Suppose $R$ is a commutative ring spectrum, and let $f,g: X \to BGL\_1(R)$ be $E\_1$-maps. Then their Thom spectra are $E\_1$ ring spectra. If $f$ and $g$ are homotopic via $E\_1$-maps, does it follow that their Thom spectra are equivalent as $E\_1$ ring spectra? (And not, for example, just as ring spectra up to homotopy?)



Invariance of Thom spectra
Yes this is true. This follows implicitly from the fact that the $E\_1$-structure on $Mf$ can be identified with the canonical $E\_1$-structure on $\mathrm{colim}\_Xf$ (see for example [here](https://arxiv.org/abs/1411.7988)) and that is invariant under equivalences of $E\_1$-map.


