Finite-rank complement theorem over compact Hausdorff bases

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 $E\oplus E^{\mathrm{\prime }}\cong X\times {\mathbb{F}}^{mn}$.