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.