Morse-Sard for Euclidean maps

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.