Relative transversality preserves a map on a closed good region

Let $V$ be the given open neighbourhood of $A$ on which $f$ is transverse. By [L4], choose smooth nonnegative functions $u,v$ with $u^{-1}(0)=A$ and $v^{-1}(0)=M\setminus V$. They have no common zero because $A\subseteq V$. Put $W:=\{p:u(p)<v(p)\}$. It is open and contains $A$, since on $A$ one has $u=0<v$. Continuity gives $\bar{W}\subseteq \{u\le v\}\subseteq V$: outside $V$ one has $v=0<u$, and the strict inequality persists in a neighbourhood disjoint from $W$. Thus the required smaller neighbourhood is constructed without an assumed global shrinking theorem. Apply [L4] to the closed set $\bar{W}$ to obtain a smooth nonnegative $\eta$ whose zero set is exactly $\bar{W}$, and put $\lambda :=\eta \mathrm{/}(1+\eta )$. Then $0\le \lambda <1$ and its zero set is $\bar{W}$.

If $M=\mathrm{\varnothing }$, the unique empty map and its empty homotopy are transverse and agree with $f$ everywhere, so the conclusion holds. Assume henceforth $M\mathrm{\ne }\mathrm{\varnothing }$. Let $\mathcal{F}:M\times B\to N$ be the perturbation family from [L2], where $B\subseteq {\mathbb{R}}^m$ and ${\mathcal{F}}_0=f$. If $m=0$, fix $p\in M$; the submersion ${\mathcal{F}}_p:B\to N$ forces $N$ to be zero-dimensional. Every map into a zero-dimensional manifold is transverse to every embedded submanifold, so in this case take the perturbed map and homotopy to be constantly $f$. Hence assume $m\ge 1$, and shrink $B$ to a ball centred at $0$.

Since $0\le \lambda <1$ and the centred ball $B$ is convex, $\lambda (p{)}^{2}a\in B$ for every $(p,a)\in M\times B$. Define $\mathcal{G}(p,a):=\mathcal{F}(p,\lambda (p{)}^{2}a)\mathrm{.}$ This is a smooth family with ${\mathcal{G}}_0=f$.

If $p\mathrm{\notin }\bar{W}$, then $\lambda (p)>0$. The derivative of $a\mapsto \mathcal{G}(p,a)$ is the surjective derivative of ${\mathcal{F}}_p$ composed with multiplication by the positive scalar $\lambda (p{)}^2$, so $\mathcal{G}$ is a submersion there. If $p\in \bar{W}$, then $\lambda (p)=0$ and $d({\lambda }^2{)}_p=0$, and the chain rule gives $d{\mathcal{G}}_{(p,a)}(v,w)=d{f}_p(v)\mathrm{.}$ Because $\bar{W}\subseteq V$ and $f⋔Z$ on $V$, the full evaluation map $\mathcal{G}$ is transverse to $Z$ on this second region as well. Thus $\mathcal{G}⋔Z$ everywhere.

Parametric transversality in [L3] makes the set of parameters whose slices are not transverse to $Z$ a null subset of $B$. Since $m\ge 1$, the dense-complement clause of [L3] makes its complement nonempty. Choose $a\in B$ there and put $g:={\mathcal{G}}_a$; then $g⋔Z$.

On $W$ one has $\lambda =0$, so $g=f$. Because the centred ball $B$ contains the whole segment from $0$ to $a$, the formula $H(p,t):=\mathcal{G}(p,ta)$ defines a homotopy from $f$ to $g$. It agrees with $f$ on $W$ for every $t\in I$. Therefore the perturbed map and the whole homotopy coincide with $f$ on the smaller neighbourhood $W$ of $A$.