Compatible tubular charts realize a prescribed normal identification

By [F1] fix a tubular chart ${\mathrm{\Phi }}_0$, and by [F2] fix a Riemannian metric $g$ on $M$; then [F3] exhibits $\nu (S)$ as the smooth quotient bundle identified with $C$ and makes ${\pi }^{\perp }$ smooth. Let $\mathrm{\Psi }:\mathrm{\Omega }\subseteq F\to M$ be any chart of a smooth bundle $F\to S$ of rank equal to $\mathrm{codim}S$ with $\mathrm{\Psi }(0_s)=i(s)$. Write $z$ for the zero section and identify $T_{(s,0)}F=d{z}_s(T_{s}S)\oplus F_s$, where $F_s=\ker d{\pi }_F$ is the vertical subspace. Since $\mathrm{\Psi }\circ z=i$, one has $d{\mathrm{\Psi }}_{(s,0)}(d{z}_s(u))=d{i}_s(u)$; hence $d{\mathrm{\Psi }}_{(s,0)}$ sends the horizontal summand isomorphically onto $d{i}_s(T_{s}S)$ and $F_s$ isomorphically onto a complement of it. The quotient class ${\beta }_s(v):=[d{\mathrm{\Psi }}_{(s,0)}(v)]\in \nu (S{)}_s$ is therefore a well-defined linear map $F_s\to \nu (S{)}_s$.

With $j:C\to \nu (S)$ the identification of [F3] and ${\tilde{\beta }}_s:={\pi }^{\perp }\circ d{\mathrm{\Psi }}_{(s,0)}{\mathrm{\mid }}_{F_s}$ one has $\beta =j\circ \tilde{\beta }$, because $d\mathrm{\Psi }(v)-{\pi }^{\perp }d\mathrm{\Psi }(v)$ lies in $d{i}_s(T_{s}S)$. In local frames of $F$ and $TM{\mathrm{\mid }}_S$ the components of $d{\mathrm{\Psi }}_{(s,0)}{\mathrm{\mid }}_{F_s}$ are smooth functions of $s$, since $\mathrm{\Psi }$ is smooth, and ${\pi }^{\perp }$ has smooth local matrices by [F3]; so [F4] makes $\tilde{\beta }$ and $\beta$ smooth bundle maps over ${\mathrm{i}\mathrm{d}}_S$. Fibrewise, ${\pi }^{\perp }d\mathrm{\Psi }(v)=0$ forces $d\mathrm{\Psi }(v)\in d{i}_s(T_{s}S)$, hence $v=0$ by the splitting and injectivity of $d\mathrm{\Psi }$; since $\mathrm{rank}F=\mathrm{codim}S=\mathrm{rank}C$, each ${\tilde{\beta }}_s$ and ${\beta }_s$ is bijective. By [F5], $\beta$ is a smooth bundle isomorphism.

Apply step 2.1 to $F=\nu (S)$ and $\mathrm{\Psi }={\mathrm{\Phi }}_0$: the induced map ${\beta }_0$ is a smooth bundle automorphism of $\nu (S)$. Put $\gamma ={\beta }_0^{-1}\circ \alpha :E\to \nu (S)$ and $\mathrm{\Phi }={\mathrm{\Phi }}_0\circ \gamma :{\gamma }^{-1}({\mathrm{\Omega }}_0)\to U_0$. Since $\gamma$ is a smooth bundle isomorphism over ${\mathrm{i}\mathrm{d}}_S$, the set ${\gamma }^{-1}({\mathrm{\Omega }}_0)$ is open and contains $0_S$, and $\mathrm{\Phi }$ is a diffeomorphism with $\mathrm{\Phi }(0_s)={\mathrm{\Phi }}_0(0_s)=i(s)$. For $v\in E_s$ one has $d{\gamma }_{(s,0)}(v)={\gamma }_s(v)$, because $\gamma$ is fibrewise linear over ${\mathrm{i}\mathrm{d}}_S$; hence by the chain rule [F6] the induced map of $\mathrm{\Phi }$ at $s$ is ${\beta }_{0,s}\circ {\gamma }_s={\beta }_{0,s}\circ {\beta }_{0,s}^{-1}\circ {\alpha }_s={\alpha }_s$.

Taking $E=\nu (S)$ and $\alpha =\mathrm{i}\mathrm{d}$ in step 3.1 gives the chart ${\mathrm{\Phi }}_0\circ {\beta }_0^{-1}$, which induces the identity, so the normal bundle admits a chart in the specified compatible class. If $S=\mathrm{\varnothing }$ then $E$, $\nu (S)$, ${\mathrm{\Omega }}_0$ and $U_0$ are empty and the condition is vacuous; if $E$ has rank zero then $\mathrm{codim}S=0$ and ${\beta }_s$ is the unique isomorphism between zero spaces, so step 3.1 still applies. The isomorphism $\gamma$ is determined by the supplied chart and $\alpha$, so no object is selected beyond [A1]; the metric of [F2] only exhibits the smooth structure and does not enter ${\beta }_0$.