Relative smooth triangulation and bundle fitting — examples

The cells $\sigma \times \tau$, with $\sigma$ a simplex of $D$ and $\tau$ a simplex of $L$, form a finite convex cell complex: the intersection $({\sigma }_1\times {\tau }_1)\cap ({\sigma }_2\times {\tau }_2)=({\sigma }_1\cap {\sigma }_2)\times ({\tau }_1\cap {\tau }_2)$ is a product of common faces, hence a common face of ${\sigma }_1\times {\tau }_1$. By [F1] there is a finite compatible triangulation $K$ of $\mathrm{\mid }D\mathrm{\mid }\times \mathrm{\mid }L\mathrm{\mid }$ subdividing these cells. Define $F(x,y):=(h(x),g(y))$ and $r(x,y):=x$. Every simplex $s$ of $K$ lies in a product cell $\sigma \times \tau$, so $F{\mathrm{\mid }}_s$ is the restriction of the smooth map $(h{\mathrm{\mid }}_{\sigma })\times (g{\mathrm{\mid }}_{\tau })$ and $r{\mathrm{\mid }}_s$ is the restriction of the affine projection ${\mathrm{pr}}_1$ onto $\sigma$; on common faces the two restrictions agree, so $F$ is piecewise smooth and $r$ is affine on each simplex into a simplex of the original $D$. Pointwise $\pi F(x,y)=h(x)=h(r(x,y))$, so $\pi F=hr$ with the original $h$ unchanged.

The map $F$ is a homeomorphism onto $B\times N$ because $h$ and $g$ are homeomorphisms onto their images and $F$ is their product; its differential on the tangent star at $(x,y)$ is $(u,v)\mapsto (d{h}_x(u),d{g}_y(v))$, where $(u,v)$ runs through the tangent cone of a refined simplex contained in $\sigma \times \tau$, so $u$ lies in the cone $\sigma -x$ of the tangent star of $D$ at $x$ and $v$ in $\tau -y$. If two tangent-star vectors $(u,v)$ and $(u\mathrm{\prime },v\mathrm{\prime })$ have the same product differential, then $d{h}_x(u)=d{h}_x(u\mathrm{\prime })$ and $d{g}_y(v)=d{g}_y(v\mathrm{\prime })$. Whole-star injectivity of the two factors gives $u=u\mathrm{\prime }$ and $v=v\mathrm{\prime }$; hence the product differential is injective on the whole tangent star, even after the product-cell subdivision. The corner-face fans of the product are products of those of the factors, and their zero sets are correspondingly products, so the product triangulation is corner adapted; moreover every product face $\sigma \times \tau$ is a union of cells of the triangulation by [F1], so product corner faces are subpolyhedra. Thus $F:K\to B\times N$ is a finite corner-adapted whole-star smooth triangulation with $\pi F=hr$ and $h$ unchanged, proved directly without invoking the general bundle theorem.