Variable-radius injectivity for normal addition

If $S$ is empty, its empty function is a positive smooth radius with the stated injectivity property. Suppose henceforth that $S$ is nonempty. For $p\in S$ and $a>0$ put $V_a(p)=\{(q,v)\in N^{\perp }S:\mathrm{\parallel }q-p\mathrm{\parallel }<a,\mathrm{\parallel }v\mathrm{\parallel }<a\}\mathrm{.}$ By [L2] these are open and form a neighbourhood basis at $(p,0)$. By [L1] some such set lies in a neighbourhood on which $E$ is a diffeomorphism. Thus $A_p=\{a\in (0,1]:E{\mathrm{\mid }}_{V_a(p)}\text{ is a diffeomorphism onto its image}\}$ is nonempty and bounded above by one. Define $r(p)=\sup A_p\in (0,1]$. No simultaneous choice of local radii is made, and the cap is imposed before taking the supremum.

For every $0<a<r(p)$, the supremum property supplies $b\in A_p$ with $a<b$, so restriction to the open set $V_a(p)$ shows $a\in A_p$. Any two points of $V_{r(p)}(p)$ lie in some common $V_a(p)$ with $a<r(p)$, by taking $a$ larger than their finitely many strict distance bounds. Hence $E$ is injective on $V_{r(p)}(p)$. Each point there also has an open neighbourhood in one such $V_a(p)$, on which $E$ is a local diffeomorphism. Therefore its image is open, and these local smooth inverses agree by injectivity, proving that $E{\mathrm{\mid }}_{V_{r(p)}(p)}$ itself is a diffeomorphism onto its image. This establishes the supremum endpoint rather than assuming it is eligible.

Let $p,q\in S$. For $0<a<r(p)-\mathrm{\parallel }p-q\mathrm{\parallel }$ one has $a<1$ and $V_a(q)\subseteq V_{r(p)}(p)$ by the triangle inequality. Step 2.1 and restriction imply $a\in A_q$, so $r(q)\ge a$. If $r(p)-\mathrm{\parallel }p-q\mathrm{\parallel }>0$, taking the supremum over these $a$ gives $r(q)\ge r(p)-\mathrm{\parallel }p-q\mathrm{\parallel }$; if not, the inequality follows from $r(q)>0$. Interchanging the points gives $\mathrm{\mid }r(p)-r(q)\mathrm{\mid }\le \mathrm{\parallel }p-q\mathrm{\parallel }\mathrm{.}$ Thus $r$ is positive and continuous.

Use [A1] to fix a smooth nonnegative proper exhaustion $\rho$. For integers $n\ge 0$ put $K_n=\{p:\rho (p)\le n+1\}$, a compact set. Define $a_n=\{\begin{array}{ll}\frac{1}{4}\min \{1,{\min }_{p\in K_n}r(p)\},& K_n\mathrm{\ne }\mathrm{\varnothing },\\ \frac{1}{4},& K_n=\mathrm{\varnothing }\mathrm{.}\end{array}$ When $K_n$ is nonempty its induced metric is compact, since $S$ is embedded, and [L5] and step 3.1 give a positive minimum. Thus $a_n>0$. Each coefficient is a uniquely specified real value, so defining all of them requires no additional countable selection.

For $t\in \mathbb{R}$ and integers $n\ge 0$ put ${\psi }_n(t)=\sigma (t-n+1)\sigma (n+1-t)\mathrm{.}$ These functions are smooth and nonnegative, supported in $[n-1,n+1]$, and positive for $n-1<t<n+1$. The supports are locally finite on $\mathbb{R}$: a bounded neighbourhood of any fixed $t$ meets only finitely many such integer-indexed intervals. Their positive sets cover $(-1,\mathrm{\infty })$; in particular ${\psi }_0(0)=1$, and every nonnegative integer $n$ has ${\psi }_n(n)=1$. By [L4], $B(t)={\sum }_{n\ge 0}{\psi }_n(t),C(t)={\sum }_{n\ge 0}{a}_n{\psi }_n(t)$ are smooth, with $B,C>0$ on $(-1,\mathrm{\infty })$. Therefore $F=C\mathrm{/}B$ is smooth and positive there.

Set $\delta =F\circ \rho$, a positive smooth function by [L4]. At a point $p$, if ${\psi }_n(\rho (p))>0$, then $\rho (p)<n+1$ and hence $p\in K_n$. Consequently $a_n\le r(p)\mathrm{/}4$. As $F(\rho (p))$ is the weighted average of these finitely many active coefficients with nonnegative weights summing to one, $0<\delta (p)\le r(p)\mathrm{/}4<r(p)\mathrm{/}2.$ Empty sublevels never contribute an active term. At endpoints a bump is zero, while the integer-centred bump is positive; thus no zero denominator occurs. This construction works equally on a zero-dimensional manifold.

Suppose $E(p,v)=E(q,w)$ with both points in ${\mathrm{\Omega }}_{\delta }$, and arrange $r(q)\le r(p)$. By [L2] and step 6.1, $\mathrm{\parallel }p-q\mathrm{\parallel }=\mathrm{\parallel }w-v\mathrm{\parallel }\le \mathrm{\parallel }v\mathrm{\parallel }+\mathrm{\parallel }w\mathrm{\parallel }<\delta (p)+\delta (q)\le \frac{r(p)+r(q)}{4}\le \frac{r(p)}{2}<r(p)\mathrm{.}$ Both normal-vector lengths are less than $r(p)$ as well. Thus both points lie in $V_{r(p)}(p)$, where step 2.1 proves injectivity, so they coincide. Therefore $E$ is injective on ${\mathrm{\Omega }}_{\delta }$. Together with the empty case this proves the claim.