Relative Whitney approximation for manifold-valued maps

If $M=\mathrm{\varnothing }$, take the empty maps and $W=\mathrm{\varnothing }$. Otherwise $N$ is nonempty because $F$ exists. Use [L1] to obtain $j,S,U,R$, and put $f=j\circ F$. The map $f$ is continuous on $M$ and smooth on $V$ by [L4]. All the embedding and retraction data are fixed before applying any approximation result.

If $C={\mathbb{R}}^k\setminus U$ is empty, put $\epsilon \equiv 1$. Otherwise put $\epsilon (x)=\min \{1,\frac{1}{2}d(f(x),C)\}\mathrm{.}$ This is finite and continuous by [L3] and continuity of the minimum operation. It is positive: at each $f(x)\in U$, openness supplies a ball of some positive radius contained in $U$, so every point of $C$ is at least that far away. This is a pointwise argument, not a simultaneous selection of radii. Moreover $B(f(x),\epsilon (x))\subseteq U$ for every $x$: when $C$ is nonempty, a point of $C$ in that ball would have distance to $f(x)$ less than $\epsilon (x)\le d(f(x),C)\mathrm{/}2$, contradicting the infimum lower bound. When $C$ is empty the inclusion is automatic.

Apply [L2] to $f$, $A$ and $\epsilon$. Obtain a smooth $H:M\to {\mathbb{R}}^k$ with $\mathrm{\parallel }H(x)-f(x)\mathrm{\parallel }<\epsilon (x)$ everywhere and $H=f$ on an open neighbourhood $W_0$ of $A$. Set $W=W_0\cap V$, still an open neighbourhood of $A$. Step 2.1 gives $H(M)\subseteq U$. Since $U$ is open, $H:M\to U$ is smooth in the same coordinate charts. Thus $\tilde{F}=j^{-1}\circ R\circ H$ is smooth by [L4]. For $x\in W$, $H(x)=j(F(x))$ and $R$ fixes $S$, giving $\tilde{F}(x)=F(x)$. No relatively compact neighbourhood of $A$ has been selected.

Define $z(x,t)=(1-t)f(x)+tH(x)$ for $(x,t)\in M\times [0,1]$. At every such pair, $\mathrm{\parallel }z(x,t)-f(x)\mathrm{\parallel }=t\mathrm{\parallel }H(x)-f(x)\mathrm{\parallel }<\epsilon (x),$ including $t=0$ because $\epsilon (x)>0$. Therefore the whole segment lies in the single ball from step 2.1 and hence in $U$. Each coordinate of $z$ is a sum of products of continuous functions pulled back from the two factors, so $z$ is jointly continuous. The formula $\mathcal{H}(x,t)=j^{-1}(R(z(x,t)))$ is consequently jointly continuous, using [L4] and the subspace topology on $U$. Its endpoints are $F$ and $\tilde{F}$, since $R$ fixes $S$.

On $W$ the equality $H=f$ makes $z(x,t)=f(x)$ for every $t$, and hence $\mathcal{H}(x,t)=F(x)$ throughout $W\times [0,1]$. Thus [F1] gives a homotopy fixed on the open neighbourhood $W$, in particular relative to $A$. The proof also covers $A=\mathrm{\varnothing }$ by permitting its open neighbourhood to be empty, and the empty-source case was settled in step 1.1. This proves every claimed conclusion.