Let $M$ be a $C^\ast$-algebra, $A, B$ be closed, two-sided ideals of $M$ such that $A+B=M$. Define $T=\{f\in C([0, 1], M):f(0) \in A, f(1) \in B\}$. Why is that the inclusion map of $C([0, 1], A\cap B)$ into $T$ induces isomorphism on $K$-theory?
I have tried the two common strategy for proving isomorphism to no avail. For one thing, I couldn't find a well-defined map from $T$ to the algebra $C([0, 1], A\cap B)$. (I was trying to claim that the said map is a homotopy equivalent)
The other strategy is via exact sequence, I couldn't find a contractible algebra to map $T$ surjectively onto. Any suggestion / hint would be greatly appreciated!
Let $C(D)$ denote the cone on a C* algebra $D$. Recall that $C(D)$ is contractible.
Consider the natural quotient map $\pi : T\to C(M/B)$, then $$ \ker(\pi) = \{f \in T : f(t) \in B \quad\forall t\} = \{f \in C([0,1],B) : f(0) \in A\cap B\} $$ Then one has $$ 0 \to C([0,1],A\cap B) \to \ker(\pi) \to C(B/A\cap B)\to 0 $$ And inclusion map $C([0,1],A\cap B) \to T$ is the composition of the two inclusion maps $$ C([0,1],A\cap B) \to \ker(\pi) \to T $$ each of which induces an isomorphism in K-theory.