Let $f:X\to BGL_1(\mathbb{S})$ be a morphism of $E_n$-spaces and determine a principle $GL_1(\mathbb{S})$-bundle over $X$. Then it can be shown in the classical case that there is always a Thom isomorphism $Mf\wedge Mf\to Mf\wedge X_+$ (Mahowald shows this, for instance here, modulo the concerns about operadic structure).

I'm interested in understanding how this map is produced in the modern $\infty$-categorical framework developed in work of Ando, Blumberg, Gepner and others. What this comes down to is showing that there is an orientation of $f$. This can be described as an equivalence of functors between $F=f\circ(-\wedge Mf):X\to BGL_1(\mathbb{S})\to BGL_1(Mf)$ and $\ast:X\to BGL_1(Mf)$. In other words, there should be an equivalence $F\simeq p^*Mf$ in the category $Mf_X\mathrm{-line}\simeq Fun(X^{op},Mf\mathrm{-line})$ where $Mf\mathrm{-line}$ is the category of $Mf$-modules which are equivalent to $Mf$ and $p:X\to \ast$ is the terminal map (note that to simplify notation I've called the composite map $F$). Such an equivalence would, upon applying $p_!$, yield an equivalence $Mf\wedge Mf=p_!F\simeq p_!p^\ast Mf=Mf\wedge X_+$.

One way to try to obtain such an orientation is to notice that the multiplication map $Mf\wedge Mf\to Mf$ is actually a map $p_!F\to Mf$ in $Mf\mathrm{-mod}$, which yields a map (simply by noting that $p_!$ is left adjoint to $p^\ast$) $F\to p^\ast Mf$ in $Mf_{X}\mathrm{-mod}$.

My question is the following: why is the map so obtained an equivalence? If it is not, is there another map $MF\to Mf$ that does give us this equivalence?

  • $\begingroup$ Is there a reason you didn't include the tag homotopy-theory? $\endgroup$ – Sean Tilson May 20 '15 at 8:38
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    $\begingroup$ Haha, no, I just can't remember all the tags sometimes. $\endgroup$ – Jonathan Beardsley May 20 '15 at 12:11

The updated version of A simple universal property of Thom ring spectra has a new section on multiplicative orientations. Corollary 3.17 proves that any $n$-fold loop map $f$ has an $E_{n-1}$ $Mf$-orientation.

  • $\begingroup$ Also, this version uses your paper Relative Thom spectra via operadic Kan extensions to give a new proof of the construction of $H\mathbb{Z}$ as an $E_2$-ring spectrum. See section 5.2. $\endgroup$ – Omar Antolín-Camarena Jun 14 '17 at 0:19
  • $\begingroup$ Hooray! Thanks for writing that down! :) $\endgroup$ – Jonathan Beardsley Jun 14 '17 at 0:20

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