# Lifting triangles in K-theory to KL-groups

Let $X$ and $Y$ be finite simplicial complexes (or $CW$-complexes) so that $Y\subseteq X$. Let $s\colon C(X)\to C(Y)$ be the map given by restriction. In particular $K_{*}(C(X))$ and $K_{*}(C(Y))$ are finitely generated. Let $\gamma\colon KL(\cdot,\cdot) \to \operatorname{Hom}(K_{*}(\cdot),K_{*}(\cdot))$ be the surjective map given by the UCT.

For what simple, separable $C^{*}$-algebras $A$ is it true that: for any $\kappa\in KL(C(X),A)$ and $h'\in \operatorname{Hom}(K_*(C(Y)),K_*(A))$ such that $\gamma(\kappa) = h'\circ K_*(s)$, there exists $\kappa'\in KL(C(Y),A)$ such that $\kappa = \kappa'\circ KL(s)$, where $\circ$ is the $KK$-product?

I would like this to be true for general simple, separable $C^*$-algebras or those with tracial rank less than one, but any insight would be appreciated. This is trivially true if $K_*(C(X))$ is torsion-free, but beyond this, I have no intuition on bivariant $K$-theory of any type. I had thought I was on my way to a proof, but I made an elementary group theory mistake.

• What do you need Gelfand representation for? Any map $Y \to X$ induces a map $C(X) \to C(Y)$ by pullback. Gelfand representation is for going the other way. – Qiaochu Yuan Nov 3 '13 at 8:23
• You're right. I just cite Gelfand whenever I use the fact that they're equivalent categories without thinking whether it's necessary. – minimalrho Nov 3 '13 at 17:22