In the top answer to the question "Is there a map of spectra implementing the Thom isomorphism?" it is explained (with a reference to Rudyaks book) that from a rank $r$ vector bundle $\mu:V\to X$, a spectrum $E$ with multiplication $m:E\wedge E\to E$ and a Thom class $t:X^\mu\to \Sigma^r E$ one gets the Thom isomorphism:

The (cohomological) Thom isomorphism sends a map $x:X_+\to E^n$ of spectra to the composition $$ X^\mu\to X_+\wedge X^\mu\xrightarrow{x\wedge t}E^n\wedge E^r\xrightarrow{m}E^{n+r}. $$ The induced isomorphism on homotopy groups is then $E^{n}(X_+)\cong E^{n+r}(X^\mu)$.

Is it possible to give an inverse map, i.e. a map sending a $X^\mu\to E^{n+r}$ to an $X_+\to E^n$ on the level of spectra, too?

I am not asking for a proof of the Thom isomorphism. Even with the existence of such a map one has to prove that the correspondence induces an isomorphism on homotopy groups.


1 Answer 1


You find the inverse Thom isomorphism if you generalize the construction from my answer to the old question to stable vector bundles. The point is that if you have two stable vector bundles $\eta$ and $\mu$ over $X$, then you get a diagonal map

$$X^{\eta\oplus\mu} \to X^{\eta} \wedge X^{\mu}.$$

Now if there is a Thom class $X^{\mu} \to \Sigma^n E$, then there is a Thom class $X^{-\mu} \to \Sigma^{-n} E$; and the inverse to the Thom isomorphism maps an element $X^{\mu}\to \Sigma^k E$ to the composition

$$X \to X^{\mu} \wedge X^{-\mu} \to \Sigma^k E \wedge \Sigma^{-n} E \to \Sigma^{k-n} E.$$


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