Producing $K$-homology cycles from $KK$-cycles For two unital (separable) $C^*$-algebras $A$ and $B$, let $(H,\rho,F)$ be a $KK$-cycle in the sense of Kasparov, or in the sense of Wikipedia :)
I wonder if there us a natural way to "forget" the right Hilbert $B$-module structure, and "project" in some sense to a Fredholm module over $A$.
This is inspired by the following construction: Looking at the KK-theory level instead, we have the Kasparov product
$$
KK(A,B) \times K(B,\mathbb{C}) \to K(A,C).
$$
Which by pairing with a $K$-theory class, produces a $K$-homology class.
QUESTION: Is there some natural way to lift this Kasparav contraction to a simple operation to turn $KK$-cycles into $K$-homology classes? 
Edit: As Ulrich points out in the comments below, one way to do this might be to use a trace, or state, on $B$.
 A: Here are just some trivial observations that came to mind after thinking about this a little longer: You are essentially asking for a canonical class in the $K$-homology group $K^0(B) = KK(B,\mathbb{C})$. In general this does not exist except in very special situations. For example, if $\text{Ext}^1_{\mathbb{Z}}(B,\mathbb{Z}) = 0$, then $K^0(B) \cong \hom(K_0(B),\mathbb{Z})$, ie. any group homomorphism $K_0(B) \to \mathbb{Z}$ lifts uniquely to a $K$-homology class. A trace on $B$ produces such a group homomorphism.
An interesting situation, which produces a class in $K^0(B)$ is when $B$ is the completion of an algebra $\mathcal{B}$ that is part of a spectral triple $(\mathcal{B}, \mathcal{H}, D)$. In case $B$ is commutative, this boils down to the statement that $K$-homology classes arise from Dirac operators on spin-manifolds. For details about the $K$-homology classes associated to spectral triples see this article (in particular Prop. 4.4)
https://ro.uow.edu.au/cgi/viewcontent.cgi?referer=https://www.google.com/&httpsredir=1&article=1745&context=eispapers
by Carey, Phillips and Rennie. 
