Strong Whitney approximation by transverse maps

If $M=\mathrm{\varnothing }$, its unique map is already transverse and lies in $\mathcal{U}$. Otherwise $N\mathrm{\ne }\mathrm{\varnothing }$. Fix $e,T,R$ as in [L1]. The ambient dimension supplied by the embedding theorem is positive, so take $d\ge 1$.

Choose a finite intersection of the basic sets of [F1] contained in $\mathcal{U}$. List its compact chart constraints $(K_i,{\phi }_i,{\psi }_i,r_i,b_i)$. Their nonempty source chart domains form an at-most-countable family: in a countable basis assign to each domain the least nonempty basis member contained in it. Each basis member is assigned only finitely many domains, since a point of that member belongs to all of them and the family is locally finite. Countable Choice enumerates these finite fibres. Add a locally finite compact coordinate cover subordinate to the open sets $f^{-1}(V)$ for target chart domains $V$, using [L2], with derivative order zero and tolerance one. The combined constraints are still countable and locally finite; their compact sets now cover $M$. They only impose extra restrictions, so satisfying all of them implies membership in $\mathcal{U}$.

For one nonempty $K_i$ consider, in source coordinates, $E_i(x,v):={\psi }_i(R(e(f({\phi }_i^{-1}(x)))+v))\mathrm{.}$ This is smooth on an open neighbourhood of ${\phi }_i(K_i)\times \{0\}$. Compactness gives a positive $t_i$ and a compact neighbourhood of ${\phi }_i(K_i)$ within its coordinate domain such that $E_i$ is defined on that neighbourhood times $\{\mathrm{\mid }v\mathrm{\mid }\le t_i\}$, with the retraction argument in $T$ and the result in $V_i$. Every finite collection of derivatives of $E_i$ is bounded and uniformly continuous on this compact product by [L3].

There is $a_i>0$ such that any smooth $u:M\to {\mathbb{R}}^d$ with ${\max }_{\mathrm{\mid }\alpha \mathrm{\mid }\le r_i}{\sup }_{x\in {\phi }_i(K_i)}\mathrm{\mid }{D}^{\alpha }(u\circ {\phi }_i^{-1})(x)\mathrm{\mid }<a_i$ makes $R(e\circ f+u)$ satisfy constraint $i$. Here the assertion only concerns points of $K_i$ where the expression is defined; the zeroth-order bound ensures that it is defined there. To verify the derivative estimate, repeated chain and product rules express each $D^{\alpha }(E_i(x,u(x)))$ as a finite sum of derivatives $D_x^{\beta }{D}_v^{\gamma }{E}_i(x,u(x))$ multiplied by products of derivatives $D^{\nu }u(x)$. At $u=0$ and all its derivatives zero the sum equals $D^{\alpha }{E}_i(x,0)$. As the finitely many derivative variables tend to zero, terms having a positive-order derivative of $u$ tend uniformly to zero by the compact derivative bounds of step 3.1; terms without such a factor converge uniformly by uniform continuity there. Choose $a_i<t_i$ small enough to make all these finitely many differences less than $b_i$. Take the first number $2^{-q}$, $q\ge 1$, for which these conclusions hold, so choosing all $a_i$ makes no new family choice. Empty $K_i$ may be omitted.

Choose the partition $({\eta }_j)$ from [L2]. For each $i,j$ put $A_{ij}:={\max }_{\mathrm{\mid }\alpha \mathrm{\mid }\le r_i}{\sup }_{x\in {\phi }_i(K_i)}\mathrm{\mid }{D}^{\alpha }({\eta }_j\circ {\phi }_i^{-1})(x)\mathrm{\mid }\mathrm{.}$ These numbers are finite by [L3]. For each $j$, the set $I_j:=\{i:K_i\cap \mathrm{supp}{\eta }_j\mathrm{\ne }\mathrm{\varnothing }\}$ is finite: every point of the compact support has a neighbourhood meeting only finitely many $K_i$, and a finite subcover gives finitely many indices in total. Define $c_j:=\min (\{1\}\cup \{\frac{a_i{2}^{-j-1}}{1+A_{ij}}:i\in I_j\})>0,\epsilon :={\sum }_{j\ge 1}{c}_j{\eta }_j\mathrm{.}$ The sum is locally finite and smooth, and $\epsilon >0$ because the nonnegative partition sums to one and every $c_j>0$. All coefficients are explicit finite minima.

For every $i$, the compact $K_i$ meets only finitely many partition supports, by the same compactness argument. Outside a support a smooth function and all its derivatives vanish. Therefore differentiation of the locally finite sum and the triangle inequality give ${\max }_{\mathrm{\mid }\alpha \mathrm{\mid }\le r_i}{\sup }_{{\phi }_i(K_i)}\mathrm{\mid }{D}^{\alpha }(\epsilon \circ {\phi }_i^{-1})\mathrm{\mid }\le {\sum }_{j:i\in I_j}{c}_j{A}_{ij}\le a_i{\sum }_{j\ge 1}{2}^{-j-1}=\frac{a_i}{2}<a_i\mathrm{.}$ This controls the entire prescribed finite jet on every compact, with tolerances allowed to shrink independently towards infinity.

Let $B=\{a\in {\mathbb{R}}^d:\mathrm{\mid }a\mathrm{\mid }<1\}$ and define $\mathcal{F}(x,a):=R(e(f(x))+\epsilon (x)a)\mathrm{.}$ For every $a\in B$, each derivative of $u_a(x):=\epsilon (x)a$ in constraint $i$ is bounded by $\mathrm{\mid }a\mathrm{\mid }$ times the corresponding bound of step 6.1. Since the $K_i$ cover $M$, step 4.1 makes the retraction argument lie in $T$ at every $x$, so $\mathcal{F}$ is a globally defined smooth map. It also makes every slice ${\mathcal{F}}_a$ satisfy every original neighbourhood constraint, hence ${\mathcal{F}}_a\in \mathcal{U}$. Finally ${\mathcal{F}}_0=R\circ e\circ f=f$.

The parameter differential at $(x,a)$ is $D_a\mathcal{F}(x,a)=\epsilon (x)d{R}_{e(f(x))+\epsilon (x)a}:{\mathbb{R}}^d⟶T_{\mathcal{F}(x,a)}N\mathrm{.}$ It is surjective because $\epsilon (x)>0$ and $R$ is a submersion. Thus the total evaluation map is a submersion and therefore transverse to $Z$: its differential already spans the whole target tangent space.

By [L4], the bad slice parameters are null in $B$. They cannot exhaust $B$. Indeed $B$ contains a closed positive-side cube $Q$. If $Q$ were null, compact-null/content-zero equivalence in [L4] would give finite cube covers of arbitrarily small total volume. Every finite cube cover of $Q$ has total volume at least $\mathrm{vol}(Q)>0$: subdivide a containing rectangle at all cube endpoints, and assign each positive-volume grid cell of $Q$ to the least covering cube that contains its midpoint. Such a cell lies wholly in that cube; finite grid additivity gives the stated volume bound. This contradiction proves that some $a\in B$ is good. No countable sequence of good parameters is selected.

Choose this one good $a$ and set $g={\mathcal{F}}_a$. It is smooth and transverse to $Z$ by step 9.1, and lies in $\mathcal{U}$ by step 7.1. This proves the full strong smooth density, including noncompact $M$. Empty source was treated in step 1.1; zero-dimensional source or target is allowed throughout, and no positive target dimension was needed for surjectivity onto its tangent spaces. Exactly [A1] is inherited through the stated suppliers; the jet radii and amplitude coefficients were defined canonically, and the final parameter selection is a single existential choice.