We have extended this square very slightly by a certain shearing map \(\sigma\) defined by \(\sigma(x, y) = (xy^{-1}, y)\). It is evident that \(\sigma\) is a homotopy equivalence, since just as we can de-scale the first coordinate by \(y\) we can re-scale by it---indeed, this is the observation that \(BO\) is a torsor for itself. We can calculate directly the behavior of the long composite: \[J_{\R} \circ \mu \circ \sigma(x, y) = J_{\R} \circ \mu(xy^{-1}, y) = J_{\R}(xy^{-1}y) = J_{\R}(x).\] It follows that the second coordinate plays no role, and that the bundle classified by the long composite can be written as \(J_{\R} \times 0\).\footnote{This factorization does \emph{not} commute with the rest of the diagram, just with the little lifting triangle it forms.} We are now in a position to see the Thom isomorphism:
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