Whitney Embedding Tubular Neighbourhoods and Approximation

Let $\mathcal{T}$ be the family of all triples $((U,x),V,\varphi )$ where $(U,x)$ is a coordinate chart, $V⋐U$ is open, and $\varphi :M\to [0,1]$ is a smooth bump supported in $U$ and equal to $1$ on $V$. By [L1], the family of their open sets $V$ covers $M$. Compactness gives finitely many such sets $V_i$ covering $M$; retain a witnessing chart $(U_i,x_i)$ and bump ${\varphi }_i$ for each of these finitely many sets. This uses only finite choices after the subcover is found.

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.

Let $\mathcal{V}$ consist of all smooth coordinate balls $V$ for which there exists a smooth coordinate ball $(U,x)$ with $\bar{V}$ compact and contained in $U$. This is an open cover of $M$: at a point $p$, take a smooth chart $(W,x)$ and concentric Euclidean balls $B(x(p),r)\subset B(x(p),R)$ whose outer closed ball lies in $x(W)$. Their inverse images give the required $V$ and $U$ by [L1]. By [L2], choose a countable subcover $(V_j)$. For each $j$, use countable choice to select a witnessing smooth coordinate ball $(U_j,x_j)$, and then apply [L3] and countable choice to select a smooth ${\varphi }_j:M\to [0,1]$ supported in $U_j$ and equal to $1$ on an open neighbourhood of $\overline{V_j}$. Thus $V_j⋐U_j$ and ${\varphi }_j=1$ on $V_j$. If $M$ is empty, take the family to be empty.

Define the coordinate blocks $B_j$ by the piecewise formula in the statement. The chart-coordinate formula is smooth on $U_j$. Its coordinate products vanish on a neighbourhood of each point outside $U_j$, because $\mathrm{supp}{\varphi }_j\subseteq U_j$ is closed. Hence extension by zero is smooth on all of $M$.

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] under countable choice 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 gives a finite covering list of data $(U_{mj},x_{mj},{\varphi }_{mj},V_{mj})$, $1\le j\le r_m$. Countable choice selects one such list for each nonempty band $K_m$.

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 $M=\mathrm{\varnothing }$, take the empty maps; all conclusions hold. Otherwise choose a countable smooth atlas as follows. Fix a countable topological basis and form all smooth chart tuples. For each nonempty basis member contained in a chart domain, [A1] selects one such tuple. These chart domains cover $M$ because every point lies in a basis member inside some chart. Enumerate them as $(U_j,x_j)$, repeating one if the family is finite. This construction records chart witnesses rather than assuming a countable-atlas theorem with missing choices.

We first prove a Euclidean estimate for later use. Let $f:D\to {\mathbb{R}}^b$ be $C^1$, where $D\subseteq {\mathbb{R}}^a$ is open and $0<a<b$. Its domain is covered by the countable family of positive-side rational closed cubes $Q\subseteq D$, by [L4]. For a fixed such cube choose a slightly larger closed cube $Q^{\mathrm{\prime }}$ with $Q\subseteq \mathrm{int}{Q}^{\mathrm{\prime }}\subseteq Q^{\mathrm{\prime }}\subseteq D$. Compactness and continuity of the finitely many derivative entries bound the derivative norm on $Q^{\mathrm{\prime }}$ by some finite $L\ge 1$. Apply [L7] on the convex OPEN set $\mathrm{int}{Q}^{\mathrm{\prime }}$ to see that $f{\mathrm{\mid }}_Q$ is $L$-Lipschitz. If $Q$ has side $\mathrm{\ell }$, subdividing into $s^a$ cubes gives an image cover by $s^a$ closed $b$-cubes of side $2L\sqrt{a}\mathrm{\ell }\mathrm{/}s$, centred at images of the prescribed lower corners. Its total volume is $(2L\sqrt{a}\mathrm{\ell }{)}^b{s}^{a-b},$ which tends to zero. Thus $f(Q)$ has content zero.

A nonempty open subset of ${\mathbb{R}}^b$, $b\ge 1$, is not null. Indeed it contains a positive-side closed cube $Q_0$. If this cube were null, [L6] would give finite cube covers with arbitrarily small total volume. Every finite cube cover of $Q_0$ has total volume at least $\mathrm{vol}(Q_0)$: subdivide a bounding rectangle at all cover-cube and $Q_0$ endpoint coordinates. Each positive-volume cell in $Q_0$ lies in some covering cube, as is seen at its midpoint, and can be assigned to the least such index. By the finite volume additivity in [L5], the sum of volumes assigned to one covering cube is at most its volume, while all cells of $Q_0$ have total volume $\mathrm{vol}(Q_0)>0$. This contradicts content zero.

Choose $\rho \ge 0$ from [L1]. For each integer $m\ge 1$ set $K_m={\rho }^{-1}([m-1,m+2]),O_m={\rho }^{-1}((m-5\mathrm{/}4,m+9\mathrm{/}4))\mathrm{.}$ The sets $K_m$ are compact by [L1] and [L2], they cover $M$, and $K_m\subseteq O_m$. The open sets $O_m$ and $O_{m^{\mathrm{\prime }}}$ are disjoint when $m\mathrm{\ne }{m}^{\mathrm{\prime }}$ and $m\equiv m^{\mathrm{\prime }}(\mathrm{m}\mathrm{o}\mathrm{d}4)$. The entire family $(O_m)$ is locally finite: the open neighbourhood ${\rho }^{-1}((\rho (p)-1,\rho (p)+1))$ can meet $O_m$ only for integers $m$ in a bounded real interval, hence only finitely many $m$.

Taking the countable union over those rational cubes and using [L5] shows that $f(D)$ is null. If $a=0<b$, $D$ is empty or a singleton and the same conclusion follows from a degenerate zero-volume cube. In particular this estimate applies to any $C^1$ map on an open subset of ${\mathbb{R}}^{2n}$ with values in ${\mathbb{R}}^{d-1}$ when $d>2n+1$. It uses only Euclidean cube nullity, not atlas invariance of manifold null sets or an equidimensional Lipschitz theorem on the wrong domain.

For a fixed nonempty $K_m$, take all tuples $(U,x,\chi ,V)$ in which $(U,x)$ is a relatively compact smooth coordinate ball with $\bar{U}\subseteq O_m$, $\chi :M\to [0,1]$ is smooth with support in $U$, and $V\subseteq U$ is open with $\chi =1$ on $V$. Their $V$-sets cover $K_m$: at a fixed point choose a small chart and apply [L3] to the compact singleton in its domain. By [L2] there is a finite ordered list of such tuples whose $V$-sets cover $K_m$. This is a finite subcover of the set of ALL eligible data, not a choice of data at every point.

We now prove the required projection assertion locally. Let $K\subseteq M$ and let $h:M\to {\mathbb{R}}^d$ be smooth, injective on $K$, with $d{h}_p$ injective for every $p\in K$, where $d>2n+1$. Write $h=(h^{\mathrm{\prime }},h_d)$ with $h^{\mathrm{\prime }}$ taking values in ${\mathbb{R}}^{d-1}$. For a parameter $v\in {\mathbb{R}}^{d-1}$ define $L_v(z^{\mathrm{\prime }},z_d)=z^{\mathrm{\prime }}-v{z}_d$. Bad secant parameters are values of $\frac{h^{\mathrm{\prime }}(q)-h^{\mathrm{\prime }}(p)}{h_d(q)-h_d(p)}$ wherever the denominator is nonzero. In every chart pair from step 1.1 this is a smooth map on an open subset of ${\mathbb{R}}^{2n}$. By steps 2.1 and [L5], its image, united over countably many chart pairs, is null.

In a source chart with coordinates $x$, write $h_j=h\circ x_j^{-1}$. Bad tangent parameters are values of $\frac{D{h}_j^{\mathrm{\prime }}(x)w}{D{h}_{d,j}(x)w}$ on the open subset of $(x_j(U_j))\times {\mathbb{R}}^n$ where the denominator is nonzero. This is smooth because $h$ is smooth, and its domain has dimension $2n$. Steps 2.1 and [L5] make the union of these images over all chart indices null. Thus the union of bad secant and tangent parameters is null in ${\mathbb{R}}^{d-1}$ and cannot be all of it by step 1.3. Choose $v$ outside that union.

For one such finite list, define each coordinate-bump block on $U$ by $(\chi ,\chi x^1,\dots ,\chi x^n)$ and by zero outside $U$. It is smooth: on the open cover $U$ and $M\setminus \mathrm{supp}\chi$ the formulas are smooth and agree; continuity follows by the same open-cover argument. Concatenate the finitely many blocks into $B_m:M\to {\mathbb{R}}^{q_m}$. Its support is contained in a finite union of compact supports inside $O_m$. If $B_m(p)=B_m(q)$ with $p,q\in K_m$, a block whose $\chi$ equals one near $p$ also has $\chi (q)=1$, and equality of its chart coordinates gives $p=q$. At $p\in K_m$ that block's differential includes the invertible chart differential, so $d{B}_{m,p}$ is injective.

The map $L_{v}h$ is injective on $K$. For if its values at $p,q\in K$ agree and $h_d(p)\mathrm{\ne }{h}_d(q)$, the quotient in step 3.1 equals $v$, a contradiction. If $h_d(p)=h_d(q)$, the equality also gives $h^{\mathrm{\prime }}(p)=h^{\mathrm{\prime }}(q)$, hence $p=q$. Similarly, if $d(L_{v}h{)}_p(w)=0$ for $p\in K$ and a tangent vector $w$, a nonzero last component of $d{h}_p(w)$ would make $v$ a bad tangent parameter. A zero last component forces all of $d{h}_p(w)$ to be zero, so $w=0$. This proves the immersion property on $K$. Notice that $h$ was never assumed to be a global embedding.

The continuous map $B_m$ is bounded. To see this without assuming a metric on $M$, its compact support is covered by the increasing open sets $\{p:\mathrm{\parallel }{B}_m(p)\mathrm{\parallel }<r\}$, $r\ge 1$ an integer; a finite subcover gives one bound, and $B_m=0$ off the support. Choose the least integer $c_m\ge 1$ strictly bounding its norm and set $H_m=2^{-m}{c}_m^{-1}{B}_m$. Then $\mathrm{\parallel }{H}_m\mathrm{\parallel }\le 2^{-m}$, its support lies in $O_m$, and it retains both separation properties on $K_m$. If $K_m$ is empty take $H_m=0$ into $\mathbb{R}$, with the properties vacuous.

Replace $L_v$ by $T_v=(1+\mathrm{\parallel }v\mathrm{\parallel }{)}^{-1}{L}_v$. This nonzero rescaling preserves both properties and is a contraction, since $\mathrm{\parallel }{L}_v(z^{\mathrm{\prime }},z_d)\mathrm{\parallel }\le \mathrm{\parallel }{z}^{\mathrm{\prime }}\mathrm{\parallel }+\mathrm{\parallel }v\mathrm{\parallel }\mathrm{\mid }{z}_d\mathrm{\mid }\le (1+\mathrm{\parallel }v\mathrm{\parallel })\mathrm{\parallel }(z^{\mathrm{\prime }},z_d)\mathrm{\parallel }\mathrm{.}$ Repeat the proved assertion a finite number of times until the target dimension is $D=2n+1$; if the initial dimension is smaller, pad by zero coordinates. Thus every such $h$ has a linear contraction $T$ into ${\mathbb{R}}^D$ preserving injectivity and injective differential on $K$. Each finite list of admissible parameters exists by finite induction; no countably dependent selection has been used.

Use [A1] on the countable family of nonempty sets of finite covering lists from step 2.2 to obtain the maps $H_m$ simultaneously. For each resulting $H_m$, step 5.1 supplies a nonempty set of finite projection certificates; a second application of [A1] selects one certificate for each $m$. Let $T_m$ be its contraction into the fixed ${\mathbb{R}}^D$, and put $C_m=T_m{H}_m$. Then $C_m$ is smooth, supported in $O_m$, bounded by $2^{-m}$, injective on $K_m$, and has injective differential there. This explicitly accounts for the countably many variable-length finite projection constructions.

For $r=0,1,2,3$ put $G_r={\sum }_{\begin{array}{c}m\ge 1\\ m\equiv r(4)\end{array}}{C}_m,G=(G_0,G_1,G_2,G_3)\mathrm{.}$ Local finiteness from step 1.4 and [L3] make these maps smooth componentwise. In each colour at most one term is nonzero at any point, by the disjointness of the $O_m$. Thus $\mathrm{\parallel }{G}_r\mathrm{\parallel }\le 1$ and $\mathrm{\parallel }G\mathrm{\parallel }\le 2$ on all of $M$. Define $J=(G,\rho )$.

If $J(p)=J(q)$, the equality $\rho (p)=\rho (q)$ puts both points in some common $K_m$, since the intervals defining the $K_m$ cover $[0,\mathrm{\infty })$. On $O_m$ the colour $m<mrow><mi\; mathvariant="normal">m</mi><mi\; mathvariant="normal">o</mi><mi\; mathvariant="normal">d</mi></mrow>4$ component of $G$ is exactly $C_m$, all other terms of that colour vanishing there. Hence $C_m(p)=C_m(q)$ and step 6.1 gives $p=q$. At a point $p\in K_m$, the same component agrees with $C_m$ on the open neighbourhood $O_m$, so its differential is injective. Therefore $d{J}_p$ is injective. This proves that $J$ is an injective immersion.

The map $J$ is proper. For a compact subset $B$ of its Euclidean target, compactness bounds the last coordinate above by some $R$ and makes $B$ closed. Thus $J^{-1}(B)$ is a closed subset of the compact sublevel set $\{\rho \le R\}$, hence compact by [L1] and [L2].

Finally $J$ is a homeomorphism onto its image. Indeed, for a closed subset $A\subseteq M$ and a point $z\mathrm{\notin }J(A)$, take the closed Euclidean unit ball $B$ about $z$. By step 8.2 the set $A\cap J^{-1}(B)$ is compact. Its continuous image $C$ is compact: pull back any ambient open cover, take a finite subcover, and map it forward. Thus $C$ is closed by [L2]. The open neighbourhood $\mathrm{int}B\setminus C$ of $z$ misses $J(A)$, showing that $J(A)$ is closed in the ambient Euclidean space. Hence the continuous bijection $J:M\to J(M)$ is closed and has continuous inverse. Together with step 8.1, [F2] makes $J$ a smooth embedding. The bounds in step 7.1 and the exhaustion from step 1.4 give all the claimed bounded-plus-proper conclusions, including the empty case.

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] under the stated Countable Choice 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 [L3], the union of the two bad sets is null. By [L2], its complement 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 kernel of the orthogonal projection $P_u$ is $\mathbb{R}u$. If distinct points $p,q$ collapsed, then $F(p)-F(q)$ would be a nonzero scalar multiple of $u$, making $u$ a secant direction up to sign, contrary to the hypothesis. If the differential killed a nonzero tangent vector $v$, then $d{F}_p(v)$ would lie in $\mathbb{R}u$; since $F$ is an immersion, this vector is nonzero and makes $u$ a tangent direction up to sign. Thus $P_u\circ F$ is injective and immersive directly from [F1], with no generic-existence theorem invoked.

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 ${\mathbb{R}}^{2n+1}$.