Mayer-Vietoris sequence in topological K-theory

Question

In topological K-theory, we define functors $K^{-n}$ on the category of compact Hausdorff spaces. With this theory we have the Mayer-Vietoris exact sequence: if $X = A \cup B$, we have an exact sequence $$ \dotsb \to K^{-n}(X) \to K^{-n}(A) \oplus K^{-n}(B) \to K^{-n}(A\cap B) \to K^{-n+1}(X) \to \dotsb.$$ I'm unsure about the details about the subspaces $A$ and $B$ of $X$, however. For since the functors $K^{-n}$ are only defined on the category on compact Hausdorff spaces, I would expect these subspaces to have to be closed. In practise however, I see the Mayer-Vietoris sequence being used with open subspaces, which (in general) are not compact spaces.

I know that the definition of the groups $K(X)$ works for any topological space $X$ (i.e., the Grothendieck completion of the semigroup of vector bundles over $X$). However, since we restrict $K^{-n}$ to compact Hausdorff spaces to ensure that it is a cohomology theory, I'm not sure if it would be right to 'just use the definition' for open subsets of $X$.

Answer 1

I believe we just ask that the interiors of the spaces cover $X$ to apply Mayer-Vietoris which means we can still have a meaningful Mayer-Vietoris theorem even if we ask that the sets are closed.

If you are worried that this stops K-theory from really being a cohomology theory, first you really need to define what a cohomology theory is when we are considering only compact, Hausdorff spaces. Such a thing will always require any subspaces we consider to be compact since how else could we talk about their cohomology?

If what you really don't like is that we aren't defining K-theory in the case of noncompact objects. There is another problem that is even more significant, compact spaces or not. We don't know the positive K-groups! Surely any cohomology theory needs positive groups.

The solution to both is Bott periodicty. It first allows us to define the higher K-groups by taking homotopy classes of maps into the K-theory spectrum, but at this point there is no longer a reason to restrict to compact (or even Hausdorff) spaces. So we've arrived at a fully fledged cohomology theory.

Answer 2

In fact you have some sort of a dilemma. The traditional approach to K-theory is restricted to compact spaces as in Atiyah's book. As you say, you can define K-groups $K(X)$ for all spaces $X$, but then it seems that relative groups $K(X,A)$ are only defined for closed $A \subset X$. See for example

Husemoller, Dale. Fibre bundles. Vol. 5. New York: McGraw-Hill, 1966.

For a survey concerning K-theory of non-compact spaces see

https://mathoverflow.net/questions/234489/k-theory-of-non-compact-spaces

Landweber, Gregory D. "K-theory and elliptic operators." arXiv preprint math/0504555 (2005).

http://math.mit.edu/~rbm/18.199-S08/Chapter10.pdf

By the way, it seems that in https://arxiv.org/abs/1811.02592 there is no need to work with open $U_i \subset S^2$. It seems to me that one could also take closed $G$-neighbourhoods of the north and south pole which cover $S^2$ and are $G$-contractible.

Answer 3

There are (at least) two ways to define K-theory of a space $X$. One is as the Grothendieck group $\mathcal{K}(X)$ on the semiring of vector bundles over $X$, and the other is as the set of free homotopy classes $K(X)=[X,BU\times\mathbb{Z}]_{free}$, where $BU=colim_nBU_n$ is the classifying space of the infinite unitary group. The point is that these are both equally valid definitions, and that they both agree on the category of compact Hausdorff spaces: $\mathcal{K}(X)=K(X)$ if $X$ is compact Hausdorff.

In either case you get an extension of the theory to a larger category of spaces, although the two ensuing extensions may not be the same.

I'd imagine with the Grothendieck group definition, proving excision for $\mathcal{K}(X)$ is simply a case of making sure that it is possible to coherently glue vector bundles over open subsets, and this seems like something you can do for any reasonable spaces.

On the other hand, getting a Mayer-Vietoris sequence for the represented functor $K(X)$ simply requires that for an open covering $U,V\subseteq X$, the pushout diagram $\require{AMScd}$ \begin{CD} U\cap V@>>>U\\ @VVV@VVV \\ V@>>>U\cup V=X \end{CD} is also a homotopy pushout. Although the diagram is far from cofibrant this should actually hold in any 'convenient' category of spaces in which one would normally do homotopy theory,say, that of compactly generated, weak Hausdorff spaces, which contains all CW complexes. See for instance Cubical Homotopy Theory, pg. 152.