Sard Theorem and Transversality

Let $f:=\psi \circ F\circ {\phi }^{-1}$. By [F1], $f$ is $C^1$ on the open set $\phi (U)\subseteq {\mathbb{R}}^m$. For each $x\in \phi (K)$ choose an open Euclidean ball $B_x$ with $\overline{B_x}$ compact and contained in $\phi (U)$.

The derivative norm $\mathrm{\parallel }Df\mathrm{\parallel }$ is continuous on each compact ball $\overline{B_x}$, so [L1] gives a finite bound $L_x$ there. Since $\overline{B_x}$ is convex, [L2] makes $f$ $L_x$-Lipschitz on $\overline{B_x}$.

Put $W_x:={\phi }^{-1}(B_x)$. Then $W_x$ is an open neighbourhood of ${\phi }^{-1}(x)$ with $\overline{W_x}\subseteq U$, and $f$ is Lipschitz on $\phi (\overline{W_x})=\overline{B_x}$. Since $x$ was arbitrary in $\phi (K)$, the claim follows.

By [F1], shrink around $p$ to a neighbourhood $U_0$ on which $F$ is a diffeomorphism onto an open set $V_0$.

If $\dim M=0$, then $U_0$ and $V_0$ are $0$-manifolds. [F2, step 1.1, cases, algebra] By [F2], a subset of $U_0$ is null exactly when it is empty, and the same holds in $V_0$; because $F{\mathrm{\mid }}_{U_0}$ is bijective, $A\subseteq U_0\text{ is null }⟺A=\mathrm{\varnothing }⟺F(A)=\mathrm{\varnothing }⟺F(A)\text{ is null}\mathrm{.}$ So the claim is proved in this case. Assume henceforth that $\dim M>0$, and let $A\subseteq U_0$.

Cover $A$ by relatively compact source-chart neighbourhoods $W\subseteq U_0$ whose images lie in target charts on $V_0$. [L1, L2, step 2.1, algebra] By [L1], the coordinate representatives of $F{\mathrm{\mid }}_W$ and $(F{\mathrm{\mid }}_W{)}^{-1}$ are Lipschitz on smaller compact closures. Therefore [L2] implies $B\subseteq W\text{ is null }⟺F(B)\subseteq F(W)\text{ is null}$ for each such piece $B$.

The manifold definition of nullity checks exactly these chart images, so [step 3.1] the equivalence in step 3.1 globalizes over $U:=U_0$. Hence $A$ is null in $M$ exactly when $F(A)$ is null in $N$.

Assume $E$ is $\mathcal{A}$-null. Fix a chart $(V,\psi )\in \mathcal{B}$. The sets $V\cap U$ with $(U,\phi )\in \mathcal{A}$ cover $V$, so [L2] gives a countable cover of $V$ by relatively compact coordinate balls $W_j\subseteq V\cap U_j$.

On each $W_j$, the transition map $\psi \circ {\phi }_j^{-1}$ is a local diffeomorphism between Euclidean chart domains. Since $E$ is $\mathcal{A}$-null, [F1] makes ${\phi }_j(E\cap W_j)$ null. Applying [L1] to the transition map shows that $\psi (E\cap W_j)$ is null for every $j$.

The set $\psi (E\cap V)$ is the countable union of the null sets $\psi (E\cap W_j)$, hence is null. Since $(V,\psi )$ was arbitrary, $E$ is $\mathcal{B}$-null by [F1]. The reverse implication is symmetric.

If $\dim M=0$, [F1] says that $E$ is null exactly when $E=\mathrm{\varnothing }$. Because the chart domains cover $M$, this is equivalent to every $E\cap U_j$ being empty, hence to every chart image being empty.

Assume $\dim M\ge 1$. If $E$ is null, then every chart image ${\phi }_j(E\cap U_j)$ is null by definition.

Conversely, the given countable atlas is itself a smooth atlas, so [L2] says that being null with respect to this atlas is the same as being null with respect to any other. Therefore the displayed chartwise condition implies that $E$ is null.

Hence this countable chart cover detects manifold null sets in every dimension.

Fix a countable smooth atlas detecting nullity as in [L1]. For each chart $(U_j,{\phi }_j)$, the set ${\phi }_j(A\cap U_j)$ is contained in the null set ${\phi }_j(E\cap U_j)$, so [F1] makes it null. Thus $A$ is null.

Fix a chart index $j$ and $\epsilon >0$. For each $m\ge 1$, nullity of ${\phi }_j(E_m\cap U_j)$ supplies cube covers with total volume at most $\epsilon 2^{-m}$. By [A1], choose these covers simultaneously. By [L2], their doubly indexed union is a single countable cube cover, and its total volume is at most ${\sum }_{m\ge 1}\epsilon 2^{-m}=\epsilon$. Hence ${\bigcup }_{m\ge 1}{\phi }_j(E_m\cap U_j)$ is null.

Since ${\phi }_j\left(\right({\bigcup }_{m\ge 1}{E}_m)\cap U_j)={\bigcup }_{m\ge 1}{\phi }_j(E_m\cap U_j),$ step 1.2 and [L1] show that ${\bigcup }_{m\ge 1}{E}_m$ is null in $M$.

Therefore subsets and countable unions of manifold null sets are null.

Suppose $M\setminus E$ were not dense. Then some nonempty open set $O\subseteq M$ would satisfy $O\subseteq E$. Choose a chart $(U,\phi )$ with $U\subseteq O$ and $\phi (U)\subseteq {\mathbb{R}}^m$, where $m=\dim M\ge 1$.

Since $U\subseteq E$, the chart image $\phi (U)$ would be a null subset of ${\mathbb{R}}^m$ by [L1]. But $\phi (U)$ is a nonempty open subset of ${\mathbb{R}}^m$, so it contains a closed cube of positive side length and therefore cannot be null. Contradiction.

Hence every nonempty open subset of $M$ meets $M\setminus E$, so $M\setminus E$ is dense.

Choose countable smooth atlases on $M$ and $N$ detecting nullity by [L1]. Refine the source atlas so that each relatively compact chart domain lies inside the inverse image of one target chart domain.

For each source chart piece $U_j$, the set ${\phi }_j(E\cap U_j)$ is null in ${\mathbb{R}}^m$. Cover $\overline{U_j}$ by finitely many smaller coordinate neighbourhoods on which the coordinate representative of $F$ is Lipschitz by [L2]. Applying [L2] on each such piece shows that the corresponding target-chart image of $F(E\cap U_j)$ is null.

The set $F(E)$ is the countable union of the sets $F(E\cap U_j)$. Since each has null image in every target chart from step 2.1, [L1] implies that $F(E)$ is null in $N$.

Choose countable smooth atlases $\{(U_i,{\phi }_i)\}$ on $P$ and $\{(V_j,{\psi }_j)\}$ on $N$ as in [L3], and replace each $U_i$ by the countable family of intersections $U_i\cap F^{-1}(V_j)$. It is enough to show that $F(U)$ is null for each resulting chart domain $U$, because those domains still cover $P$ and $F(P)$ is the countable union of the sets $F(U)$.

Fix such a chart $U$ with coordinates $\phi :U\to \mathrm{\Omega }\subseteq {\mathbb{R}}^m$ and with $F(U)\subseteq V$ for some target chart $(V,\psi )$. Define $\tilde{F}:\mathrm{\Omega }\times {\mathbb{R}}^{n-m}\to {\mathbb{R}}^n,\tilde{F}(u,z):=(\psi \circ F\circ {\phi }^{-1})(u)\mathrm{.}$ The slice $\mathrm{\Omega }\times \{0\}$ is a null subset of ${\mathbb{R}}^n$, and $\tilde{F}$ is $C^1$. By [L1], $\tilde{F}(\mathrm{\Omega }\times \{0\})=\psi (F(U))$ is null.

Therefore $F(U)$ is null in the target chart $V$. Since the atlas on $N$ from step 1.1 detects nullity by [L2], each set $F(U)$ is null in $N$, and then the countable union from step 1.1 shows that $F(P)$ is null in $N$.

By [F1], every point of $S$ has a neighbourhood in $N$ on which $S$ is the image of a smooth immersion from an $m$-manifold.

Because $m<n$, [L1] makes each such local image null in the ambient neighbourhood. Therefore $S$ is locally null, and a countable cover of $S$ by such neighbourhoods shows that $S$ is null in $N$.

Hence every positive-codimension immersed submanifold is null.

If $a=b$, then $K$ is contained in the hyperplane $\{a\}\times {\mathbb{R}}^n$, which is the image of the smooth map $y\mapsto (a,y)$ from ${\mathbb{R}}^n$ to ${\mathbb{R}}^{n+1}$; [L2] makes that hyperplane null, and hence $K$ is null. Assume henceforth that $a<b$.

Fix $\epsilon >0$ and put $L:=b-a+1$. For each $t\in [a,b]$, [F1] supplies the following covers. [F1, given, choose] There are finitely many closed $n$-cubes $Q_{t,1},\dots ,Q_{t,m_t}$ covering the compact section $K_t$ with total $n$-volume below $\epsilon \mathrm{/}(2^{n+3}L)$. Enlarge them slightly to open $n$-cubes ${\tilde{Q}}_{t,\mathrm{\ell }}\supseteq Q_{t,\mathrm{\ell }}$ so that, with $O_t:={\bigcup }_{\mathrm{\ell }=1}^{m_t}{\tilde{Q}}_{t,\mathrm{\ell }},$ one still has ${\sum }_{\mathrm{\ell }=1}^{m_t}{\mathrm{vol}}_n({\tilde{Q}}_{t,\mathrm{\ell }})<\frac{\epsilon }{2^{n+2}L}\mathrm{.}$ Thus each section has an open finite cube cover with the stated uniform volume budget.

Since $K_t\subseteq O_t$ and $O_t$ is open, compactness of $K$ gives the following interval. [step 1.2, given, contradiction] There is an open interval $I_t$ about $t$ such that $K\cap (I_t\times {\mathbb{R}}^n)\subseteq I_t\times O_t\mathrm{.}$ Otherwise one could find $(t_j,y_j)\in K$ with $t_j\to t$ and $y_j\mathrm{\notin }{O}_t$; passing to a convergent subsequence inside the compact set $K$ yields a limit point $(t,y)\in K$ with $y\in K_t\setminus O_t$, contradiction.

By [L1], the cover $\{I_t:t\in [a,b]\}$ has a Lebesgue number. Choose a finite partition of $[a,b]$ into closed intervals $J_1,\dots ,J_r$ of positive length smaller than that number, and for each $j$ choose $t_j$ with $J_j\subseteq I_{t_j}$. Then ${\sum }_j\mathrm{\mid }{J}_j\mathrm{\mid }=b-a<L$. For each prism $J_j\times {\tilde{Q}}_{t_j,\mathrm{\ell }}$, use the following subdivision. [L1, step 2.1, choose] Let ${\lambda }_j:=\mathrm{\mid }{J}_j\mathrm{\mid }$ and let $s_{j,\mathrm{\ell }}$ be the side length of ${\tilde{Q}}_{t_j,\mathrm{\ell }}$. If ${\lambda }_j\ge s_{j,\mathrm{\ell }}$, partition the interval direction into at most $\lceil {\lambda }_j\mathrm{/}{s}_{j,\mathrm{\ell }}\rceil$ pieces of length at most $s_{j,\mathrm{\ell }}$; if ${\lambda }_j\le s_{j,\mathrm{\ell }}$, partition each of the $n$ base directions into at most $\lceil s_{j,\mathrm{\ell }}\mathrm{/}{\lambda }_j\rceil$ pieces of length at most ${\lambda }_j$. In either case the prism is covered by finitely many closed $(n+1)$-cubes of total volume at most $2^n{\lambda }_j{s}_{j,\mathrm{\ell }}^n=2^n\mathrm{\mid }{J}_j\mathrm{\mid }{\mathrm{vol}}_n({\tilde{Q}}_{t_j,\mathrm{\ell }})\mathrm{.}$ These cubes cover $K$, and their total volume is at most $2^n{\sum }_{j=1}^r\mathrm{\mid }{J}_j\mathrm{\mid }{\sum }_{\mathrm{\ell }=1}^{m_{t_j}}{\mathrm{vol}}_n({\tilde{Q}}_{t_j,\mathrm{\ell }})<2^{n}L\cdot \frac{\epsilon }{2^{n+2}L}<\epsilon \mathrm{.}$ By [F1], $K$ is null.

Therefore compact null sections imply the whole compact set is null.

Fix $x\in K$. [L1, given, choose] Because $x\mathrm{\notin }{C}_{j+1}$, some partial derivative of order $j+1$ of some component of $f$ is nonzero at $x$. After reordering coordinates and components, choose a multi-index $\alpha$ with $\mathrm{\mid }\alpha \mathrm{\mid }=j$ and a component $f^{\mu }$ such that, for

one has $\mathrm{\partial }g\mathrm{/}\mathrm{\partial }{x}^1(x)\mathrm{\ne }0$. Since $g$ is $C^{r-j}$, [L1] applied to

gives a neighbourhood $W_x$ of $x$ and a $C^{r-j}$ diffeomorphism from $W_x$ onto an open set $I_x\times {\mathrm{\Omega }}_x\subseteq \mathbb{R}\times {\mathbb{R}}^{m-1}$.

Every point of $K\cap W_x$ lies in $C_j$, so $g$ vanishes there by definition of $C_j$. [step 1.1, algebra] Hence

Define

This map is $C^{r-j}$. If $q={\mathrm{\Phi }}_x^{-1}(0,u)\in K\cap W_x$, then $q\in C_j$ and $j\ge 1$, so $D{f}_q=0$. Therefore

so $u$ is a critical point of $h_x$. The induction hypothesis therefore gives that

is a null subset of ${\mathbb{R}}^n$.

Finitely many neighbourhoods $W_x$ cover the compact set $K$, so $f(K)$ is a finite union of null sets and therefore null.

If $m=0$, then $K$ has at most one point and $f(K)$ is finite, hence null in ${\mathbb{R}}^n$ by cubes of arbitrarily small side. Assume henceforth that $m\ge 1$. For each $x\in K$, choose nested closed cubes $Q_x\subseteq {\widehat{Q}}_x\subseteq U$ with $x\in \mathrm{int}(Q_x)$ and ${\widehat{Q}}_x$ having $Q_x$ in its interior. Compactness gives finitely many inner cubes $Q_1,\dots ,Q_s$ whose interiors cover $K$. It suffices to prove that each $f(K\cap Q_i)$ is null, because finite unions of Euclidean null sets are null directly from the cube-cover definition [F1].

Fix one pair $Q\subseteq \widehat{Q}$, let $K_Q:=K\cap Q$, and let $\lambda$ be the side length of $Q$. The finitely many order-$r$ partial derivatives of the components of $f$ are uniformly continuous on the compact cube $\widehat{Q}$ by [L2]. Since they vanish at every $x\in K_Q$, applying [L1] componentwise with degree $r-1$ shows that for every $\eta >0$ there is a single $\delta >0$ such that $\mathrm{\mid }f(y)-f(x)\mathrm{\mid }\le \eta \mathrm{\parallel }y-x{\mathrm{\parallel }}^r$ whenever $x\in K_Q$, $y\in Q$, and $\mathrm{\parallel }y-x\mathrm{\parallel }<\delta$.

Fix $\epsilon >0$. If $rn=m$, choose $\eta >0$ so small that $(2\eta {)}^n{m}^{rn\mathrm{/}2}{\lambda }^{rn}<\epsilon$; if $rn>m$, choose any $\eta >0$. Let $\delta$ be furnished by step 2.1, and choose $N$ so large that the congruent subcubes in the subdivision of $Q$ into $N^m$ cubes have diameter below $\delta$. Use the following target cubes. [step 2.1, choose, cases] For each subcube $Q_{\nu }$ meeting $K_Q$, choose a point $x_{\nu }\in K_Q\cap Q_{\nu }$. If $y\in K_Q\cap Q_{\nu }$, then $\mathrm{\parallel }y-x_{\nu }\mathrm{\parallel }\le \sqrt{m}\lambda \mathrm{/}N$, so step 2.1 gives $\mathrm{\mid }f(y)-f(x_{\nu })\mathrm{\mid }\le \eta (\frac{\sqrt{m}\lambda }{N}{)}^r\mathrm{.}$ Hence $f(K_Q\cap Q_{\nu })$ lies in an $n$-cube of side length at most $2\eta (\frac{\sqrt{m}\lambda }{N}{)}^r\mathrm{.}$

The union of those target cubes covers $f(K_Q)$, and its total $n$-volume has the following bound. [F1, step 3.1, cases, algebra] It is at most $N^m\left(2\eta \right(\frac{\sqrt{m}\lambda }{N}{)}^r{)}^n=(2\eta {)}^n{m}^{rn\mathrm{/}2}{\lambda }^{rn}{N}^{m-rn}\mathrm{.}$ If $rn>m$, increase $N$ until this quantity is below $\epsilon$; if $rn=m$, the choice of $\eta$ in step 3.1 already makes it smaller than $\epsilon$. In either case the total covering volume is below $\epsilon$. Therefore [F1] implies that $f(K_Q)$ is null.

Applying step 4.1 to the finite cover from step 1.1 shows that $f(K)$ is null. Thus the infinitely flat critical stratum has null image.

If $m=0$, then $U$ is at most a point, so $\mathrm{Crit}(f)$ is [F1, given, choose, cases] finite and $f(\mathrm{Crit}(f))$ is finite. Every finite subset of ${\mathbb{R}}^n$ is null, so the theorem follows. Assume henceforth that $m\ge 1$. Exhaust $U$ by countably many compact cubes $Q_{\nu }$ with interiors contained in $U$. It is enough to show that $f(\mathrm{Crit}(f)\cap Q_{\nu })$ is null for every $\nu$, because the critical value set in [F1] is the countable union of those images.

Fix one cube $Q_{\nu }$ and put [L1, step 1.1, algebra] $C_0:=\mathrm{Crit}(f)\cap Q_{\nu },C_j:=\{x\in Q_{\nu }:D^{\alpha }f(x)=0\text{ for every }1\le \mathrm{\mid }\alpha \mathrm{\mid }\le j\}(j\ge 1)\mathrm{.}$ Then $C_0=(C_0\setminus C_1)\cup {\bigcup }_{j=1}^{r-1}(C_j\setminus C_{j+1})\cup C_r\mathrm{.}$ For each $1\le j<r$ and each $\mathrm{\ell }\ge 1$, the set $K_{j,\mathrm{\ell }}:=\{x\in C_j\cap Q_{\nu }:\mathrm{dist}(x,C_{j+1})\ge 1\mathrm{/}\mathrm{\ell }\}$ is compact and contained in $C_j\setminus C_{j+1}$. Because $C_j\setminus C_{j+1}={\bigcup }_{\mathrm{\ell }\ge 1}{K}_{j,\mathrm{\ell }},$ the nonflat lemma in [L1] shows that $f(C_j\setminus C_{j+1})$ is null. Also, the set $C_r$ is compact, and the hypothesis $r>\max \{m-n,0\}$ implies $rn\ge m$; thus the flat lemma in [L1] makes $f(C_r)$ null.

It remains to show that $f(C_0\setminus C_1)$ is null. [L1, L2, step 1.1, algebra] If $n=1$, then a linear map ${\mathbb{R}}^m\to \mathbb{R}$ is surjective exactly when it is nonzero, so $C_0=C_1$ and there is nothing to prove. Assume $n>1$, and fix $x\in C_0\setminus C_1$. Some first partial derivative of some component of $f$ is nonzero at $x$; after reordering coordinates and components, assume $\frac{\mathrm{\partial }{f}^1}{\mathrm{\partial }{x}^1}(x)\mathrm{\ne }0.$ By [L2], after shrinking choose a neighbourhood $W_x$ of $x$ and a $C^r$ diffeomorphism ${\mathrm{\Phi }}_x(y):=(f^1(y),y^2,\dots ,y^m)$ from $W_x$ onto an open set $I_x\times {\mathrm{\Omega }}_x\subseteq \mathbb{R}\times {\mathbb{R}}^{m-1}$. Write $f\circ {\mathrm{\Phi }}_x^{-1}(t,u)=(t,{\tilde{f}}_x(t,u))\mathrm{.}$ Each slice map $u\mapsto {\tilde{f}}_x(t,u)$ is $C^r$, and because $r>\max \{m-n,0\}=\max \{(m-1)-(n-1),0\},$ the induction hypothesis applies to those maps. If $q={\mathrm{\Phi }}_x^{-1}(t,u)\in C_0\cap W_x$, then in these coordinates the differential of $f$ has block form $D{f}_q=\left[\begin{array}{cc}1& 0\\ \ast & D({\tilde{f}}_x{)}_t(u)\end{array}\right],$ so $q$ is critical for $f$ exactly when $u$ is a critical point of the slice $u\mapsto {\tilde{f}}_x(t,u)$. Now choose an open neighbourhood $W_x^{\mathrm{\prime }}\subseteq W_x$ of $x$ with compact closure $K_x\subseteq W_x$. The compact set $f(C_0\cap K_x)$ has sections contained in the critical value sets of the slice maps, hence null in ${\mathbb{R}}^{n-1}$ by induction. Applying the slicing lemma in [L1] shows that $f(C_0\cap K_x)$ is null in ${\mathbb{R}}^n$. A countable subcover of $C_0\setminus C_1$ by the neighbourhoods $W_x^{\mathrm{\prime }}$ therefore makes $f(C_0\setminus C_1)$ null.

Step 2.1 shows that $f(C_j\setminus C_{j+1})$ is null for every [F1, step 2.1, step 2.2, step 1.1] $1\le j<r$ and that $f(C_r)$ is null, while step 2.2 handles $f(C_0\setminus C_1)$. Hence $f(C_0)=f(\mathrm{Crit}(f)\cap Q_{\nu })$ is null. Applying step 1.1 shows that the whole critical value set of $f$ is null.

If $\dim N=0$, then every differential [F1, F2, given, cases] $d{F}_p:T_{p}M\to T_{F(p)}N=\{0\}$ is surjective, so [F1] makes every value of $F$ regular. Thus the critical value set is empty, which is null by [F2]. Assume henceforth that $\dim N>0$.

Choose countable smooth atlases $\{(U_i,{\phi }_i)\}$ on $M$ and [L1, step 1.1, given, choose] $\{(V_j,{\psi }_j)\}$ on $N$ detecting nullity by [L1], and refine the source atlas so that each $F(U_i)$ lies in some $V_{j(i)}$.

For each $i$, the coordinate representative [L2, step 2.1, algebra]

is smooth between Euclidean open sets with positive-dimensional target. A point of $U_i$ is critical for $F$ exactly when its coordinate representative is critical for $f_i$, because the chart maps have invertible differentials. By [L2], the critical value set of $f_i$ is null in ${\psi }_{j(i)}(V_{j(i)})$. Therefore ${\psi }_{j(i)}(F(\mathrm{Crit}(F)\cap U_i))$ is null for every $i$.

By [F3], the critical value set of $F$ is the countable union of the sets [F3, L1, step 3.1] $F(\mathrm{Crit}(F)\cap U_i)$, so [L1] shows that it is null in $N$.