Thom isomorphism for generalized cohomology theories.

On p. 77 of these notes, the authors give the following version of the Thom isomorphism theorem for a generalized cohomology theory. Suppose we have a ring spectrum $R$ and an $R$-oriented rank $n$ vector bundle $\xi \colon E \to B$ with orientation class $u \in R^n(\mathrm{Th}\xi)$. Then we can consider the following composite

$$ R \wedge \mathrm{Th}(\xi) \xrightarrow{1 \wedge \Delta_T} R \wedge \mathrm{Th}(\xi) \wedge B_+ \xrightarrow{1 \wedge u \wedge 1} R \wedge \Sigma^nR \wedge B_+ \xrightarrow{\mu \wedge 1} R \wedge \Sigma^nB_+. $$ It is then claimed that this induces an isomorphism on $\pi_*$ by construction. Could anyone explain to me why this is the case?

Solution 1:

@withoutfeather is correct; he is not claiming that this is a proof of the Thom isomorphism, but rather a corollary. Look at Milnor and Stasheff's proof of the Thom isomorphism; it is entirely phrased in terms of the axioms of a homology theory plus the notion of an orientation, so the same proof of the Thom isomorphism works for any ring spectrum and vector bundle oriented over that ring spectrum.

Then Bruner is claiming his composite implements the map of the Thom isomorphism, and so by the Thom isomorphism is an equivalence.