Topological Vector Bundles and Grassmannian Classification

For each $u\in X$, let $S_u=\{v\in X:uRv\}$. Every $S_u$ is nonempty because $R$ is entire. Apply [F1] to the set $\mathcal{S}=\{S_u:u\in X\}$ and let $c$ be its choice function. Define $s:X\to X$ by $s(u)=c(S_u)$. This is well-defined even when two successor sets coincide, and $uRs(u)$ for every $u\in X$.

Apply [F3] to $s$ and the prescribed $a$. It gives $x:\mathbb{N}\to X$ with $x_0=a$ and $x_{n+1}=s(x_n)$. Step 1.1 then gives $x_{n}R{x}_{n+1}$ for every $n$.

Since $X$, the entire relation $R$, and $a$ were arbitrary, the sequence in step 2.1 satisfies exactly the prescribed-initial-point formulation in [F2]. Thus AC implies DC, and every later use of this lemma spends AC only in the simultaneous choice made in step 1.1.

The cocycle with $k=i$ gives $g_{ij}{g}_{ji}=I$. Hence the displayed relation is reflexive, symmetric, and transitive: the transitive calculation is $g_{kj}(g_{ji}v)=g_{ki}v$. It therefore defines a quotient $q:{\coprod }_i{U}_i\times {\mathbb{F}}^n\to E$ and a map $p:E\to X$ by $p[x,v,i]=x$. Since $pq(x,v,i)=x$ is continuous on every summand, the quotient property [F2] makes $p$ continuous.

The quotient map $q$ is open. Indeed, if $O$ is open in the coproduct, then the part of its saturation in the $j$th summand is the union over $k$ of the images of $O\cap ((U_j\cap U_k)\times {\mathbb{F}}^n\times \{k\})$ under the homeomorphism $(x,v,k)\mapsto (x,g_{jk}(x)v,j)$; its inverse uses $g_{kj}=g_{jk}^{-1}$. Hence every such part is open. The restriction of $q$ over the saturated open set $p^{-1}(U_i)$ is therefore again a quotient map. Define ${\mathrm{\Phi }}_i[x,v,k]=(x,g_{ik}(x)v)$. The cocycle makes this independent of the representative, and its composite with the restricted quotient map is continuous on every summand, so [F2] makes ${\mathrm{\Phi }}_i$ continuous. Its inverse is $(x,w)\mapsto [x,w,i]$ and is continuous as the $i$th-summand inclusion followed by $q$. Thus ${\mathrm{\Phi }}_i:p^{-1}(U_i)\cong U_i\times {\mathbb{F}}^n$ is a fiberwise-linear chart. Its overlap from $i$ to $j$ is $g_{ji}$, so [F1] proves that $E$ is the claimed bundle.

For the primed cocycle, the maps on summands $(x,v,i)\mapsto [x,h_i(x)v,i{]}^{\mathrm{\prime }}$ respect the relation because $g_{ji}^{\mathrm{\prime }}{h}_i=h_j{g}_{ji}$. By [F2] they descend to a continuous fiberwise-linear map $E\to E^{\mathrm{\prime }}$. Replacing $h_i$ by $h_i^{-1}$ gives its continuous inverse, so it is a bundle isomorphism.

Finally, a linear atlas of a rank-$n$ bundle supplies the functions $g_{ji}$ and their cocycle law by [F1]. Sending the quotient class $[x,v,i]$ to ${\varphi }_i^{-1}(x,v)$ is well-defined, continuous by [F2], and in each chart is the identity map on $U_i\times {\mathbb{F}}^n$. It is therefore a bundle isomorphism from the reconstructed quotient to the original bundle.

The maps $(z,e)\mapsto e$ and $e\mapsto (q(e),e)$ are mutually inverse fiberwise-linear continuous maps ${\mathrm{id}}_Z^{\ast }E\rightleftarrows E$. They are continuous by the product and subspace formulas in [F1].

The composite comparison is $\alpha (x,(f(x),e))=(x,e)$, with inverse $(x,e)\mapsto (x,(f(x),e))$. Both maps are continuous restrictions of product-coordinate maps, are linear on each fiber, and preserve the defining equations, so [F1] makes them inverse bundle isomorphisms.

For a third base map, every route through the associativity comparisons deletes the same redundant base coordinates and ends at the same pair $(w,e)$. For a bundle map, both naturality routes apply that map to the same final $e$-coordinate. Thus the comparisons are coherent and natural.

The pullback section formula sends $x$ to $(x,s(g(f(x))))$ whether it is applied once along $gf$ or twice along $g$ and $f$; the identity formula similarly reduces to $s$. Hence sections respect identities and composition under the canonical comparisons.

First suppose the numeration in [F1] is supplied. Transport the standard Euclidean or Hermitian form to $E{\mathrm{\mid }}_{U_i}$ and call it $h_i$. Define on each fiber $h_x(v,w)={\sum }_i{\rho }_i(x)h_{i,x}(v,w)$, taking the $i$th term to be zero off $U_i$. Since $\mathrm{supp}{\rho }_i\subseteq U_i$, this zero extension is continuous near every point outside $U_i$, and local finiteness makes the sum continuous.

Each summand is positive semidefinite. At every $x$, some ${\rho }_i(x)>0$ because the coefficients sum to one; for $v\mathrm{\ne }0$, the corresponding $h_{i,x}(v,v)>0$. Hence $h_x(v,v)>0$. The formula is symmetric bilinear over $\mathbb{R}$ or conjugate-symmetric sesquilinear over $\mathbb{C}$, so it is the required metric.

If $X$ is paracompact Hausdorff, apply [A1] to obtain DC and then [F2] to the linear chart cover of $E$. This supplies a numeration, so steps 1.1–2.1 give a metric. AC is used only through this invocation of the published partition theorem; with supplied data those two steps make no choices.

Use [A1] to give $E$ a metric, or use the metric supplied in the more general clause. Let $C=i(E^{\mathrm{\prime }}{)}^{\perp }$. In a local frame adapted to the subbundle $i(E^{\mathrm{\prime }})$, the Gram matrix is continuous and positive definite; solving the finite linear equations $⟨v,i(e^{\mathrm{\prime }})⟩=0$ expresses $C$ as the graph of a continuous linear map. Hence $C$ is a vector subbundle of rank $\mathrm{rank}E-\mathrm{rank}{E}^{\mathrm{\prime }}$.

On every fiber, $E_x=i(E_x^{\mathrm{\prime }})\oplus C_x$. Exactness in [F1] gives $\ker q_x=i(E_x^{\mathrm{\prime }})$, so $q_x{\mathrm{\mid }}_{C_x}:C_x\to E_x^{\mathrm{\prime }\mathrm{\prime }}$ is injective and, by equal finite dimensions, bijective. In the local graph frames its matrix and inverse vary continuously, hence $q{\mathrm{\mid }}_C:C\to E^{\mathrm{\prime }\mathrm{\prime }}$ is a bundle isomorphism.

The map $E^{\mathrm{\prime }}\oplus C\to E$, $(e^{\mathrm{\prime }},c)\mapsto i(e^{\mathrm{\prime }})+c$, is a bundle isomorphism by the local direct-sum frames. Composing its $C$ summand with $(q{\mathrm{\mid }}_C{)}^{-1}$ yields $E^{\mathrm{\prime }}\oplus E^{\mathrm{\prime }\mathrm{\prime }}\cong E$. The result depends on the chosen metric, so no canonical splitting is asserted.

If $X=\mathrm{\varnothing }$ or $n=0$, the asserted $N=0$ is immediate. Otherwise, use [A1] to obtain DC and [F1] to choose a finite linear trivializing cover $U_1,\dots ,U_m$ with a subordinate partition ${\rho }_1,\dots ,{\rho }_m$. Let ${\varphi }_i:E{\mathrm{\mid }}_{U_i}\to U_i\times {\mathbb{F}}^n$ be the corresponding fiber coordinates.

Define $j:E\to X\times ({\mathbb{F}}^n{)}^m$ by $j(e)=(p(e),(\sqrt{{\rho }_i(p(e))}{\varphi }_i(e){)}_{i=1}^m)$, interpreting the $i$th coordinate as zero off $U_i$. Support containment makes every coordinate continuous. If $e\mathrm{\ne }0$ lies over $x$, some ${\rho }_i(x)>0$, so the $i$th coordinate is nonzero; hence each $j_x$ is injective.

In a local frame, $j$ is a continuous full-rank matrix $A(x)$. The matrix $A(x)(A(x{)}^{\ast }A(x){)}^{-1}A(x{)}^{\ast }$ is the continuous orthogonal projection onto $j(E_x)$. Its complementary projections therefore have locally constant rank $mn-n$, and [F2] makes their images a subbundle $E^{\mathrm{\prime }}\subseteq X\times {\mathbb{F}}^{mn}$. Fiberwise orthogonal decomposition gives $j(E)\oplus E^{\mathrm{\prime }}=X\times {\mathbb{F}}^{mn}$ and hence $E\oplus E^{\mathrm{\prime }}\cong X\times {\mathbb{F}}^{mn}$.

For each $x\in X$, compactness of $I$ gives a partition $0=t_0<\cdots <t_r=1$ and neighborhoods $U_{x,j}$ of $x$ such that $\xi$ is trivial on $U_{x,j}\times [t_{j-1},t_j]$. After shrinking to $U_x={\bigcap }_j{U}_{x,j}$, modify each later strip trivialization by its transition matrix at the common endpoint; consecutive trivializations then agree there and paste. Thus $\xi {\mathrm{\mid }}_{U_x\times I}$ is trivial.

Apply [A1] and [F2] to the cover $(U_x)$. Apply the countabilization [F3] to obtain a countable open cover $(V_m)$ and a locally finite partition $({\lambda }_m)$ with $\mathrm{supp}{\lambda }_m\subseteq V_m$, where each $V_m$ is a disjoint union of open sets contained in members of the original cover. Pasting the corresponding trivializations over those disjoint pieces makes $\xi {\mathrm{\mid }}_{V_m\times I}$ trivial.

Put ${\psi }_0=0$, ${\psi }_m={\sum }_{j\le m}{\lambda }_j$, and let $X_m\subseteq X\times I$ be the graph of ${\psi }_m$. Over $V_m\times I$, its chosen trivialization transports a vector vertically from $(x,{\psi }_m(x))$ to $(x,{\psi }_{m-1}(x))$. Outside $\mathrm{supp}{\lambda }_m$ the two graph points agree, so extending by the identity gives a continuous bundle isomorphism $h_m:\xi {\mathrm{\mid }}_{X_m}\to \xi {\mathrm{\mid }}_{X_{m-1}}$.

Near any $x$, choose $M$ so every ${\lambda }_m$ with $m>M$ vanishes there. On that neighborhood ${\psi }_M=1$, and $h_1\circ \cdots \circ h_M$ carries the restriction over the graph of $1$ to the graph of $0$. Enlarging $M$ does not change this map locally because the added $h_m$ are identities. These local formulas therefore define a continuous fiberwise-linear isomorphism $\xi {\mathrm{\mid }}_{X\times \{1\}}\to \xi {\mathrm{\mid }}_{X\times \{0\}}$. Reversing the finite local composites gives its continuous inverse.

By [F1] the two endpoint restrictions are precisely $f_1^{\ast }E$ and $f_0^{\ast }E$, so step 4.1 proves the assertion. AC was used in step 2.1 to supply DC and countabilize the arbitrary partition; the finite strip constructions are choice-free once their data are supplied. Different partitions and charts can give different endpoint maps, which is why no canonical isomorphism is claimed.

The orthonormal frames form a principal $\mathrm{O}(n)$-subbundle of $\mathrm{Fr}(\xi )$: Gram–Schmidt in a bundle chart gives local orthonormal frames, and any two differ by a unique orthogonal matrix. Given an orientation, let $Q$ consist of its positive orthonormal frames. In oriented orthonormal charts, $Q\cong U\times \mathrm{SO}(n)$, and every orthonormal frame is a positive one followed by an element of $\mathrm{O}(n)$. Thus $Q$ is an $\mathrm{SO}(n)$-reduction in the sense of [F1].

Conversely, let $Q$ be an $\mathrm{SO}(n)$-reduction. At $x$, transport the standard orientation of ${\mathbb{R}}^n$ through any $q\in Q_x$. Replacing $q$ by $qh$ with $h\in \mathrm{SO}(n)$ does not change the orientation. A local section of $Q$ makes the choice continuous, so it defines an orientation of $\xi$.

Starting from an orientation, step 2.1 applied to its positive orthonormal frames returns that orientation. Starting from $Q$, the positive orthonormal frames for the resulting orientation are exactly $Q$, since each $\mathrm{O}(n)$-fiber has precisely the determinant-positive coset. Hence the constructions are inverse.

An orientation-preserving isometry of metric bundles carries positive orthonormal frames to positive orthonormal frames and hence carries the associated reduction to the associated reduction. When $n=0$, both groups and the frame fiber are singletons, and the same conclusion holds.

The case $n=0$ is the one-point space by [F1]. Suppose $n>0$. Let $O(e_r)=e_{2r-1}$ and $P(e_r)=e_{2r}$. For $Q\in \{O,P\}$ put $L_s^Q=(1-s)I+sQ$. Every $L_s^Q$ is injective. For $s>0$ and a nonzero finite vector whose largest nonzero coordinate is $r$, coordinate $2r$ of $L_s^{P}v$ is $s{v}_r\mathrm{\ne }0$. For $Q=O$ and $r>1$, coordinate $2r-1$ is $s{v}_r\mathrm{\ne }0$, while for $r=1$ coordinate $1$ is $v_1$. The case $s=0$ is immediate.

If $u=(u_1,\dots ,u_n)$ is a frame, injectivity of $L_s^Q$ makes $L_s^Q{u}_1,\dots ,L_s^Q{u}_n$ independent. Writing their column matrix as $A_s^Q$, the Gram-normalized matrix $A_s^Q((A_s^Q{)}^{\ast }{A}_s^Q{)}^{-1\mathrm{/}2}$ is an orthonormal frame and varies continuously in $(u,s)$ because positive-definite finite matrices have continuous inverse square roots. For each $Q\in \{O,P\}$ this homotopes $u$ to $Q(u)$, since $L_0^Q=I$ and $Q$ is an isometry.

Let $E=(e_2,e_4,\dots ,e_{2n})$. The vectors of $O(u)$ occupy odd coordinates and are orthogonal to the vectors of $E$. Therefore, for $0\le \theta \le \pi \mathrm{/}2$, the columns $\cos \theta O(u_i)+\sin \theta e_{2i}$ are orthonormal: their pairings are $({\cos }^2\theta +{\sin }^2\theta ){\delta }_{ij}$. They give a homotopy from $O(u)$ to the constant frame $E$.

Each finite Stiefel stage is compact Hausdorff because it is a closed subspace of a finite product of unit spheres, and its coordinate inclusion into the next stage is closed. Thus [F1] is a $k_{\omega }$-decomposition. By [F2], $V_n({\mathbb{F}}^{\mathrm{\infty }})\times I$ has the weak topology tested on $V_n({\mathbb{F}}^N)\times I$. On that product stage, the $O$ and $P$ homotopies in step 2.1 land respectively in $V_n({\mathbb{F}}^{2N-1})$ and $V_n({\mathbb{F}}^{2N})$, while step 3.1 lands in a finite stage containing also the first $2n$ coordinates. The compatible stagewise formulas are continuous, so [F2] proves ordinary continuity of both parity homotopies and of their concatenation from the identity to the constant frame $E$. This is the required contraction and uses no choice principle.

In a local frame of $E$, write $j_x$ as a continuous full-rank $N\times n$ matrix $A(x)$. The orthogonal projection onto its image is $P(x)=A(x)(A(x{)}^{\ast }A(x){)}^{-1}A(x{)}^{\ast }$. Invertibility of the positive matrix $A^{\ast }A$ and continuity of matrix inversion make $P$ continuous. The graph chart of [F1] identifies a plane continuously from its projection, so $x\mapsto j(E_x)$ is continuous.

For supplied numerating charts ${\varphi }_s:E{\mathrm{\mid }}_{U_s}\to U_s\times {\mathbb{F}}^n$ and partition $({\rho }_s)$ from [F2], define $J(e)=(\sqrt{{\rho }_s(p(e))}{\varphi }_s(e){)}_{s\in S}$, with a zero coordinate off $U_s$. Support containment makes each zero extension continuous, and local finiteness makes $J$ locally land in a finite coordinate subspace. Some ${\rho }_s(x)>0$ at every $x$, so $J_x$ is injective. This proves the locally finite-coordinate assertion without any choice beyond the supplied data.

The map $\mathrm{\Theta }:E\to c_j^{\ast }{\gamma }_n^N$, $e\mapsto (p(e),j(e))$, is continuous and linear and bijective on every fiber. In the same local frame, its inverse on the image plane is represented by $(A^{\ast }A{)}^{-1}{A}^{\ast }$, which varies continuously. Hence $\mathrm{\Theta }$ is a bundle isomorphism.

Apply the countabilization [F3] to an arbitrary numeration. Using its countable chart domains in step 1.2 gives an embedding into ${\mathbb{F}}^{\mathrm{\infty }}$, continuous for the weak direct-limit topology because it locally lands in a finite stage. This is the exact AC use in the countable clause.

If $X$ is compact Hausdorff, [F4] gives $E\oplus E^{\mathrm{\prime }}\cong X\times {\mathbb{F}}^N$ for finite $N$. Restricting this isomorphism to the first summand gives a finite-dimensional bundle embedding, and steps 1.1–2.1 give its finite Grassmannian map and tautological pullback. The empty base and rank-zero cases use $N=0$ as in [F4].

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.