There is a simple, intuitive "construction" of twisted K-theory if we are allowed to ignore that many things only hold up to homotopy. We know that maps to $K(Z,2)$ give line bundles on a space and that $K(Z,2)$ forms a group corresponding to the tensor product of line bundles. Line bundles also act as endomorphisms of K-theory given by the tensor product. Thus, there is an action of $K(Z,2)$ on $F$ (where $F$ is the classifying space for $K^0$). $K(Z,2)$ principal bundles are classified by maps to $BK(Z,2) \cong K(Z,3)$, ie, elements of $H^3$. Choosing such a map, we get a principal $K(Z,2)$ bundle, $E$, and we can form the associated bundle $E \times_{K(Z,2)} F$. Twisted K-theory is then the homotopy classes of sections of this bundle.

The usual constructions of twisted K-theory that I have seen make the above precise by choosing representatives of the relevant objects so that all the needed relations hold on the nose. My question is whether you can avoid doing that. In other words, can you define all the various notions up to homotopy and obtain a definition of twisted K-theory that way?