= intersection product) by the element $\sigma$, i.e. $\delta_1(x)=\sigma\otimes x$.

In addition to that I suppose the map $\delta_2$ is induced by the intersection product in the following way (or similarly): There is an element $y\in KK_1(A/J,J)$ and a map $\psi :KK_{n+1}(\mathbb{C},A/J)=K_{n+1}(A/J)\to KK_{n}(\mathbb{C},J)=K_{n}(J)$ defined by $\psi (\eta)=\eta\otimes y$. Then set $\delta_2:=Hom(\psi;K_*(B))$ (i.e. hom-functor apllied to $\psi$).

There is still something wrong because the square doesn't commute, but the maps $\delta_1$ and $\delta_2$ must be defined in a similar way as above I would say. Do you know how to do this correctly?

Answer 1

Observe that the $KK$-class $\sigma \in KK_1(A/J,J)$, which you mention in your edited paragraph only depends on the extension $$ 0 \to J \to A \to A/J \to 0 $$ and not on $B$. So we have $\delta_1^n(x) = \sigma \otimes_J x$ for $x\in KK_n(J,B)$. If $\delta_3 \colon K_*(A/J) \to K_{*+1}(J)$ denotes the boundary map associated to the six-term exact sequence in $K$-theory, this can also be written as $\delta_3(z) = z \otimes_{A/J} \sigma$ using the same $\sigma$. Hence $$ \delta_2^n(\varphi)(y) = \varphi(y \otimes_{A/J} \sigma) $$ for $y \in K_{n+1}(A/J)$ and $\varphi \in \hom(K_{n}(J), K_*(B))$. Let $x\in KK_n(J,B)$, $y \in K_{n+1}(A/J)$. Then we have $$ \gamma_{n+1}(A/J,B)(\delta_1^n(x))(y) = \gamma_{n+1}(A/J,B)(\sigma \otimes_J x)(y)= y \otimes_{A/J} (\sigma \otimes_J x)\ , $$ $$ \delta_2^n(\gamma_n(J,B)(x))(y) = \gamma_n(J,B)(x)(y \otimes_{A/J} \sigma) = (y \otimes_{A/J} \sigma) \otimes_J x \ . $$ The result now follows from the associativity of the Kasparov product