Let $f$, $g$ : $({\Bbb R}^n,0)\rightarrow ({\Bbb R}^p,0)$ be two $C^\infty$ map-germs. Then $f$ and $g$ are $C^0$-equivalent if there exist homeomorphism-germs $h$ and $l$ of
$({\Bbb R}^n,0)$ and $({\Bbb R}^p,0)$ respectively such that
$g=l\circ f\circ h^{-1}$. Let
$k$ be a positive integer. A germ $f$ is
$k$-$C^0$-determined if every germ $g$ with $j^k g(0)=j^k f(0)$ is
$C^0$-equivalent to $f$. Moreover, we say that $f$ is
finitely topologically determined if $f$ is
$k$-$C^0$-determined for some finite $k$.
We prove a theorem giving a sufficient condition for a
germ to be finitely topologically
determined. We explain this condition below.
Let $N$ and $P$ be two $C^\infty$ manifolds. Consider the jet bundle $J^k(N,P)$
with fiber $J^k(n,p)$. Let $z\in J^k(n,p)$ and let $f$ be such that $z=
j^kf(0)$. Define
\[
\chi(f)=\dim_{\Bbb R} \frac{\theta (f)}
{tf(\theta (n))+f^{\ast}(m_p)\theta(f)}.
\]
Whether $\chi(f)(k$ depends only on $z$, not on $f$ . We can
therefore define the set
$W^k= W^k(n,p)=\{z\in J^k(n,p)\vert \chi(f)\ge k$
for some representative $f$ of $z$\}.
Let $W^k(N,P)$ be
the subbundle of $J^k(N,P)$ with fiber $W^k(n,p)$. Mather has
constructed a finite Whitney (b)-regular stratification
${\mathcal{S}}^k(n,p)$ of $J^k(n,p)-W^k(n,p)$ such that all strata
are semialgebraic and
$\mathcal K$-invariant, having the property that if
${\mathcal S}^k(N,P)$ denotes the corresponding stratification of
$J^k(N,P)-W^k(N,P)$ and $f\in C^\infty (N,P)$ is a $C^\infty$ map such that
$j^k f$ is multitransverse to ${\mathcal S}^k(N,P)$, $j^k f(N)\cap
W^k(N,P)=\emptyset$ and $N$ is compact (or $f$ is proper), then $f$ is
topologically stable.
For a map-germ $f:({\Bbb R}^n,0)\rightarrow ({\Bbb R}^p,0)$,
we define a certain {\L}ojasiewicz inequality.
The inequality implies that there exists a
representative $f : U\rightarrow{\Bbb R}^p$ such that
$j^kf(U-\{0\})\cap W^k({\Bbb R}^n,{\Bbb R}^p)=\emptyset$ and such that
$j^kf$ is multitransverse to
${\mathcal S}^k({\Bbb R}^n,{\Bbb R}^p)$ at any finite set of points
$S\subset U-\{0\}$.
Moreover, the inequality controls the rate $j^k f$ becomes
non-transverse as we approach 0. We show that if $f$ satisfies this
inequality, then $f$ is finitely topologically determined.
1991 Mathematics Subject Classification: 58C27.
