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 $p$ so that $F{\mathrm{\mid }}_U:U\to V$ is a diffeomorphism and both $U$ and $V$ are chart domains. The coordinate map $T=\psi \circ F\circ {\phi }^{-1}$ is a $C^1$ diffeomorphism between open subsets $D=\phi (U)$ and $E=\psi (V)$ of ${\mathbb{R}}^n$.

If $n=0$, $T$ is bijective between subsets of the one-point space ${\mathbb{R}}^0$, so a set is empty exactly when its image is empty. This gives the claim in dimension zero.

Suppose $n>0$. Choose a closed coordinate cube $Q\subseteq D$ around $\phi (p)$ and a closed coordinate cube $R\subseteq E$ around $T(\phi (p))$, both with interiors containing those points. Shrink $U$ to the preimage of $\mathrm{int}Q\cap T^{-1}(\mathrm{int}R)$ and set $V=F(U)$. The continuous derivatives of $T$ on $Q$ and $T^{-1}$ on $R$ are bounded; the mean-value inequality on these convex cubes makes the restrictions Lipschitz, as in [L1].

The coordinatewise clamps $r_Q:{\mathbb{R}}^n\to Q$ and $r_R:{\mathbb{R}}^n\to R$ are $1$-Lipschitz. Thus $T\circ r_Q$ and $T^{-1}\circ r_R$ are global Lipschitz maps; the first agrees with $T$ on $\phi (U)\subseteq Q$, and the second agrees with $T^{-1}$ on $\psi (V)\subseteq R$. Apply [L2] to each. For every $B\subseteq \phi (U)$, $B$ is null if and only if $T(B)$ is null. Taking $B=\phi (A)$ and including step 2.1 proves the claim.

Assume $E$ is $\mathcal{A}$-null and fix $(V,\psi )\in \mathcal{B}$. The overlaps $V\cap U$ with $(U,\phi )\in \mathcal{A}$ cover $V$. At each point of each overlap, [L1] gives a smaller open neighbourhood $W$ on which the coordinate transition $\psi \circ {\phi }^{-1}$ preserves Euclidean nullity in both directions. These $W$ form an open cover of $V$. Apply [L2] under the given countable choice to obtain a countable cover $(W_j)$ subordinate to these local transition neighbourhoods, and use [A1] to fix a containing local transition neighbourhood and its source chart $(U_j,{\phi }_j)$ for each $j$.

Since $E$ is $\mathcal{A}$-null, each ${\phi }_j(E\cap W_j)$ is null as a subset of ${\phi }_j(E\cap U_j)$. The chartwise conclusion of [L1], on the local neighbourhood containing $W_j$, gives that $C_j=\psi (E\cap W_j)$ is Euclidean null. The countable family $(W_j)$ covers $V$, so $\psi (E\cap V)={\bigcup }_j{C}_j$.

Fix $\epsilon >0$. Each null $C_j$ admits a sequence of closed-cube covers with total volume at most $\epsilon 2^{-j-1}$. Use [A1] to choose these sequences simultaneously. Flatten them by [L3]; the resulting sequence covers ${\bigcup }_j{C}_j$ and has total volume at most ${\sum }_j\epsilon 2^{-j-1}=\epsilon$. Thus $\psi (E\cap V)$ is null. Since $(V,\psi )$ was arbitrary, $E$ is $\mathcal{B}$-null. Interchanging $\mathcal{A}$ and $\mathcal{B}$ gives the reverse implication.

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] under the stated countable choice 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 $p\in O$ and a chart $(V,\phi )\in \mathcal{A}$ containing $p$, and put $W=V\cap O$. Then $\phi (W)$ is a nonempty open subset of ${\mathbb{R}}^m$, where $m=\dim M\ge 1$.

Since $W\subseteq E\cap V$, [L1] and [L2] make $\phi (W)$ null. But $\phi (W)$ contains a closed cube of positive side length, which has positive $m$-dimensional volume and cannot be null. Contradiction.

Hence every nonempty open subset of $M$ meets $M\setminus E$, so $M\setminus E$ is dense. Under Countable Choice, a manifold-null set has a fixed atlas for which it is $\mathcal{A}$-null by Null subsets of a smooth manifold, so the same conclusion applies.

If $m=0$, then nullity means emptiness by [L1], so $E=\mathrm{\varnothing }$ and the claim follows. Assume $m\ge 1$. By [L0] under the stated Countable Choice, take countable chart atlases $(U_j,{\phi }_j)$ of $M$ and $(V_k,{\psi }_k)$ of $N$. For each pair $(j,k)$, the set $D_{jk}:={\phi }_j(U_j\cap F^{-1}(V_k))$ is open in ${\mathbb{R}}^m$ and the coordinate map $f_{jk}:={\psi }_k\circ F\circ {\phi }_j^{-1}$ is $C^1$ on $D_{jk}$.

For every $(j,k)$, enumerate the nondegenerate closed cubes with rational endpoints whose closures lie in $D_{jk}$. Their interiors cover $D_{jk}$, and the family is countable. On each such cube $Q$, [L2] bounds $\mathrm{\parallel }D{f}_{jk}\mathrm{\parallel }$ and makes $f_{jk}{\mathrm{\mid }}_Q$ Lipschitz, since $Q$ is convex. Define ${\pi }_Q:{\mathbb{R}}^m\to Q$ by clamping each coordinate to its interval; each scalar clamp is $1$-Lipschitz, so ${\pi }_Q$ is $1$-Lipschitz in Euclidean norm and fixes $Q$. Hence $f_{jk}\circ {\pi }_Q$ is Lipschitz on all of ${\mathbb{R}}^m$. The set $A_{jkQ}:={\phi }_j(E\cap U_j)\cap Q$ is null by [L1], so [L3] makes $f_{jk}(A_{jkQ})=(f_{jk}\circ {\pi }_Q)(A_{jkQ})$ null.

Fix $k$. Every point of ${\psi }_k(F(E)\cap V_k)$ comes from some $p\in E\cap U_j\cap F^{-1}(V_k)$, and ${\phi }_j(p)$ lies in the interior of one of the cubes from step 2.1. Thus ${\psi }_k(F(E)\cap V_k)$ is contained in the countable union of the null sets $f_{jk}(A_{jkQ})$ over $j,Q$. Countable subadditivity makes it null. Since this holds for each $k$, [L1] detects $F(E)$ as null in $N$.

Under the stated Countable Choice, choose countable smooth atlases $\{(U_i,{\phi }_i)\}$ on $P$ and $\{(V_j,{\psi }_j)\}$ on $N$ as in [L3]. The countable family of open sets $U_i\cap F^{-1}(V_j)$ covers $P$. It is enough to show that the image of each is null in $N$, then use [L4] for their union.

Fix one open overlap $U=U_i\cap F^{-1}(V_j)$, write $\mathrm{\Omega }={\phi }_i(U)\subseteq {\mathbb{R}}^m$, and let $(V_j,{\psi }_j)$ be its target chart. Define $\tilde{F}:\mathrm{\Omega }\times {\mathbb{R}}^{n-m}\to {\mathbb{R}}^n,\tilde{F}(u,z):=({\psi }_j\circ F\circ {\phi }_i^{-1})(u)\mathrm{.}$ The slice $\mathrm{\Omega }\times \{0\}$ is Euclidean null: cover it by countably many bounded $m$-cubes times $\{0\}$; for each cube, finite $(n)$-cube covers of arbitrarily small total volume are obtained by choosing a sufficiently thin grid in the remaining $n-m\ge 1$ coordinates, and [F1] combines these with geometric error budgets under Countable Choice. The map $\tilde{F}$ is $C^1$. Thus [L1] makes $\tilde{F}(\mathrm{\Omega }\times \{0\})={\psi }_j(F(U))$ Euclidean null.

To check every target chart $V_k$, apply the same argument to the further open overlap $U\cap F^{-1}(V_k)$; its coordinate map has target $V_k$, so step 2.1 makes ${\psi }_k(F(U)\cap V_k)$ Euclidean null. The countable chart cover on $N$ detects $F(U)$ as null by [L2]. Finally [L4] makes the countable union $F(P)$ null.

By [F1], the immersed submanifold has a smooth injective immersion $i:S\to N$; its image is the subset of $N$ in the statement.

Since $m<n$, [L1] under the stated Countable Choice makes $i(S)$ null in $N$.

If $a=b$, compactness bounds $K$ inside $\{a\}\times [-R,R{]}^n$ for some $R>0$. Divide the $n$-dimensional base into a finite grid of cubes of side at most $\delta$ and cover each corresponding point of $K$ by an $(n+1)$-cube of side $\delta$ centred in the first coordinate at $a$. The number of grid cubes is at most $(2R\mathrm{/}\delta +2{)}^n$, so their total $(n+1)$-volume is at most $(2R+2\delta {)}^n\delta \to 0$. By [F1], $K$ 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.

The cases $A=\mathrm{\varnothing }$ and $A={\mathbb{R}}^d$ are immediate: use zero in the first case, and in the second define $H(a)=P_a(a)$; the compatibility condition is precisely the Taylor criterion for its derivatives $D^{\alpha }H(a)=D^{\alpha }{P}_a(a)$. Assume $A$ is nonempty and proper. Subdivide the standard integer-translated dyadic grid into closed cubes. Retain every dyadic cube $Q\subset {\mathbb{R}}^d\setminus A$ satisfying $4\sqrt{d}\mathrm{\ell }(Q)\le \mathrm{dist}(Q,A)$ which is maximal under dyadic parent inclusion. Every $z\mathrm{\notin }A$ belongs to the closure of one of these cubes: arbitrarily small cubes about $z$ satisfy the inequality, while sufficiently large ancestors fail it. Their interiors are disjoint. A retained cube has $4\sqrt{d}\mathrm{\ell }(Q)\le \mathrm{dist}(Q,A)\le 10\sqrt{d}\mathrm{\ell }(Q)$: the upper bound follows because its parent fails the retention test and every point of the parent lies within $2\sqrt{d}\mathrm{\ell }(Q)$ of $Q$. Consequently cubes whose fixed small enlargements meet have comparable side lengths, and only a dimension-dependent bounded number of those enlargements meet any one point. All cubes are obtained from a countable, explicitly ordered grid.

Fix one nonnegative $C^{\mathrm{\infty }}$ bump on the unit cube, equal to one on that cube and supported in its concentric $9\mathrm{/}8$ enlargement. Rescale it to each retained cube to get ${\varphi }_Q$. Set ${\psi }_Q={\varphi }_Q\mathrm{/}{\sum }_R{\varphi }_R$ on ${\mathbb{R}}^d\setminus A$. The denominator is at least one because the cubes cover the complement, and the sum is locally finite by step 1.1. Thus ${\sum }_Q{\psi }_Q=1$ and $\mathrm{\mid }{D}^{\alpha }{\psi }_Q\mathrm{\mid }\le C_{\alpha }\mathrm{\ell }(Q{)}^{-\mathrm{\mid }\alpha \mathrm{\mid }}$; the constants are locally uniform because intersecting enlarged cubes have comparable sizes and bounded overlap. For each cube choose the nearest point $a_Q\in A$ to its centre, breaking ties by successive minimum coordinates on the compact nearest-point set. This specifies $a_Q$ without arbitrary selections. Define $H(z)={\sum }_Q{\psi }_Q(z)P_{a_Q}(z)$ off $A$, and $H(a)=P_a(a)$ on $A$.

Fix $a\in A$ and let $z\to a$ through the complement. Write $t=\mathrm{dist}(z,A)$ and choose its lexicographically first nearest point $a_z$. If ${\psi }_Q(z)\mathrm{\ne }0$, step 1.1 gives $\mathrm{\ell }(Q)\asymp t$ and $\mathrm{\mid }{a}_Q-a_z\mathrm{\mid }=O(t)$. Taylor expansion of the polynomial $P_{a_Q}-P_{a_z}$ about $a_z$, combined with the assumed compatibility of every derivative order, gives $D^{\beta }(P_{a_Q}-P_{a_z})(z)=o(t^{r-\mathrm{\mid }\beta \mathrm{\mid }})(\mathrm{\mid }\beta \mathrm{\mid }\le r),$ locally uniformly as $z\to a$. Differentiating the partition sum and using ${\sum }_Q{\psi }_Q=1$ and the derivative bound in step 2.1 yields $D^{\alpha }H(z)-D^{\alpha }{P}_{a_z}(z)={\sum }_Q{\sum }_{\beta \le \alpha }\left(\genfrac{}{}{0px}{}{\alpha }{\beta }\right)D^{\alpha -\beta }{\psi }_Q(z)D^{\beta }(P_{a_Q}-P_{a_z})(z)=o(t^{r-\mathrm{\mid }\alpha \mathrm{\mid }})$ for $\mathrm{\mid }\alpha \mathrm{\mid }\le r$. There are only boundedly many terms at $z$, and each has the displayed order.

Compatibility again gives $D^{\alpha }{P}_{a_z}(z)-D^{\alpha }{P}_a(z)=o(\mathrm{\mid }z-a{\mathrm{\mid }}^{r-\mathrm{\mid }\alpha \mathrm{\mid }})$: $\mathrm{\mid }{a}_z-a\mathrm{\mid }\le 2\mathrm{\mid }z-a\mathrm{\mid }$, and the polynomial Taylor sum converts every jet discrepancy into that bound. Since $t\le \mathrm{\mid }z-a\mathrm{\mid }$, step 3.1 implies $D^{\alpha }H(z)-D^{\alpha }{P}_a(z)=o(\mathrm{\mid }z-a{\mathrm{\mid }}^{r-\mathrm{\mid }\alpha \mathrm{\mid }})$ as $z\to a$ through the complement. The same estimate at points of $A$ is exactly the given compatibility. Starting with $\mathrm{\mid }\alpha \mathrm{\mid }=0$, these estimates prove by induction that each derivative of order at most $r$ extends continuously across $A$ with value $D^{\alpha }{P}_a(a)$: for $\mathrm{\mid }\alpha \mathrm{\mid }<r$ subtract the linear part of $P_a$ and divide by $\mathrm{\mid }z-a\mathrm{\mid }$ to verify differentiability, and for $\mathrm{\mid }\alpha \mathrm{\mid }=r$ use the zero-order continuity estimate. Thus $H\in C^r$ and has the required jets. Every selection in the construction was by an ordered grid or a compact-set coordinate minimum.

Put $\theta =r-s\ge 1$. Work first on a compact $K\subseteq A^{\ast }$ contained in a relatively compact ball of $W$. For $x\in K$ let $F_x$ be the order-$r$ Taylor polynomial of $f$ at $g(x)$ and $G_x$ the order-$\theta$ Taylor polynomial of $g$ at $x$. Define $P_x$ to be the degree-at-most-$r$ Taylor polynomial at $x$ of the polynomial composition $F_x\circ G_x$. Since $G_x(x)=g(x)$ and the derivatives of $f$ of orders $1,\dots ,s$ vanish at $g(x)\in A$, $P_x(x)=f(g(x)),D^j{P}_x(x)=0(1\le j\le s)\mathrm{.}$ Although $g$ has only $\theta$ derivatives, $P_x$ has a well-defined order-$r$ jet: every nonconstant term uses at least $s+1$ factors from $G_x-g(x)$, so a coefficient of degree at most $r$ cannot use a derivative of $g$ above order $r-s=\theta$.

We verify the exact Whitney condition in [F1] using only function-value Taylor estimates. Uniformly for $x$ in a compact part of $A^{\ast }$ and small $h$, [F2] gives $g(x+h)-G_x(x+h)=o(\mathrm{\mid }h{\mathrm{\mid }}^{\theta })$ and $G_x(x+h)-g(x)=O(\mathrm{\mid }h\mathrm{\mid })$. Since $Df$ and its first $s-1$ derivatives vanish at $g(x)$, [F2] gives $\mathrm{\mid }Df(v)\mathrm{\mid }=O(\mathrm{\mid }h{\mathrm{\mid }}^s)$ on the segment joining $g(x+h)$ and $G_x(x+h)$. The mean-value integral therefore gives $f(g(x+h))-f(G_x(x+h))=o(\mathrm{\mid }h{\mathrm{\mid }}^{s+\theta })=o(\mathrm{\mid }h{\mathrm{\mid }}^r)\mathrm{.}$ Taylor expansion of $f$ at $g(x)$ gives $f(G_x(x+h))-F_x(G_x(x+h))=o(\mathrm{\mid }h{\mathrm{\mid }}^r)$, and truncating the polynomial composition to degree $r$ costs $O(\mathrm{\mid }h{\mathrm{\mid }}^{r+1})$. Consequently $f(g(x+h))-P_x(x+h)=o(\mathrm{\mid }h{\mathrm{\mid }}^r)$ uniformly on compact parts of $A^{\ast }$. The same estimate holds when $s=0$: then $Df$ is merely bounded and $\theta =r$.

Take $x,y\in K$, $d=\mathrm{\mid }x-y\mathrm{\mid }$, and let $z$ range over the ball $B(y,d)$. Then $\mathrm{\mid }z-x\mathrm{\mid }\le 2d$ and $\mathrm{\mid }z-y\mathrm{\mid }\le d$, so step 2.1 gives $\mathrm{\mid }{P}_x(z)-P_y(z)\mathrm{\mid }=o(d^r)$ uniformly throughout that ball. The difference is a polynomial of degree at most $r$. Rescale $z=y+dw$: on the finite-dimensional space of degree-$r$ polynomials, each coefficient functional is bounded by a constant times the supremum on the unit ball. One can see this directly by evaluating on a fixed finite tensor-product grid of distinct points and inverting its fixed Vandermonde matrix. Thus for every multi-index $\mathrm{\mid }\alpha \mathrm{\mid }\le r$, $D^{\alpha }{P}_x(y)-D^{\alpha }{P}_y(y)=o(d^{r-\mathrm{\mid }\alpha \mathrm{\mid }})\mathrm{.}$ This is the compatibility hypothesis of [F1], locally uniformly.

To pass from compact $K$ to relatively closed $A^{\ast }$, choose a compact exhaustion $K_j\subset \mathrm{int}{K}_{j+1}$ of $W$, namely $K_j=\{x:\mathrm{\mid }x\mathrm{\mid }\le j,\mathrm{dist}(x,{\mathbb{R}}^d\setminus W)\ge 1\mathrm{/}j\}$ (with distance to the empty set interpreted as infinite). Choose smooth bumps $b_j$ supported in $\mathrm{int}{K}_{j+1}$, equal to one on $K_j$, by convolving distance cutoffs in the positive gap between $K_j$ and ${\mathbb{R}}^d\setminus \mathrm{int}{K}_{j+1}$. Set ${\chi }_j=1-{\prod }_{i\le j}(1-b_i)$, so ${\chi }_j$ is supported in $\mathrm{int}{K}_{j+1}$, equals one on $K_j$, and $0\le {\chi }_j\le {\chi }_{j+1}\le 1$. Put ${\rho }_1={\chi }_1$, ${\rho }_j={\chi }_j-{\chi }_{j-1}$ for $j>1$. The nonnegative ${\rho }_j$ form a locally finite smooth partition of unity on $W$, and $\mathrm{supp}{\rho }_j\subseteq K_{j+1}$. Apply [F1] on the closed compact set $A^{\ast }\cap K_{j+1}$ to obtain an extension $H_j$ of its jets. The locally finite sum $H={\sum }_j{\rho }_j{H}_j$ is $C^r$ on $W$. At any $x\in A^{\ast }$ with ${\rho }_j(x)\mathrm{\ne }0$, the full order-$r$ jet of $H_j$ is $P_x$. Leibniz's rule and ${\sum }_j{\rho }_j=1$ show that the full jet of $H$ at $x$ is also $P_x$. Step 1.1 therefore gives $H=f\circ g$ and $D^{j}H=0$ for $1\le j\le s$ on $A^{\ast }$.