On finite regular and holomorphic mappings

Let $X,Y$ be smooth algebraic varieties of the same dimension. Let $f, g : X \to Y$befinitepolynomialmappings\mathrm{.}Wesaythat$f, g$ are equivalent if there exists a regular automorphism $\mathrm{\Phi }\in Aut(X)$ such that $f = g\circ \Phi$\mathrm{.}Ofcourseif$f, g$ are equivalent, then they have the same discriminant and the same geometric degree. We show, that conversely there is only a finite number of non-equivalent proper polynomial mappings $f:X\to Y$, such that $D(f)=V$ and $\mu (f)=k\mathrm{.}$ We prove the same statement in the local holomorphic situation. In particular we show that if $f : (\Bbb C^n, 0) \to (\Bbb C^n, 0)$ is a proper and holomorphic mapping of topological degree two, then there exist biholomorphisms $P,Q:({\mathbb{C}}^n,0)\to ({\mathbb{C}}^n,0)$ such that $P\circ f\circ Q(x_1,x_2,\mathrm{.}\mathrm{.}\mathrm{.},x_n)=(x_1^2,x_2,\mathrm{.}\mathrm{.}\mathrm{.},x_n)$. Moreover, for every proper holomorphic mapping $f:({\mathbb{C}}^n,0)\to ({\mathbb{C}}^n,0)$ with smooth discriminant there exist biholomorphisms $P,Q : (\Bbb C^n, 0) \to (\Bbb C^n, 0)$suchthat$P\circ f\circ Q (x_1, x_2,…, x_n) = (x_1^k, x_2, …, x_n)$,where$k = \mu(f).$