Whitney approximation for manifold-valued maps

If $M$ is empty, the empty smooth map and empty homotopy prove the claim. Otherwise $N$ is nonempty because $F$ exists. Choose by [L1] a proper smooth embedding $j:N\to {\mathbb{R}}^k$ and put $S=j(N)$. Its corestriction $j:N\to S$ has smooth inverse. Set $f=j\circ F$, which is continuous. Only some finite $k$ is needed, not the dimension-optimal Whitney embedding theorem.

We construct the needed Euclidean neighbourhood directly. On the normal bundle of $S$, for $p\in S$ and $a>0$ set $V_a(p)=\{(q,v):\mathrm{\parallel }q-p\mathrm{\parallel }<a,\mathrm{\parallel }v\mathrm{\parallel }<a\}\mathrm{.}$ These open sets form a neighbourhood basis at $(p,0)$ by [L2]. Thus [L3] makes $A_p=\{a\in (0,1]:E{\mathrm{\mid }}_{V_a(p)}\text{ is a diffeomorphism onto an open image}\}$ nonempty. It is bounded above by one. Define $r(p)=\sup A_p\in (0,1]$ by [L4]. The cap is inside the set before the supremum, and no family of local radii is selected.

For $0<a<r(p)$ there is $b\in A_p$ with $a<b$, so restriction to $V_a(p)$ shows $a\in A_p$. Any two points of $V_{r(p)}(p)$ lie in a common $V_a(p)$ for some $a<r(p)$, by taking $a$ above their finitely many strict distance bounds. Hence $E$ is injective on $V_{r(p)}(p)$. Each point there also lies in one such open $V_a(p)$, where $E$ is a local diffeomorphism. The image of $V_{r(p)}(p)$ is therefore open, and its local smooth inverses agree by injectivity. Thus $E$ is a diffeomorphism on this whole supremum-endpoint domain.

For $p,q\in S$ and $0<a<r(p)-\mathrm{\parallel }p-q\mathrm{\parallel }$, the triangle inequality gives $V_a(q)\subseteq V_{r(p)}(p)$ and $a<1$. Step 3.1 implies $a\in A_q$. If $r(p)-\mathrm{\parallel }p-q\mathrm{\parallel }>0$, taking the supremum over such $a$ gives $r(q)\ge r(p)-\mathrm{\parallel }p-q\mathrm{\parallel }$; otherwise this inequality is automatic from positivity. Interchanging $p,q$ gives $\mathrm{\mid }r(p)-r(q)\mathrm{\mid }\le \mathrm{\parallel }p-q\mathrm{\parallel }\mathrm{.}$ In particular $r$ is positive and continuous.

The set $\mathrm{\Omega }=\{(p,v)\in N^{\perp }S:\mathrm{\parallel }v\mathrm{\parallel }<r(p)\mathrm{/}2\}$ is open, contains the zero section, and at every point $E{\mathrm{\mid }}_{\mathrm{\Omega }}$ is a local diffeomorphism: $(p,v)$ lies in $V_{r(p)}(p)$, where step 3.1 applies. If $E(p,v)=E(q,w)$ with both pairs in $\mathrm{\Omega }$, arrange $r(q)\le r(p)$. Then $\mathrm{\parallel }p-q\mathrm{\parallel }=\mathrm{\parallel }w-v\mathrm{\parallel }\le \mathrm{\parallel }v\mathrm{\parallel }+\mathrm{\parallel }w\mathrm{\parallel }<\frac{r(p)+r(q)}{2}\le r(p)\mathrm{.}$ Both vector lengths are also less than $r(p)$, so both pairs lie in $V_{r(p)}(p)$ and step 3.1 makes them equal. This proves injectivity on $\mathrm{\Omega }$.

Consequently $U=E(\mathrm{\Omega })$ is open and contains $S$. The bijection $E:\mathrm{\Omega }\to U$ has smooth inverse: every point admits a local smooth inverse by step 5.1, and these agree with the global inverse. Define $R:U⟶S,R=\pi \circ (E{\mathrm{\mid }}_{\mathrm{\Omega }}{)}^{-1}\mathrm{.}$ This is smooth, and $R(p)=p$ on $S$ because $E(p,0)=p$. This construction needs only the continuous radius $r\mathrm{/}2$, not a smooth minorant, a partition of unity, or a general ambient tubular-neighbourhood theorem.

Put $C={\mathbb{R}}^k\setminus U$. If $C$ is empty, set $\epsilon \equiv 1$ on $M$. Otherwise put $\epsilon (x)=\min \{1,\frac{1}{2}d(f(x),C)\}\mathrm{.}$ By [L4] and continuity of $f$, this is finite and continuous; the minimum operation is continuous since $\mathrm{\mid }\min (1,a)-\min (1,b)\mathrm{\mid }\le \mathrm{\mid }a-b\mathrm{\mid }$. It is positive: openness of $U$ gives, at each $f(x)$ separately, a ball of some positive radius in $U$, so every point of $C$ is at least that far from $f(x)$. No simultaneous choice of these radii is required. In either case the entire open ball $B(f(x),\epsilon (x))$ lies in $U$, since a point in $C$ would have distance from $f(x)$ at least $d(f(x),C)>\epsilon (x)$. The empty-complement case has this inclusion trivially.

Apply [L5] to obtain a smooth $H:M\to {\mathbb{R}}^k$ with $\mathrm{\parallel }H(x)-f(x)\mathrm{\parallel }<\epsilon (x)$ for all $x$. Its image lies in $U$ by step 7.1, and as a map into this open subset it is smooth in the same coordinates. Therefore $\tilde{F}=j^{-1}\circ R\circ H:M\to N$ is smooth by [L1] and [L6].

For $x\in M$ and $t\in [0,1]$ let $z(x,t)=(1-t)f(x)+tH(x)$. The estimate $\mathrm{\parallel }z(x,t)-f(x)\mathrm{\parallel }=t\mathrm{\parallel }H(x)-f(x)\mathrm{\parallel }<\epsilon (x)$ holds also at $t=0$, since $\epsilon (x)>0$. Thus every segment point lies in the same ball from step 7.1 and hence in $U$. The map $z$ is jointly continuous on the product: each of its finitely many coordinates is a sum of products of continuous coordinate functions pulled back from the two factors. Consequently $\mathcal{H}(x,t)=j^{-1}(R(z(x,t)))$ is jointly continuous by [L6] and the subspace topology of $U$. Since $R$ fixes $S$, its endpoints are $\mathcal{H}(x,0)=F(x)$ and $\mathcal{H}(x,1)=\tilde{F}(x)$. This is the homotopy required by [F1], completing the proof including the empty case.