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The answer is no. Let $G$ be a cyclic group of order $n$ not divisible by $2$, let $V$ be an irreducible $2$-dimensional representation of $G$, and consider the associated vector bundle $EG\times_G V …

The answer should probably go something like this: Both $q_!$ and $ind$ are $K(A)$-module maps. Since $q_!$ is an isomorphism, to check that these are the same maps, it suffices to check they are th …

I'll write $U(X)=[X,BU_\otimes]$. So $U(X)\subset K(X)=[X,Z\times BU]$ is the multiplicatively closed subset corresponding to "virtual bundles of rank 1". Likewise, I'll write $I(X)=[X,BU]\subset K(X …

To see why $K$-theory should be Brauer graded, it may help to see how, for a superalgebra $A$, the associated $K$-group only depends on the category $\mathcal{M}_A$ of finite dimensional $Z/2$-graded …

There are many spaces $X$ whose $K$ theory is trivial (isomorphic to that of a point), but whose ordinary cohomology is not. Famously, there is a 4-cell complex obtained by taking the homotopy cofibe …

Perhaps the deeper story you want involves the notion of "$E_{\infty}$ product". The cup product in cohomology, and the sum (and for that matter, the product) in K-theory are commutative (and associa …

This is not really answering your question. But it's worth pointing out that periodicity of Clifford algebras (closely tied to Bott periodicity) already gives you the "periodicity" in the explicit lo …

I think the answer is yes, it commutes. The identities $rc=2$, $\psi^k c=c\psi^k$, and additivity of these operations already implies that $\epsilon:= \psi^k r-r\psi^k$ satisfies $2\epsilon =0$ in $[ …

Here is one conceptual description of the relationship. (I wouldn't call it an explanation; I don't really know why it's true, except that it's because Bott periodicity is true.) KO-theory is the fi …