The homotopy relation of Kasparov-Cycles is definied in Blackadar's book in 17.2.2. It is an equivalence relation. However, I really don't see a good argument for transitivity and can't find any proof in the literature. It is always ommitted, so I assume it's pretty straight forward. Could somebody roughly outline the argument for me or maybe give me a good source.
Thank you
It is really the same thing as for ordinary homotopy. Mainly a concatenation.
I will write for a space $X$, $XB$ the algebra of continous functions from $X$ to $B$. All the other notations follow those from Blackadar. Suppose you have
$(E_{1 2}, \phi_{1 2}, F_{1 2}) \in E(A, [1, 2]B) $ connecting $(E_{1 }, \phi_{1 }, F_{1 }) $ and $(E_{2}, \phi_{ 2}, F_{ 2}) $ in $E(A, B) $ $(E_{23}, \phi_{23}, F_{23}) \in E(A, [2,3]B) $ connecting $(E_{2 }, \phi_{2 }, F_{2 }) $ and $(E_{3}, \phi_{ 3}, F_{ 3}) $ in $E(A, B) $
You have two maps $ev_2 : E_{12} \to E_2$ and $ev_2 : E_{23} \to E_2$ and you can consider the $(A, [1,2]B \times_B [2,3]B)$ bimodule $E_{12} \times_{E_2}E_{23}$. It is the sub $ [1,2]B \times [2,3]B$ module of $E_{12} \times E_{23} $ given by $\{(x, y) \in E_{12} \times E_{23}, f(x) = g(y) \}$.
It is in fact a $[1,2]B \times_B [2,3]B$ module. But this algebra is nothing more than $[1, 3]B$
$(E_{12} \times_{E_2}E_{23}, \phi_{12} \times_{E_2}\phi_{23}, F_{12} \times_{E_2}F_{23})$ then connects $(E_{1 }, \phi_{1 }, F_{1 }) $ and $(E_{3}, \phi_{ 3}, F_{ 3}) $ and showing that it is indeed in $ E(A, IB) $ is straightforward
