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)](https://mathoverflow.net/questions/241884/commutative-diagram-with-k-i1a-to-k-ia-rtimes-rho-mathbbr-for-c) :


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?



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


