On semi-equivalence of generically-finite polynomial mappings

Abstract

Let $f,g:X \rightarrow Y$ be continuous mappings. We say that f is topologically equivalent to g if there exist homeomorphisms $\Phi : X\rightarrow X$ and $\Psi : Y\rightarrow Y$ such that $\Psi \circ f\circ \Phi =g.$ Moreover, we say that f is topologically semi-equivalent to g if there exist open, dense subsets $U, V\subset X$ and homeomorphisms $\Phi : U\rightarrow V$ and $\Psi : Y\rightarrow Y$ such that $\Psi \circ f\circ \Phi |_U=g|_{U}.$ Let X, Y be smooth irreducible affine complex varieties. We show that every algebraic family $F: M\times X\ni (m, x)\mapsto F(m, x)=f_m(x)\in Y$ of polynomial mappings contains only a finite number of topologically non-equivalent proper mappings and only a finite number of topologically non-semi-equivalent generically-finite mappings. In particular there are only a finite number of classes of topologically non-equivalent proper polynomial mappings $f:{\mathbb {C}}^n\rightarrow {\mathbb {C}}^m$ of a bounded (algebraic) degree. The same is true for a number of classes of topologically non-semi-equivalent generically-finite polynomial mappings $f:{\mathbb {C}}^n\rightarrow {\mathbb {C}}^m$ of a bounded (algebraic) degree.

1 Introduction

Let $f,g:X \rightarrow Y$ be continuous mappings. We say that f is topologically equivalent to g if there exist homeomorphisms $\Phi : X\rightarrow X$ and $\Psi : Y\rightarrow Y$ such that $\Psi \circ f\circ \Phi =g.$ Moreover, we say that f is topologically semi-equivalent to g if there exist open, dense subsets $U, V\subset X$ and homeomorphisms $\Phi : U\rightarrow V$ and $\Psi : Y\rightarrow Y$ such that $\Psi \circ f\circ \Phi |_U=g|_{U}.$

In the case $X={\mathbb {C}}^n$ and $Y={\mathbb {C}}$ René Thom stated a Conjecture that there are only finitely many topological types of polynomials $f: X\rightarrow Y$ of bounded degree. This Conjecture was confirmed by Fukuda [2]. Also a more general problem was considered: how many topological types are there in the family P(n, m, k) of polynomial mapping $f:{\mathbb {C}}^n\rightarrow {\mathbb {C}}^m$ of degree bounded by k? Aoki and Noguchi [1] showed that there are only a finite number of topologically non-equivalent mappings in the family P(2, 2, k). Finally Nakai [8] showed that each familiy P(n, m, k), where $n, m ,k > 3,$ contains infinitely many different topological types even if we consider only generically-finite mappings. Hence the General Thom Conjecture is not true even for generically-finite mappings. However, we show in this paper that there are only a finite number of classes of topologically semi-equivalent generically-finite polynomial mappings $f: {\mathbb {C}}^n\rightarrow {\mathbb {C}}^m$ of a bounded (algebraic) degree. As a by product of our considerations we give a simple proof of the following interesting fact: for every n, m and k there are only a finite number of topological types of proper polynomial mappings $f: {\mathbb {C}}^n\rightarrow {\mathbb {C}}^m$ of (algebraic) degree bounded by k. Hence we can say that Thom Conjecture is true for proper polynomial mappings. We show also that if $n\le m$ and $\Omega _n(d_1,...,d_m)$ denotes the family of all polynomial mappings $F=(f_1,...,f_m):{\mathbb {C}}^n\rightarrow {\mathbb {C}}^m$ of a multi-degree bounded by $(d_1,...,d_m),$ then any two general member of this family are topologically equivalent.

In fact we prove more: if X, Y are smooth affine irreducible varieties, then every algebraic family ${\mathcal {F}}$ of polynomial mappings from X to Y contains only a finite number of topologically non-semi-equivalent (non-equivalent) generically-finite (proper) mappings. Moreover, if a family ${\mathcal {F}}$ is irreducible, then two generic members of ${\mathcal {F}}$ are in the same equivalence class.

Let us recall here, that a mapping $f:X\rightarrow Y$ is generically finite, if for general $x\in X$ the set $f^{-1}(f(x))$ is finite. Our proof goes as follows. Let M be a smooth affine irreducible variety and let ${\mathcal {F}}$ be a family of polynomial mappings induced be a regular mapping $F: M\times X \rightarrow Y,$ i.e., ${\mathcal {F}}:=\{f_m: X\ni x\mapsto F(m,x)\in Y, \ m\in M\}.$ Let us recall that if $f: X\rightarrow Z$ is a generically finite polynomial mapping of affine varieties, then the bifurcation set B(f) of f is the set $\{ z\in Z: z\in Sing(Z) \ or\ \# f^{-1}(z)\not =\mu (f)\}$, where $\mu (f)$ is the topological degree of f. The set B(f) is always closed in Z. We show that there exists a Zariski open, dense subset U of M such that

• (1) for every $m\in U$ we have $\mu (f_m)=\mu ({\mathcal {F}}),$ where we treat $f_m$ as a mapping $f_m: X\rightarrow Z_m:=\overline{f_{m}(X)},$

• (2) for every $m_1, m_2\in U$ the pairs $(\overline{f_{m_1}(X)}, B(f_{m_1}))$ and $(\overline{f_{m_2}(X)}, B(f_{m_2}))$ are equivalent via a homeomorphism, i.e., there is a homeomorphism $\Psi : Y\rightarrow Y$ such that $\Psi (\overline{f_{m_1}(X)})=\overline{f_{m_2}(X)}$ and $\Psi (B(f_{m_1}))=B(f_{m_2}).$

In particular the group $G=\pi _1(\overline{f_{m}(X)}{\setminus } B(f_m))$ does not depend on $m\in U.$ Using elementary facts from the theory of topological coverings, we show that the number of topological semi-types (types) of generically-finite (proper) mappings in the family ${\mathcal {F}}_{|U}$ is bounded by the number of subgroups of G of index $\mu ({\mathcal {F}})$, hence it is finite. Then we conclude the proof by induction. Finally, the case of arbitrary M can be easily reduced to the smooth, irreducible, affine case.

Remark 1.1

In this paper we use the term “polynomial mapping” for every regular mapping $f: X\rightarrow Y$ of affine varieties.

2 Bifurcation set

Let X, Z be affine irreducible varieties of the same dimension and assume that X is smooth. Let $f: X\rightarrow Z$ be a dominant polynomial mapping. It is well known that there is a Zariski open non-empty subset U of Z such that for every $x_1,x_2\in U$ the fibers $f^{-1}(x_1), f^{-1}(x_2)$ have the same number $\mu (f)$ of points. We say that $\mu (f)$ is the topological degree of f. Recall the following (see [5, 6]).

Definition 2.1

Let X, Z be as above and let $f : X \rightarrow Z$ be a dominant polynomial mapping. We say that f is finite at a point $z \in Z$ if there exists an open neighborhood U of z such that the mapping $f_{|f^{-1}(U)} :f^{-1} (U)\rightarrow U$ is proper.

It is well-known that the set $S_f$ of points at which the mapping f is not finite is either empty or it is a hypersurface (see [5, 6]). We say that $S_f$ is the set of non-properness of f.

Definition 2.2

Let X be a smooth affine n-dimensional variety and let Z be an affine variety of the same dimension. Let $f : X\rightarrow Z$ be a generically finite dominant polynomial mapping of geometric degree $\mu (f).$ The bifurcation set of f is

$$\begin{aligned} B(f)=\{z\in Z : z\in Sing(Z) \ or \ \# f^{-1}(z)\not =\mu (f)\}. \end{aligned}$$

Remark 2.3

The same definition makes sense for those continuous mapping $f: X\rightarrow Z$, for which we can define the topological degree $\mu (f)$ and singularities of Z. In particular if $Z_1, Z_2$ are affine algebraic varieties, $f: X\rightarrow Z_1$ is a dominant polynomial mapping and $\Phi : Z_1\rightarrow Z_2$ is a homeomorphism which preserves singularities, then we can define $B(\Phi \circ f)$ as $\Phi (B(f)).$ Moreover, the mapping $\Phi \circ f$ behaves topologically as an analytic covering. We will use this facts in the proof of Theorem 3.5.

We have the following theorem (see also [7]).

Theorem 2.4

Let X, Z be affine irreducible complex varieties of the same dimension and suppose X is smooth. Let $f: X \rightarrow Z$ be a polynomial dominant mapping. Then the set B(f) is closed and $B(f)=K_0\cup S_f\cup Sing(Z).$

Proof

Let us note that outside the set $S_f\cup Sing(Z)$ the mapping f is a (ramified) analytic covering of degree $\mu (f).$ By Lemma 2.5 below, if $z\not \in Sing(Z)$ we have $\# f^{-1}(z)\le \mu (f).$ Moreover, since f is an analytic covering outside $S_f\cup Sing(Z)$ we see that for $y \not \in S_f\cup Sing(Z)$ the fiber $f^{-1}(z)$ has exactly $\mu (f)$ points counted with multiplicity. Take $X_0:=X {\setminus } f^{-1}(Sing(Z)\cup S_f).$ If $z\in K_0(f_{|X_0})$, the set of critical values of $f_{|X_0},$ then $\# f^{-1}(z)<\mu (f).$

Now let $z\in S_f{\setminus } Sing(Z).$ There are two possibilities:

• (a) $\# f^{-1}(z)=\infty .$

• (b) $\# f^{-1}(z)<\infty .$

In case (b) we can assume that $f^{-1}(z)\not =\emptyset .$ Let U be an affine neighborhood of z disjoint from Sing(z) over which the mapping f has finite fibers. Let $V=f^{-1}(U).$ By the Zariski Main Theorem in the version given by Grothendieck (see [3]), there exists a normal variety $\overline{V}$ and a finite mapping $\overline{f}:\overline{ V}\rightarrow U$ such that

• (1) $V\subset \overline{V}$,

• (2) $\overline{f}_{|V}=f.$

Since $z\in \overline{f}(\overline{V}{\setminus } V)$, it follows from Lemma 2.5 below that $\# f^{-1}(z)<\mu (f).$ Consequently, if $z\in S_f,$ we have $\# f^{-1}(z)<\mu (f).$ Finally, we have $B(f)= K_0(f_{|X_0})\cup S_f\cup Sing(Z).$ However, the set $K_0(f_{|X_0})$ is closed in $Z{\setminus } (S_f\cup Sing(Z)).$ Hence B(f) is closed in Z. $\square$

Lemma 2.5

Let X, Z be affine normal varieties of the same dimension. Let $f:X\rightarrow Z$ be a finite mapping. Then for every $z\in Z$ we have $\# f^{-1}(z)\le \mu (f).$

Proof

Let $\# f^{-1}(z)=\{ x_1, \ldots , x_r\}.$ We can choose a function $h\in {\mathbb {C}}[X]$ which separates all $x_i$ (in particular we can take as h the equation of a general hyperplane section). Since f is finite, the minimal equation of h over the field ${\mathbb {C}}(Z)$ is of the form:

$$\begin{aligned} T^s+a_1(f)T^{{s-1}}+\cdots +a_s(f)\in f^*{\mathbb {C}}[Z][T], \end{aligned}$$

where $s\le \mu (f).$ If we substitute $f=z$ into this equation we get the desired result. $\square$

3 Main result

We start with the following:

Lemma 3.1

Let $f: X^k\rightarrow Y^l$ be a dominant polynomial mapping of affine irreducible varieties. There exists a Zariski open non-empty subset $U\subset Y$ such that for any $y\in U$ we have $Sing(f^{-1}(y))=f^{-1}(y)\cap Sing(X).$

Proof

We can assume that Y is smooth. Since there exists a mapping $\pi : Y^l\rightarrow {\mathbb {C}}^l$ which is generically etale, we can assume that $Y={\mathbb {C}}^l.$ Let us recall that if Z is an algebraic variety, then a point $z\in Z$ is smooth if and only if the local ring ${\mathcal {O}}_z(Z)$ is regular, or equivalently $\dim _{\mathbb {C}}{{\mathfrak {m}}}/{{\mathfrak {m}}}^2=\dim Z,$ where ${\mathfrak {m}}$ denotes the maximal ideal of ${\mathcal {O}}_z(Z)$.

Let $y=(y_1,...,y_l)\in {\mathbb {C}}^l$ be a sufficiently generic point. Then by Sard’s Theorem the fiber $Z=f^{-1}(y)$ is smooth outside Sing(X) and $\dim Z=\dim X - l=k-l.$ Note that the generic (scheme-theoretic) fiber F of f is reduced. Indeed, this fiber $F=Spec({\mathbb {C}}(Y)\otimes _{{\mathbb {C}}[Y]} {\mathbb {C}}[X])$ is the spectrum of a localization of ${\mathbb {C}}[X]$ and so a domain. Since we are in characteristic zero, the reduced ${\mathbb {C}}(Y)$-algebra ${\mathbb {C}}(Y)\otimes _{{\mathbb {C}}[Y]} {\mathbb {C}}[X]$ is necessarily geometrically reduced (i.e. stays reduced after extending to an algebraic closure of ${\mathbb {C}}(Y)$). Since the property of fibres being geometrically reduced is open on the base, i.e. on Y, thus the fibres over an open subset of Y will be reduced. Consequently, there is a Zariski open, non-empty subset $U\subset Y$ such that for $y\in U$ the fiber $f^{-1}(y)$ is reduced. Hence we can assume that Z is reduced. It is enough to show that every point $z\in Z\cap Sing(X)$ is singular on Z.

Assume that $z\in Z\cap Sing(X)$ is smooth on Z. Let $f: X\rightarrow {\mathbb {C}}^l$ be given as $f=(f_1,...,f_l)$, where $f_i\in {\mathbb {C}}[X].$ Then ${\mathcal {O}}_z(Z)={\mathcal {O}}_z(X)/(f_1-y_1,...,f_l-y_l).$ In particular if ${\mathfrak {m}}'$ denotes the maximal ideal of ${\mathcal {O}}_z(Z)$ and ${\mathfrak {m}}$ denotes the maximal ideal of ${\mathcal {O}}_z(X)$ then ${\mathfrak {m}}'={\mathfrak {m}}/(f_1-y_1,...,f_l-y_l).$ Let $\alpha _i$ denote the class of the polynomial $f_i-y_i$ in ${{\mathfrak {m}}}/{{\mathfrak {m}}}^2.$ Let us note that

$$\begin{aligned} {{\mathfrak {m}}'}/{{\mathfrak {m}}'}^2 ={{\mathfrak {m}}}/({{\mathfrak {m}}}^2+(\alpha _1,...,\alpha _l)). \end{aligned}$$

(1)

Since the point z is smooth on Z we have $\dim _{\mathbb {C}}{{\mathfrak {m}}'}/{{\mathfrak {m}}'}^2=\dim Z=\dim X-l.$ Take a basis $\beta _1,...,\beta _{k-l}$ of the space ${{\mathfrak {m}}'}/{{\mathfrak {m}}'}^2$ and let $\overline{\beta _i}\in {{\mathfrak {m}}}/{{\mathfrak {m}}}^2$ correspond to $\beta _i$ under the correspondence (1). Note that the vectors $\overline{\beta _1},...,\overline{\beta _{k-l}}, \alpha _1,..., \alpha _l$ generate the space ${{\mathfrak {m}}}/{{\mathfrak {m}}}^2.$ This means that $\dim _{\mathbb {C}}{{\mathfrak {m}}}/{{\mathfrak {m}}}^2\le k-l+l=k=\dim X.$ Hence the point z is smooth on X, a contradiction. $\square$

We have:

Lemma 3.2

Let X, Y be smooth complex irreducible algebraic varieties and $f:X\rightarrow Y$ a regular dominant mapping. Let $N\subset W\subset X$ be closed subvarieties of X. Then there exists a non-empty Zariski open subset $U\subset Y$ such that for every $y_1, y_2\in U$ the triples $(f^{-1}(y_1), W\cap f^{-1}(y_1), N\cap f^{-1}(y_1))$ and $(f^{-1}(y_2), W\cap f^{-1}(y_2), N\cap f^{-1}(y_2))$ are homeomorphic.

Proof

Let $X_1$ be an algebraic completion of X and let $\overline{Y}$ be a smooth algebraic completion of Y. Take $X_1':=\overline{graph(f)}\subset X_1\times \overline{Y}$ and let $X_2$ be a desingularization of $X_1'.$

We can assume that $X\subset X_2.$ We have an induced mapping $\overline{f}: X_2\rightarrow \overline{Y}$ such that $\overline{f}_{|X}=f.$ Let $Z=X_2{\setminus } X.$ Denote by $\overline{N}, \overline{W}$ the closures of N and W in $X_2.$ Let ${\mathcal {R}}=\{\overline{N}\cap Z, \overline{W}\cap Z, \overline{N}, \overline{W}, Z\} ,$ a collection of algebraic subvarieties of $X_2.$ There is a Whitney stratification ${\mathcal S}$ of $X_2$ which is compatible with ${\mathcal {R}}.$

For any smooth strata $S_i\in \mathcal S$ let $B_i$ be the set of critical values of the mapping $\overline{f}_{|S_i}$ and denote $B=\overline{ \bigcup B_i}.$ Take $X_3=X_2{\setminus } \overline{f}^{-1}(B).$ The restriction of the stratification $\mathcal S$ to $X_3$ gives a Whitney stratification which is compatible with the family ${{\mathcal {R}}}':={{\mathcal {R}}}\cap X_3.$ We have a proper mapping $f':=\overline{f}_{|X_3} : X_3\rightarrow \overline{Y}{\setminus } B$ which is a submersion on each stratum. By the Thom first isotopy theorem there is a trivialization of $f'$ which preserves the strata. It is an easy observation that this trivialization gives a trivialization of the mapping $f: X{\setminus } f^{-1}(B)\rightarrow Y{\setminus } B:=U.$ In particular the fibers $f^{-1}(y_1)$ and $f^{-1}(y_2)$ are homeomorphic via a stratum preserving homeomorphism. This means that the triples $(f^{-1}(y_1), W\cap f^{-1}(y_1), N\cap f^{-1}(y_1))$ and $(f^{-1}(y_2), W\cap f^{-1}(y_2), N\cap f^{-1}(y_2))$ are homeomorphic. $\square$

We also need the following:

Definition 3.3

Let X, Y be smooth affine varieties. By a family of regular mappings ${\mathcal {F}}_M(X,Y,F):={\mathcal {F}}$ we mean a regular mapping $F: M\times X\rightarrow Y$, where M is an algebraic variety. The members of a family ${\mathcal {F}}$ are the mappings $f_m: X\ni x\rightarrow F(m,x)\in Y.$ Let

$$\begin{aligned} G: M\times X\ni (m,x)\mapsto (m, F(m,x))\in Z=\overline{G(M\times X)}\subset M\times Y. \end{aligned}$$

If G is generically finite, then by the topological degree $\mu ({\mathcal {F}})$ we mean the number $\mu (G)$. Otherwise we put $\mu ({\mathcal {F}})=0.$

Later we will sometimes identify the mapping $f_m$ with the mapping $G(m,\cdot )=(m, f_m): X\rightarrow m\times Y.$ The following lemma is important:

Lemma 3.4

Let X, Y be smooth affine complex varieties. Let M be a smooth affine irreducible variety and let ${\mathcal {F}}$ be the family induced by a mapping $F: M\times X \rightarrow Y,$ i.e., ${\mathcal {F}}=\{ f_m: X\ni x\mapsto F(m,x)\in Y, \ m\in M\}.$ Assume that $\mu ({\mathcal {F}})>0.$ Take $Z=\overline{G(M\times X)}$ and put $Z_m= (m\times Y)\cap Z.$

Then

• (1) There is an open non-empty subset $U_1\subset M$ such that for every $m\in U_1$ we have $\mu (f_m)=\mu ({\mathcal {F}});$

• (2) There is a non-empty open subset $U_2\subset U_1$ such that for every $m\in U_2$ we have $\overline{f_m(X)}=Z_m:=(m\times Y)\cap Z$ and $B(f_m)=B(G)_m:=(m\times Y)\cap B(G);$

• (3) There is a non-empty open subset $U_3\subset U_2$ such that for every $m_1, m_2\in U_3$ the pairs $(\overline{f_{m_1}(X)}, B(f_{m_1}))$ and $(\overline{f_{m_2}(X)}, B(f_{m_2}))$ are equivalent by means of a homeomorphism, i.e., there is a homeomorphism $\Psi : Y\rightarrow Y$ such that $\Psi (\overline{f_{m_1}(X)})=\overline{f_{m_2}(X)}$ and $\Psi (B(f_{m_1}))=B(f_{m_2}).$

Proof

• (1) Take $G: M\times X\ni (m,x)\mapsto (m, F(m,x))\in Z.$ The mapping $G: M\times X \ni (m,x)\mapsto (m, F(m,x))\in Z$ has a constant number of points in the fibers outside the bifurcation set $B(G)\subset Z.$ Take $U=Z{\setminus } B(G).$ By Theorem 2.4 the set U is open. Let $\pi : Z \ni (m, y)\mapsto m\in M$ be the projection. We show that the constructible set $\pi (U)$ is dense in M. Indeed, assume that $\overline{\pi (U)}=N$ is a proper subset of M. Since U is dense in Z, we have $\pi (Z)\subset N$, i.e., $Z\subset N\times Y.$ This is a contradiction. In particular the set $\pi (U)$ is dense in M and it contains a Zariski open, non-empty subset $U_1\subset M.$ Of course $\mu (f_m)=\mu ({\mathcal {F}})$ for $m\in U_1.$

• (2) Consider the projection $\pi : Z\ni (m,y)\mapsto m\in M.$ As we know from (1), the mapping $\pi$ is dominant. By a well known result, after shrinking $U_1$ we can assume that every fiber $Z_m$ of $\pi$ ($m\in U_2\subset U_1$) is of pure dimension $d=\mathrm{dim}\ Z- \mathrm{dim}\ M=\mathrm{dim}\ X.$ However, $Z_m =\overline{f_m(X)}\cup B(G)_m.$ Generically the dimension of $B(G)_m$ is less than d. Hence if we possibly shrink $U_2$, we get $Z_m=\overline{f_m(X)}$ for $m\in U_2.$ Moreover, by Lemma 3.1 (after shrinking $U_2$ if necessary), we can assume that $Sing(Z_m)=Sing(Z)_m:=(m\times Y)\cap Sing(Z).$ Now it is easy to see that $B(f_m)=B(G)_m.$

• (3) We have $\overline{f_m(X)}=Z_m$ and $B(f_m)= B(G)_m$ for $m\in U_2$. Now apply Lemma 3.2 with $X=U_2\times Y,$ $W=(U_2\times Y)\cap Z$, $N=(U_2\times Y)\cap B(G)$ and $f: U_1\times Y\ni (m,y)\mapsto m\in U_1.$

$\square$

Now we are ready to prove our main result:

Theorem 3.5

Let X, Y be smooth affine irreducible varieties. Every algebraic family ${\mathcal {F}}$ of polynomial mappings from X to Y contains only a finite number of topologically non-semi-equivalent (non-equivalent) generically-finite (proper) mappings.

Proof

The proof is by induction on $\dim M.$ We can assume that M is affine, irreducible and smooth. Indeed, M can be covered by a finite number of affine subsets $M_i$, and we can consider the families ${\mathcal {F}}_{|M_i}$ separately. For the same reason we can assume that M is irreducible. Finally dim $M{\setminus } Reg(M)<$ dim M and we can use induction to reduce the general case to the smooth one.

Assume that M is smooth and affine. If $\mu ({\mathcal {F}})=0$, then ${\mathcal {F}}$ does not contain any generically-finite mapping. Hence we can assume that $\mu ({\mathcal {F}})=k>0.$ By Lemma 3.4 there is a non-empty open subset $U\subset M$ such that for every $m_1, m_2\in U$ we have

• (1) $\mu (f_{m_1})=\mu (f_{m_2})=k,$

• (2) The pairs $(\overline{f_{m_1}(X)}, B(f_{m_1}))$ and $(\overline{f_{m_2}(X)}, B(f_{m_2}))$ are equivalent by means of a homeomorphism, i.e., there is a homeomorphism $\Psi : Y\rightarrow Y$ such that $\Psi (\overline{f_{m_1}(X)})=\overline{f_{m_2}(X)}$ and $\Psi (B(f_{m_1}))=B(f_{m_2}).$

Fix a pair $Q=\overline{f_{m_0}(X)}, B=B(f_{m_0})$ for some $m_0\in U_3.$ For $m\in U_3$ the mapping $f_m: X\rightarrow Y$ is topologically equivalent to the continuous mapping $f'_m=\Psi _m\circ f_m$ with $\overline{f'_{m}(X)}=Q$ and $B(f'_m)=B$ (Lemma 3.4). Every mapping $f'_m$ induces a topological covering $f'_m :X{\setminus } {f'_m}^{-1}(B)=P_{f'_m}\rightarrow R=Q{\setminus } B$. Take a point $a\in R$ and let $a_{f'_m}\in {f'_m}^{-1}(a).$ We have an induced homomorphism

$$\begin{aligned} f_* :\pi _1(P_{f'_m}, a_{f'_m})\rightarrow \pi _1(R,a). \end{aligned}$$

Denote $H_f=f_*(\pi _1(P_f, a_f))$ and $G=\pi _1(R,a).$ Hence $[G:H_f]=k.$ It is well known that the fundamental group of a smooth algebraic variety is finitely generated. In particular the group $G:=\pi _1(Q{\setminus } B, a)$ is finitely generated. Let us recall the following result of M. Hall (see [4]):

Lemma 3.6

Let G be a finitely generated group and let k be a natural number. Then there are only a finite number of subgroups $H\subset G$ such that $[G:H]=k.$

By Lemma 3.6 there are only a finite number of subgroups $H_1,..., H_r\subset G$ with index k. Choose generically-finite (proper) mappings $f_i=f'_{m_i}=\Psi _i\circ f_{m_i} : X\rightarrow Y$ such that $H_{f_i}=H_i$ (of course only if such a mapping $f_i$ does exist). We show that every generically-finite (proper) mapping $f'_m$ ($m\in U$) is semi-equivalent (equivalent) to one of mappings $f_i.$

Indeed, let $H_{f'_m}=H_{f_i}$ (here $f'_m=\Psi _m\circ f_m$). We show that $f'_m:=f$ is equivalent to $f_i.$ Let us consider two coverings $f: (P_f, a_f)\rightarrow (R,a)$ and $f_i: (P_{f_i}, a_{f_i})\rightarrow (R,a).$ Since $f_*(\pi _1(P_f, a_f))={f_i}_*(\pi _1(P_{f_i}, a_{f_i}))$ we can lift the covering f to a homeomorphism $\phi : P_f\rightarrow P_{f_i}$ such that following diagram commutes:

Hence for generically-finite mappings we have

$$\begin{aligned} (\Psi _i)^{-1}\circ \Psi _m\circ f_{m}\circ \phi ^{-1}|_U=f_{m_i}|_U, \end{aligned}$$

where $V=X{\setminus } f_{m}^{-1}(B(f_{m}))$ and $U=X{\setminus } f_{m_i}^{-1}(B(f_{m_i}))$. Hence $f_{m}$ is semi-equivalent to $f_{m_i}.$

In the case of proper mappings we show additionally that the mapping $\phi$ can be extended to a continuous mapping $\Phi$ on the whole of X. Indeed, take a point $x\in f^{-1}(B)$ and let $y=f(x).$ The set $f_i^{-1}(y)=\{ b_1,..., b_s\}$ is finite. Take small open disjoint neighborhoods $W_i(r)$ of $b_i$, such that $W_i(r)$ shrinks to $b_i$ as r tends to 0. We can choose an open neighborhood V(r) of y so small that $f_i^{-1}(V(r))\subset \bigcup ^s_{j=1} W_i(r).$ Now take a small connected neighborhood $P_x(r)$ of x such that $f(P_x(r))\subset V(r).$ The set $P_x(r){\setminus } f^{-1}(B)$ is still connected and it is transformed by $\phi$ into one particular set $W_{i_0}(r).$ We take $\Phi (x)=b_{i_0}.$ It is easy to see that the mapping $\Phi$ so defined is a continuous extension of $\phi .$ In fact $\phi (P_x(r){\setminus } f^{-1}(B))$ shrinks to $b_{i_0}$ if r goes to 0. Moreover, we still have $f=f_i\circ \Phi .$

In a similar way the mapping $\Lambda$ determined by $\phi ^{-1}$ is continuous. It is easy to see that $\Lambda \circ \Phi =\Phi \circ \Lambda =identity$, hence $\Phi$ is a homeomorphism. Consequently, the mapping $f_i\circ \Phi =\Psi _i\circ f_{m_i}\circ \Phi$ is equal to $f=\Psi _m\circ f_{m}.$ Finally, we get

$$\begin{aligned} (\Psi _i)^{-1}\circ \Psi _m\circ f_m\circ \Phi ^{-1}=f_{m_i}. \end{aligned}$$

This means that the family ${\mathcal {F}}_{|U}$ contains only a finite number of topologically non-semi-equivalent (non-equivalent) generically-finite (proper) mappings. In fact, the number of topological semi-types (types) of generically-finite (proper) mappings in ${\mathcal {F}}_{|U}$ is bounded by the number of subgroups of G of index $\mu ({\mathcal {F}}).$

Let $T=M{\setminus } U.$ Hence dim $T<$ dim M. By the induction the family ${\mathcal {F}}_{|T}$ also contains only a finite number of topologically non-semi-equivalent (non-equivalent) generically-finite (proper) mappings. Consequently so does ${\mathcal {F}}$. $\square$

Corollary 3.7

There is only a finite number of topologically non-semi-equivalent (non-equivalent) generically-finite (proper) polynomial mappings $f: {\mathbb {C}}^n\rightarrow {\mathbb {C}}^m$ of a bounded algebraic degree. $\square$

4 Families of proper mappings

In this section we slightly extend our previous result in the case of irreducible families of proper (or generically-finite) mappings. First we prove a following lemma:

Lemma 4.1

Let $Y={\mathbb {R}}^{n}$ and let $Z\subset Y$ be a linear subspace of Y. Fix $\epsilon >0$ and take $\eta <\epsilon .$ Let $B( 0, \eta )$ be a ball of radius $\eta .$ Let $\gamma : I\ni t\mapsto \gamma (t)\in B( 0, \eta )\cap Z$ be a smooth path. Then there exists a continuous family of diffeomorphisms $\Phi _t : Y\rightarrow Y$, $t\in [0,1]$ such that

• (1) $\Phi _t(\gamma (t))=\gamma (0) \ \mathrm{{ and}}\ \ \Phi _t(z)=z \ \ \mathrm{{ for}}\ \ \Vert z\Vert \ge \epsilon .$

• (2) $\Phi _0=\mathrm{identity}.$

• (3) $\Phi _t(Z)=Z.$

Proof

Let $v_t=\gamma (0)-\gamma (t)\in T{\mathbb {R}}^n.$ We construct a family of diffeomorphisms $\Phi _t$, which are interpolation between translation $x\rightarrow x+v_t$ and identity.

Let $\sigma : Y\rightarrow [0, 1]$ be a differentiable function such that $\sigma =1$ on $B(0, \eta )$ and $\sigma =0$ outside $B(0, \epsilon ).$ Define a vector field $V(x)=\sigma (x)v_t.$ Integrating this vector field we get desired diffeomeorphisms $\Phi _t,$ for any t. $\square$

Corollary 4.2

Let Y be a smooth manifold and Z be a smooth submanifold. For every point $a\in Z$ and every open neighborhood $V_a$ of the point a, there is an open connected neighborhood $U_a$ of the point a, such that:

• (a) $\overline{U_a}\subset V_a$,

• (b) if $\gamma : I\ni t\mapsto \gamma (t)\in U_a\cap Z$ is a smooth path, then there is a continuous family of diffeomorphism $\psi _t : Y\rightarrow Y$, $t\in [0,1]$ such that

• (1) $\psi _t(\gamma (t))= \gamma (0),$

• (2) $\psi _t(x)=x$ for $x\not \in V_a$ and $\psi _0=\mathrm{identity},$

• (3) $\psi _t(Z)=Z.$

Now we are in a position to prove:

Theorem 4.3

Let X, Y be smooth affine irreducible varieties. Let ${\mathcal {F}}: M\times X\rightarrow Y$ be an algebraic family of proper polynomial mappings from X to Y. Assume that M is an irreducible variety. Then there exists a Zariski open dense subset $U\subset M$ such that for every $m,m'\in U$ mappings $f_m$ and $f_{m'}$ are topologically equivalent.

Proof

We follow the proof of Theorem 3.5 and we use here the same notation. By Lemma 3.4 there is a non-empty open subset $U\subset M$ such that for every $m_1, m_2\in U$ we have

• (1) $\mu (f_{m_1})=\mu (f_{m_2})=k,$

• (2) The pairs $(\overline{f_{m_1}(X)}, B(f_{m_1}))$ and $(\overline{f_{m_2}(X)}, B(f_{m_2}))$ are equivalent by means of a homeomorphism, i.e., there is a homeomorphism $\Psi : Y\rightarrow Y$ such that $\Psi (\overline{f_{m_1}(X)})=\overline{f_{m_2}(X)}$ and $\Psi (B(f_{m_1}))=B(f_{m_2}).$

Fix a pair $(Q=\overline{f_{m_0}(X)}, B=B(f_{m_0}))$ for some $m_0\in U.$ For $m\in U$ the mappings $f_m$ and $f_{m_0}$ can be connected by a continuous path $f_t, f_0=f_{m_0}, f_1=f_m.$ Moreover we have also a continuous family of homeomorphisms $\Psi _t : Y\rightarrow Y$ such that $\Psi _t(\overline{f_{t}(X)})=\overline{f_{0}(X)}$ and $\Psi (B(f_{t}))=B(f_{0}).$ It is enough to prove that mappings $F_t=\Psi _t\circ f_t$ are locally (in the sense of parameter t) equivalent.

(1) First step of the proof. Let $C_t\subset X$ denotes the preimage by $F_t$ of the set B (in fact $C_t=f_t^{-1}(B(f_t)$) and put $X_t=X{\setminus } C_t.$ Put $Q':=Q{\setminus } B.$Assume that for all mappings $F_t$ there is a point $a\in (X{\setminus } \bigcup _{t\in I} C_t)$ such that for all $t\in I$ we have $F_t(a)=b.$

We have an induced homomorphism ${G_t}_*: \pi _1(X_t,a)\rightarrow \pi _1(Q',b)$. We show that the subgroup ${F_t}_*( \pi _1(X_t,a))\subset \pi _1(Q',b)$ does not depend on t.

Indeed let $\gamma _1,..., \gamma _s$ be generators of the group $\pi _1(X_{t_0}, a).$ Let $U_i$ be an open relatively compact neighborhoods of $\gamma _i$ such that $\overline{U_i}\cap C_{t_0}=\emptyset .$ For sufficiently small number $\epsilon >0$ and $t\in (t_0-\epsilon ,t_0+\epsilon )$ we have $\overline{U_i}\cap C_{t}=\emptyset .$ Let $t\in (t_0-\epsilon ,t_0+\epsilon )$. Note that the loop $F_t(\gamma _i)$ is homotopic with the loop $F_{t_0}(\gamma _i).$ In particular the group ${F_{t_0}}_*( \pi _1(X_{t_0},a))$ is contained in the group ${F_t}_*( \pi _1(X_t,a)).$ Since they have the same (finite!) index in $\pi _1(Y',b)$ they are equal. This means that the subgroup ${G_t}_*( \pi _1(X_t,a))\subset \pi _1(Y',b)$ is locally constant, hence it is constant.

Let us consider two coverings $F_t: (X_t, a)\rightarrow (Q',b)$ and $F_0: (X_0, a)\rightarrow (Q',b)$. Since ${F_t}_* \pi _1(X_t, a)={F_0}_*\pi _1 (X_0, a)$ we can lift the covering $F_t$ to a homeomorphism $\phi _t: X_t\rightarrow X_0.$ As before we can extend the homeomorphism $\phi _t$ to the homeomorphism $\Phi _t: X\rightarrow X$, such that $F_0\circ \Phi _t =F_t.$

(2) The general case. Now we can prove Theorem 4.3. Since in general there is no a point $a\in (X{\setminus } \bigcup _{t\in I} C_t)$ such that for all $t\in I$ we have $F_t(a)=b,$ we have to modify our construction.

First we prove that for every $t_0\in I$ there exists $\epsilon >0$ and a family of homeomorphisms $\Phi _t: X\rightarrow X$, $t\in (t_0-\epsilon , t_0+\epsilon )$ such that $F_t=F_{t_0}\circ \Phi _t$ for $t\in (t_0-\epsilon , t_0+\epsilon ).$ Take a point $a\in X_{t_0}$ and choose $\epsilon >0$ so small that $a\in X_t$ for $t\in (t_0-\epsilon , t_0+\epsilon )$. Put $\gamma (t)\ni t\mapsto F_t(a)\in Y'.$ We can take $\epsilon$ so small that the hypothesis of Corollary 4.2 is satisfied. Applying Corollary 4.2 with $Y'=Y{\setminus } B$ and $Z=Q{\setminus } B$ we have a continuous family of diffeomeorphisms $\psi _t: Y\rightarrow Y$ which preserves Q and B, $t\in (t_0-\epsilon , t_0+\epsilon )$ such that $\psi _t(F_t(a))=F_{0}(a).$ Take $G_t=\psi _t\circ F_t.$ Arguing as in the first part of our proof all $G_t$ are topologically equivalent for $t\in (t_0-\epsilon , t_0+\epsilon )$. Hence also all $F_t$ are topologically equivalent for $t\in (t_0-\epsilon , t_0+\epsilon )$. Since $F_t$ are locally topologically equivalent, they are topologically equivalent for every $t\in I$. $\square$

Corollary 4.4

Let $n\le m$ and let $\Omega _n(d_1,...,d_m)$ denotes the family of all polynomial mappings $F=(f_1,...,f_m):{\mathbb {C}}^n\rightarrow {\mathbb {C}}^m$ of a multi-degree bounded by $(d_1,...,d_m).$ Then any two general members of this family are topologically equivalent.

Proof

Indeed, it is enough to note that a generic mapping $f\in \Omega _n(d_1,...,d_m)$ is proper. $\square$

Using the same method we can prove:

Theorem 4.5

Let X, Y be smooth affine irreducible varieties. Let ${\mathcal {F}}: M\times X\rightarrow Y$ be an algebraic family of generically-finite polynomial mappings from X to Y. Assume that M is an irreducible variety. Then there exists a Zariski open dense subset $U\subset M$ such that for every $m,m'\in U$ the mappings $f_m$ and $f_{m'}$ are topologically semi-equivalent.