Whitney Embedding Tubular Neighbourhoods and Approximation

For each $p\in M$, choose a coordinate chart $(U_p,x_p)$ and an open set $V_p⋐U_p$ containing $p$. By [L1] there is a smooth function ${\varphi }_p$ supported in $U_p$ and equal to $1$ on $V_p$. Compactness gives finitely many such $V_i$ covering $M$, with associated charts $(U_i,x_i)$ and bumps ${\varphi }_i$.

Define the coordinate-bump blocks $B_i(q):=({\varphi }_i(q),{\varphi }_i(q)x_i^1(q),\dots ,{\varphi }_i(q)x_i^n(q))\in {\mathbb{R}}^{n+1},$ and let $F:=(B_1,\dots ,B_m)$. The map is smooth because each block is smooth on $U_i$ and vanishes off $U_i$.

To prove immersion, fix $p\in M$ and choose $i$ with $p\in V_i$. Because ${\varphi }_i$ is identically $1$ on the open set $V_i$, its differential vanishes there. Thus on $V_i$ the last $n$ coordinates of $B_i$ are just the chart coordinates $x_i^1,\dots ,x_i^n$, whose differentials form an isomorphism $T_{p}M\to {\mathbb{R}}^n$. Hence $d{F}_p$ is injective.

To prove injectivity, suppose $F(p)=F(q)$. Choose $i$ with $p\in V_i$. Then ${\varphi }_i(p)=1$, hence ${\varphi }_i(q)=1$ as well because the first coordinates of $B_i(p)$ and $B_i(q)$ agree. Therefore $q\in U_i$, and the equalities ${\varphi }_i(p)x_i^a(p)={\varphi }_i(q)x_i^a(q)$ give $x_i^a(p)=x_i^a(q)$ for every $a$. Since $x_i$ is injective on $U_i$, one gets $p=q$.

Steps 3.1 and 2.2 show that $F$ is an injective immersion. By [L2], $F$ is a smooth embedding.

Apply [L1] to the trivial cover $\{M\}$ to obtain countably many relatively compact coordinate balls $U_j$ covering $M$. Shrinking each one slightly inside itself, choose open sets $V_j⋐U_j$ that still cover $M$. By [L2] there is a smooth bump ${\varphi }_j$ supported in $U_j$ and equal to $1$ on $V_j$.

Define the coordinate blocks $B_j$ as in the statement. Each $B_j$ is smooth because it equals the smooth chart-coordinate formula on $U_j$ and vanishes off $U_j$.

If $p\mathrm{\ne }q$, choose $j$ with $p\in V_j$. If $B_j(p)=B_j(q)$, then ${\varphi }_j(p)=1$, hence ${\varphi }_j(q)=1$, so both points lie in $U_j$. Equality of the last $n$ coordinates of $B_j$ then gives $x_j(p)=x_j(q)$, contradicting the injectivity of the chart map. Therefore some block separates $p$ and $q$.

Fix $p\in M$ and choose $j$ with $p\in V_j$. On $V_j$ one has ${\varphi }_j\equiv 1$, so the last $n$ coordinates of $B_j$ are exactly the chart coordinates $x_j^1,\dots ,x_j^n$. Their differential is an isomorphism at $p$, so this single block already detects every nonzero tangent vector at $p$. Thus the family $(B_j)$ separates tangent vectors as claimed.

Choose a nonnegative smooth proper exhaustion $\rho$ from [L1]. Let $K_m:={\rho }^{-1}([m-1,m+2])(m\ge 1)\mathrm{.}$ Each $K_m$ is compact and the family $(K_m)$ covers $M$. If $m\equiv m^{\mathrm{\prime }}(\mathrm{m}\mathrm{o}\mathrm{d}4)$ and $m\mathrm{\ne }{m}^{\mathrm{\prime }}$, the defining intervals are separated by a positive gap.

Put $O_m:={\rho }^{-1}((m-5\mathrm{/}4,m+9\mathrm{/}4))$. Then $K_m\subseteq O_m$, and $O_m\cap O_{m^{\mathrm{\prime }}}=\mathrm{\varnothing }$ for distinct congruent indices modulo $4$. If $K_m=\mathrm{\varnothing }$, take $Q_m=1$ and $H_m=0$; all four requirements for this index are then immediate. Henceforth suppose $K_m\mathrm{\ne }\mathrm{\varnothing }$.

For every $p\in K_m$, a chart from [F1] can be shrunk over a Euclidean ball to a coordinate domain $(U,x)$ with $p\in U$, compact closure, and $\bar{U}\subseteq O_m$. Applying [L2] to $\{p\}\subseteq U$ gives a smooth $\varphi :M\to [0,1]$ supported in $U$ and equal to $1$ on an open neighbourhood $V$ of $p$. The collection of all plateau neighbourhoods $V$ obtainable in this way covers $K_m$, so compactness selects finitely many data $(U_{mj},x_{mj},{\varphi }_{mj},V_{mj})$, $1\le j\le r_m$, whose $V_{mj}$ cover $K_m$.

For each selected datum define a global block $B_{mj}:M\to {\mathbb{R}}^{n+1}$ by $B_{mj}(q):=\{\begin{array}{ll}({\varphi }_{mj}(q),{\varphi }_{mj}(q)x_{mj}(q)),& q\in U_{mj},\\ 0,& q\mathrm{\notin }{U}_{mj}\mathrm{.}\end{array}$ On the open cover $U_{mj}\cup (M\setminus \mathrm{supp}({\varphi }_{mj}))$ the two formulas are smooth and agree on the overlap, so [L3] makes $B_{mj}$ smooth. Set $H_m^{\mathrm{\prime }}:=(B_{m1},\dots ,B_{m{r}_m}):M\to {\mathbb{R}}^{r_m(n+1)}$. Its support lies in the finite union of the compact sets $\mathrm{supp}({\varphi }_{mj})\subseteq O_m$.

The map $H_m^{\mathrm{\prime }}$ separates points of $K_m$: if $H_m^{\mathrm{\prime }}(p)=H_m^{\mathrm{\prime }}(q)$, choose $j$ with $p\in V_{mj}$. Equality of the first coordinate of the $j$th block gives ${\varphi }_{mj}(q)=1$, and equality of the remaining coordinates gives $x_{mj}(p)=x_{mj}(q)$, whence $p=q$. It also separates tangent vectors: for $p\in V_{mj}$, the function ${\varphi }_{mj}$ is locally constant with value $1$, so the last $n$ components of $d{B}_{mj,p}$ are $d{x}_{mj,p}$, an isomorphism. Thus $d{H}_{m,p}^{\mathrm{\prime }}$ is injective for every $p\in K_m$.

Compact support makes $H_m^{\mathrm{\prime }}$ bounded. Choose $C_m\ge 1$ with $\mathrm{\parallel }{H}_m^{\mathrm{\prime }}(q)\mathrm{\parallel }\le C_m$ for every $q\in M$, and put $H_m:=2^{-m}{C}_m^{-1}{H}_m^{\mathrm{\prime }}\mathrm{.}$ This positive rescaling preserves support and both separation properties, and it gives $\mathrm{\parallel }{H}_m\mathrm{\parallel }\le 2^{-m}$ everywhere.

For a nonempty band, steps 4.1 and 6.1 put $\mathrm{supp}(H_m)$ inside $O_m$; for an empty band, step 2.1 gives empty support. The sets $O_m$ are disjoint for distinct congruent indices modulo $4$, so the corresponding supports are disjoint. Together with steps 1.1, 5.1, and 6.1, this proves all four stated properties.

If $M$ is compact, [L1] already gives a smooth embedding of $M$ into some finite-dimensional Euclidean space.

If $M$ is noncompact, [F1] gives an embedding $(G,\rho )$ with $G$ bounded and $\rho$ proper.

Combining the compact case from step 1.1 and the noncompact case from step 1.2 proves the theorem.

By [L1], each point $p\in M$ has a neighbourhood $U_p$ such that $F{\mathrm{\mid }}_{U_p}$ is an embedding onto an embedded submanifold of $N$. In particular, the image of a closed subset of $U_p$ is closed in $F(U_p)$.

By [L2], $F$ is continuous and $M$ is locally compact while $N$ is Hausdorff. A proper continuous map from a locally compact Hausdorff space to a Hausdorff space is closed, so $F$ sends closed sets in $M$ to closed sets in $N$. Therefore the corestriction $F:M\to F(M)$ is a closed continuous bijection.

A closed continuous bijection onto a subspace is a homeomorphism. Thus the corestriction $F:M\to F(M)$ is a homeomorphism, while step 1.1 already gives the local embedded-submanifold model coming from the immersion. By [F1], $F$ is a smooth embedding.

The manifold $(M\times M)\setminus {\mathrm{\Delta }}_M$ has dimension $2n$, and $TM\setminus 0_M$ also has dimension $2n$. Since $S^{N-1}$ has dimension $N-1>2n$, [L1] shows that both images ${\sigma }_f((M\times M)\setminus {\mathrm{\Delta }}_M)$ and ${\tau }_f(TM\setminus 0_M)$ are null subsets of $S^{N-1}$.

By [L2], the complement of the union of those two bad sets is dense in $S^{N-1}$. Fix $u$ in that complement.

If $P_u(f(p))=P_u(f(q))$, then $f(q)-f(p)$ is parallel to $u$. Because $f$ is injective, either $p=q$ or $u=\pm {\sigma }_f(p,q)$. The second alternative is impossible by step 2.1, so $p=q$. Thus $P_u\circ f$ is injective.

If $d(P_u\circ f{)}_p(v)=0$ for some $v\mathrm{\ne }0$, then $d{f}_p(v)$ is parallel to $u$, so $u=\pm {\tau }_f(p,v)$. This again contradicts step 2.1. Hence $d(P_u\circ f{)}_p$ is injective for every $p$, and $P_u\circ f$ is an immersion.

Therefore every $u$ outside the secant and tangent images gives an injective immersion after projection, and such $u$ form a dense set.

The injectivity and immersion assertions follow exactly as in the generic-projection lemma recorded in [F1]: since $u$ is not a secant direction, distinct points cannot collapse under $P_u$, and since $u$ is not a tangent direction, no nonzero tangent vector lies in the kernel of $d(P_u\circ F)$.

Let $e=(0,1)$ be the last-coordinate unit vector and put $e^{\mathrm{\prime }}:=P_u(e)$. The hypothesis that $u$ is not parallel to $e$ is exactly $e^{\mathrm{\prime }}\mathrm{\ne }0$. Decompose $u^{\perp }=\mathbb{R}{e}^{\mathrm{\prime }}\oplus (e^{\mathrm{\prime }}{)}^{\perp }$. Since $P_u(F(p))=P_u(g(p),0)+\rho (p)e^{\mathrm{\prime }},$ its $(e^{\mathrm{\prime }}{)}^{\perp }$-component is bounded. Its scalar component along $e^{\mathrm{\prime }}\mathrm{/}\mathrm{\parallel }{e}^{\mathrm{\prime }}\mathrm{\parallel }$ is ${\rho }^{\mathrm{\prime }}(p)=\mathrm{\parallel }{e}^{\mathrm{\prime }}\mathrm{\parallel }\rho (p)+b(p),$ where $b$ is bounded.

The function ${\rho }^{\mathrm{\prime }}$ is proper. Indeed, if $J\subseteq \mathbb{R}$ is compact and $\mathrm{\mid }b\mathrm{\mid }\le B$, then ${\rho }^{\mathrm{\prime }}(p)\in J$ forces $\rho (p)$ into a bounded closed interval because $\mathrm{\parallel }{e}^{\mathrm{\prime }}\mathrm{\parallel }>0$. Thus $({\rho }^{\mathrm{\prime }}{)}^{-1}(J)$ is a closed subset of the inverse image under the proper map $\rho$ of a compact interval.

If $K\subseteq u^{\perp }$ is compact, its image under the linear coordinate along $e^{\mathrm{\prime }}$ is compact. Hence $(P_u\circ F{)}^{-1}(K)$ is a closed subset of the compact set $({\rho }^{\mathrm{\prime }}{)}^{-1}({\mathrm{pr}}_{e^{\mathrm{\prime }}}K)$ and is compact. Therefore $P_u\circ F$ is proper. Together with step 1.1, it is a proper injective immersion.

Choose the embedding $F:M\to {\mathbb{R}}^d$ from [L1]. If $M$ is compact, $F$ is automatically proper. If $M$ is noncompact, use the supplied form $F=(G,\rho )$ relative to a decomposition ${\mathbb{R}}^d=e^{\perp }\oplus \mathbb{R}e$, with $G$ bounded and $\rho$ proper.

If $d\le 2n+1$, compose $F$ with a linear isometric inclusion ${\mathbb{R}}^d\hookrightarrow {\mathbb{R}}^{2n+1}$. This composite is still a proper smooth embedding, so the theorem is proved in this case. Hence assume $d>2n+1$.

The generic-projection lemma in [L2] gives a dense set of directions $u\in S^{d-1}$ for which $P_u\circ F$ is an injective immersion. If $M$ is compact, choose any such $u$; the projected map is proper because its source is compact. If $M$ is noncompact, the set $S^{d-1}\setminus \{e,-e\}$ is a nonempty open set, so it meets the dense good-direction set. Choose $u$ in that intersection. Then $u$ is not parallel to the proper-coordinate axis, and the corrected properness lemma in [L2] makes $P_u\circ F$ proper. In either case [L3] upgrades the proper injective immersion to a smooth embedding into $u^{\perp }\cong {\mathbb{R}}^{d-1}$.

In the noncompact case put $e^{\mathrm{\prime }}:=P_u(e)\mathrm{\ne }0$ and decompose $u^{\perp }=\mathbb{R}{e}^{\mathrm{\prime }}\oplus (e^{\mathrm{\prime }}{)}^{\perp }$. The component of $P_u\circ F=P_u(G)+\rho e^{\mathrm{\prime }}$ perpendicular to $e^{\mathrm{\prime }}$ is bounded. Its scalar component along $e^{\mathrm{\prime }}\mathrm{/}\mathrm{\parallel }{e}^{\mathrm{\prime }}\mathrm{\parallel }$ is $\mathrm{\parallel }{e}^{\mathrm{\prime }}\mathrm{\parallel }\rho +b$ with $b$ bounded, and the argument in the properness lemma shows this function is proper. Thus the projected embedding again has bounded-plus-proper form, now with proper-coordinate unit vector $e^{\mathrm{\prime }}\mathrm{/}\mathrm{\parallel }{e}^{\mathrm{\prime }}\mathrm{\parallel }$.

If the new ambient dimension is still greater than $2n+1$, repeat steps 3.1-4.1. Step 3.1 restores the embedding hypothesis after each projection; compactness preserves properness in the compact case, and step 4.1 preserves the bounded-plus-proper form in the noncompact case. After the finite number $d-(2n+1)$ of projections, the ambient dimension is $2n+1$ and the resulting map is a proper smooth embedding.

The low-dimensional branch is step 2.1, and the projection branch is step 5.1. Therefore every smooth $n$-manifold admits a proper smooth embedding into ${\mathbb{R}}^{2n+1}$.

By [F1], the smooth manifold $M$ admits a smooth immersion into ${\mathbb{R}}^{2n}$.

This is exactly the claimed statement.

At a zero vector $(p,0)$, the tangent space of the normal bundle splits as $T_{(p,0)}(N^{\perp }S)\cong T_{p}S\oplus N_p^{\perp }S\mathrm{.}$ With the formula in [F1], the differential sends $(u,w)$ to $u+w\in {\mathbb{R}}^m$.

Because $T_{p}S$ and $N_p^{\perp }S$ are orthogonal complementary subspaces of ${\mathbb{R}}^m$, the map $(u,w)\mapsto u+w$ is a linear isomorphism. Hence $d{E}_{(p,0)}$ is invertible.

Apply [L1] at each $(p,0)$. The map $E$ is a local diffeomorphism along the zero section.

By [L1], for every $y\in S$ there is some ${\epsilon }_y>0$ such that the normal addition map is a diffeomorphism on $V_{{\epsilon }_y}(y):=\{(y^{\mathrm{\prime }},v^{\mathrm{\prime }})\in N^{\perp }S:\mathrm{\parallel }{y}^{\mathrm{\prime }}-y\mathrm{\parallel }<{\epsilon }_y,\mathrm{\parallel }{v}^{\mathrm{\prime }}\mathrm{\parallel }<{\epsilon }_y\}\mathrm{.}$ Define $r(y):=\min (1,\sup \{\epsilon >0:E{\mathrm{\mid }}_{V_{\epsilon }(y)}\text{ is a diffeomorphism}\})\mathrm{.}$ Then $r(y)>0$ for every $y\in S$.

The function $r$ is continuous. Indeed, if $\mathrm{\parallel }y-y^{\mathrm{\prime }}\mathrm{\parallel }<r(y)$ and $0<\epsilon <r(y)-\mathrm{\parallel }y-y^{\mathrm{\prime }}\mathrm{\parallel }$, then $V_{\epsilon }(y^{\mathrm{\prime }})\subseteq V_{r(y)}(y)$, so $E{\mathrm{\mid }}_{V_{\epsilon }(y^{\mathrm{\prime }})}$ is also a diffeomorphism. Hence $r(y^{\mathrm{\prime }})\ge r(y)-\mathrm{\parallel }y-y^{\mathrm{\prime }}\mathrm{\parallel }$. Swapping $y$ and $y^{\mathrm{\prime }}$ gives $\mathrm{\mid }r(y)-r(y^{\mathrm{\prime }})\mathrm{\mid }\le \mathrm{\parallel }y-y^{\mathrm{\prime }}\mathrm{\parallel }$.

By [L2], choose a smooth positive function $\delta :S\to (0,\mathrm{\infty })$ with $\delta (y)\le r(y)\mathrm{/}2$ for every $y\in S$.

Suppose $E(y,v)=E(y^{\mathrm{\prime }},v^{\mathrm{\prime }})$ with both points in ${\mathrm{\Omega }}_{\delta }$, and assume without loss of generality that $r(y^{\mathrm{\prime }})\le r(y)$. Then $\mathrm{\parallel }y-y^{\mathrm{\prime }}\mathrm{\parallel }=\mathrm{\parallel }v-v^{\mathrm{\prime }}\mathrm{\parallel }\le \mathrm{\parallel }v\mathrm{\parallel }+\mathrm{\parallel }{v}^{\mathrm{\prime }}\mathrm{\parallel }<\delta (y)+\delta (y^{\mathrm{\prime }})\le \frac{r(y)}{2}+\frac{r(y^{\mathrm{\prime }})}{2}\le r(y)\mathrm{.}$ Also $\mathrm{\parallel }v\mathrm{\parallel }<\delta (y)\le r(y)\mathrm{/}2<r(y)$ and $\mathrm{\parallel }{v}^{\mathrm{\prime }}\mathrm{\parallel }<\delta (y^{\mathrm{\prime }})\le r(y^{\mathrm{\prime }})\mathrm{/}2\le r(y)\mathrm{/}2<r(y)$. Therefore both $(y,v)$ and $(y^{\mathrm{\prime }},v^{\mathrm{\prime }})$ lie in $V_{r(y)}(y)$, where $E$ is injective by step 1.1. Hence $(y,v)=(y^{\mathrm{\prime }},v^{\mathrm{\prime }})$. So $E$ is injective on ${\mathrm{\Omega }}_{\delta }$.

If $S=\mathrm{\varnothing }$, take the unique function $\delta :S\to (0,\mathrm{\infty })$. Then ${\mathrm{\Omega }}_{\delta }=\mathrm{\varnothing }$, and $E$ is a diffeomorphism from the empty manifold onto the open neighbourhood $\mathrm{\varnothing }$ of $S$. Hence assume $S\mathrm{\ne }\mathrm{\varnothing }$.

Let $W$ be the union of all normal-bundle neighbourhoods on which [L2] makes $E$ a local diffeomorphism. It is open and contains the zero section. By shrinking bundle trivializations around their base points, choose an open cover $(U_i)$ of $S$ and numbers $r_i>0$ such that $\{(p,v):p\in U_i,\mathrm{\parallel }v\mathrm{\parallel }<r_i\}\subseteq W\mathrm{.}$ By [L3], choose a locally finite smooth partition $({\varphi }_i)$ subordinate to this cover.

Define $r(p):={\left({\sum }_i\frac{{\varphi }_i(p)}{r_i}\right)}^{-1}\mathrm{.}$ The locally finite sum is smooth and positive. At each $p$, the finite nonempty set $I(p):=\{i:{\varphi }_i(p)>0\}$ has an index $i_0$ with $r_{i_0}={\max }_{i\in I(p)}{r}_i$. Since $r(p)\le r_{i_0}$ and $p\in \mathrm{supp}({\varphi }_{i_0})\subseteq U_{i_0}$, the whole fibre ball $\mathrm{\parallel }v\mathrm{\parallel }<r(p)$ over $p$ lies in $W$.

Let ${\delta }_0:S\to (0,\mathrm{\infty })$ be the positive smooth injectivity radius supplied by [L2], and put $\delta (p):=\frac{{\delta }_0(p)r(p)}{{\delta }_0(p)+r(p)}\mathrm{.}$ This function is positive and smooth, with $\delta <{\delta }_0$ and $\delta <r$.

The set ${\mathrm{\Omega }}_{\delta }$ is open in $N^{\perp }S$ because $(p,v)\mapsto \mathrm{\parallel }v\mathrm{\parallel }-\delta (p)$ is continuous. Since $\delta <r$, step 3.1 gives ${\mathrm{\Omega }}_{\delta }\subseteq W$, so $E$ is a local diffeomorphism at every point of ${\mathrm{\Omega }}_{\delta }$. Since $\delta <{\delta }_0$, [L2] also makes $E$ injective on ${\mathrm{\Omega }}_{\delta }$.

A local diffeomorphism is open. Hence $U:=E({\mathrm{\Omega }}_{\delta })$ is open and contains $S$ because $E(p,0)=p$. The injective local diffeomorphism $E:{\mathrm{\Omega }}_{\delta }\to U$ is a homeomorphism, and its local smooth inverses agree and assemble to a smooth global inverse. Thus $E$ is the required diffeomorphism. Together with the empty case in step 1.1, this proves the theorem.

Let $E:{\mathrm{\Omega }}_{\delta }\to U$ be the tubular diffeomorphism from [L1]. The bundle projection $\pi :{\mathrm{\Omega }}_{\delta }\to S,\pi (p,v)=p,$ is smooth.

Define $r:=\pi \circ E^{-1}:U\to S$. This map is smooth, and for $p\in S$ one has $E^{-1}(p)=(p,0)$, so $r(p)=p$. Hence $r$ is a smooth retraction.

Write $x=E(p,v)=p+v$ in the tubular coordinates from [L1]. Because $v$ is orthogonal to $T_{p}S$, the function $q\mapsto \mathrm{\parallel }x-q{\mathrm{\parallel }}^2$ has vanishing first derivative at $q=p$. Its Hessian on the tangent directions equals the Euclidean metric plus terms that go to zero with $\mathrm{\parallel }v\mathrm{\parallel }$. Therefore, after shrinking the radius if necessary, $q=p$ is a strict local minimizer on each normal fibre.

On each compact piece of $S$, the radius can be shrunk once more so that this local minimizer is the only point of $S$ at the same or smaller distance from $x$. Applying this on a locally finite cover yields a still smaller tubular neighbourhood on which every point has a unique nearest point in $S$.

In the tubular coordinates, that unique nearest point is exactly the base point $p$ of the normal vector $v$. But [L2] defines the tubular retraction by sending $x=E(p,v)$ to $p$. Hence the nearest-point projection and the tubular retraction agree on the shrunken tube.

By [F1], the embedded submanifold $S$ has a tubular neighbourhood in $M$.

By [L1], this means there is an open neighbourhood $\mathrm{\Omega }$ of the zero section in the normal bundle and a diffeomorphism from $\mathrm{\Omega }$ onto an open neighbourhood of $S$ in $M$ that restricts to the inclusion on the zero section. This is exactly the claimed statement.

Let $\pi :\mathrm{\Omega }\to S$ be the bundle projection, and define $r:=\pi \circ {\mathrm{\Phi }}^{-1}:U\to S\mathrm{.}$ This map is smooth because $\pi$ and ${\mathrm{\Phi }}^{-1}$ are smooth.

For $p\in S$, one has ${\mathrm{\Phi }}^{-1}(p)=(p,0)$, so $r(p)=p$. Therefore $r$ is a smooth retraction of $U$ onto $S$.