Relative Whitney approximation for Euclidean-valued maps

Use [L1] to obtain $Q$ and [L2] to obtain $\chi$. Let $W$ be an open neighbourhood of $A$ on which $\chi =1$. Define on all of $M$ $\tilde{F}=\chi F+(1-\chi )Q\mathrm{.}$ This is a continuous Euclidean-valued map, since $F,Q,\chi$ are continuous and the formula uses finite scalar sums and products. These are two applications of the explicitly assumed countable-choice existence results, not a compact-neighbourhood selection.

On $U$, the formula for $\tilde{F}$ is smooth by [L4], since $F{\mathrm{\mid }}_U$, $Q{\mathrm{\mid }}_U$ and $\chi {\mathrm{\mid }}_U$ are smooth. On the open set $M\setminus \mathrm{supp}(\chi )$ it equals $Q$, hence is smooth. These two open sets cover $M$ because $\mathrm{supp}(\chi )\subseteq U$. Therefore [L3] proves that $\tilde{F}$ is globally smooth; no differentiability of $F$ outside $U$ is used.

On $W$, $\chi =1$ and hence $\tilde{F}=F$. At every $p\in M$, $\mathrm{\parallel }\tilde{F}(p)-F(p)\mathrm{\parallel }=(1-\chi (p))\mathrm{\parallel }Q(p)-F(p)\mathrm{\parallel }\le \mathrm{\parallel }Q(p)-F(p)\mathrm{\parallel }<\epsilon (p)\mathrm{.}$ The weak inequality before the final strict inequality includes the case $\chi (p)=1$. Thus both claims hold. The same construction includes empty $A$ (its neighbourhood can be empty), empty $M$, and noncompact closed $A$.