Stack Exchange network consists of 183 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers.
You are right. The equation you write holds for every c.p.c. map. At this point, the assumption of "order zero" is not needed. However, at the part of the paper by Winter-Zacharias that you are reading, it is shown that a c.p.c. order zero $\varphi\colon A\to B$ induces a map $\mathrm{Cu}(\varphi)\colon\mathrm{Cu}(A)\to\mathrm{Cu}(B)$ between the respective Cuntz semigroups. A general c.p.c. map need not preserves Cuntz (sub)equivalence, which is why the assumption of order zero is needed.
@WillBrian: OK, I guess that is a matter of definition. The wikipedia article you linked also sais: "However, many authors use the term dyadic space with the same meaning as dyadic compactum." I don't really have a strong opinion about that. When I wrote my answer I must have thought that "dyadic space" means quotient of the Cantor set. But you are right that I only address closed subspaces.
Yes, with inductive colimit I mean the colimit of a system of groups indexed over a directed set, for example a sequential system $G_1\to G_2\to G_3\to\ldots$.
It was not clear to me that every group is an intersection of two perfect groups, which seems like a useful fact to know. About the projective limits, I meant projective systems indexed over a directed set, for example sequential systems like $\ldots G_3\to G_2\to G_1$. But I guess one can also ask the question for arbitrary (co)filtered limits.
@YCor: Yes, you are right. I was not aware of "locally free groups". Also your argument that inductive limits of free groups are locally free seems like a perfect answer to me.
With injective connecting homomorphisms the question will be very different. In fact, it seems to me that an inductive limit of free groups with injective connecting homomorphisms is again a free group.
Every C*-algebra is the directed union of its separable sub-C*-algebras, and K-theory is a continuous functor, that is, it commutes with direct limits (=colimits). Therefore, $K_0(A)$ is the colimit of $K_0(A')$ over all separable sub-C*-algebras $A'$, ordered by inclusion. I think the grading will not cause any problems. One only needs to use that every graded C*-algebra is also the directed union of its graded, separable sub-C*-algebras.
Dear Mateusz, thank you for the thorough answer. This satisfyingly settles the case of "internal" weak*-norm-continuous operators (i.e., thus given by mutliplication by an element of $M$.) The situation might be different if one considers maps $T\colon M\to M$. In particular, if $F\subseteq M$ is finite-dimensional, and if $T\colon M\to F$ is normal, then $T$ is also weak*-norm-continuous. Thus, there tend to be many such maps, and I guess that if $M$ is hyperfinite then there are enough to approximate the identity on $M$ in some sense.
