Homotopic Grassmannian maps classify isomorphic bundles and conversely

If $H:X\times I\to {\mathrm{Gr}}_n({\mathbb{F}}^{\mathrm{\infty }})$ joins $f_0$ to $f_1$, then $H^{\ast }{\gamma }_n$ restricts to $f_0^{\ast }{\gamma }_n$ and $f_1^{\ast }{\gamma }_n$. By [F1] these endpoint bundles are isomorphic. This proves the forward implication, with AC used exactly as in [F1].

Conversely, suppose $\mathrm{\Phi }:f_0^{\ast }{\gamma }_n\to f_1^{\ast }{\gamma }_n$ is an isomorphism. With coordinates indexed by $r\ge 0$, let $O(e_r)=e_{2r+1}$ and $P(e_r)=e_{2r+2}$. The injective paths $(1-s)I+sO$ and $(1-s)I+sP$, with Gram normalization as in [F3], homotope $f_0$ to $O{f}_0$ and $f_1$ to $P{f}_1$ as image-plane maps. The formulas are stagewise continuous and hence continuous in the stable weak topology.

View the tautological inclusions as embeddings $j_i:f_i^{\ast }{\gamma }_n\to X\times {\mathbb{F}}^{\mathrm{\infty }}$. For $0\le \theta \le \pi \mathrm{/}2$, the fiber map $v\mapsto \cos \theta O(j_{0}v)+\sin \theta P(j_1\mathrm{\Phi }v)$ is injective: its two terms lie in orthogonal odd and even coordinate subspaces, so its squared norm is ${\cos }^2\theta \mathrm{\parallel }{j}_{0}v{\mathrm{\parallel }}^2+{\sin }^2\theta \mathrm{\parallel }{j}_1\mathrm{\Phi }v{\mathrm{\parallel }}^2>0$ for $v\mathrm{\ne }0$. Its image planes therefore give a continuous homotopy from $O{f}_0$ to $P{f}_1$.

Concatenate the first displacement homotopy, the interpolation of step 2.1, and the reverse of the second displacement homotopy. This proves $f_0\simeq f_1$, the reverse implication. If two embeddings classify one bundle, [F2] identifies both pullbacks with that bundle, so this reverse implication applies. For $n=0$ every Grassmannian is a point, and for $X=\mathrm{\varnothing }$ there is one map and one bundle; both implications remain valid.