# Injectivity of the Composition Operators of Étale Mappings

Authors: Ronen Peretz
Journal: Algebra
 (2014)

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## Abstract

LetX be a topological space. The semigroup of all the étale mappings of X (the local homeomorphisms X→X) is denoted by et(X). If G∈et(X), then the G-right (left) composition operator on et(X) is defined by RG  LG:et(X)→et(X), RGF=F∘G(LGF=G∘F). When are the composition operators injective?  The Problem originated in a new approach to study étale polynomial mappings C2→C2 and in particular the two-dimensional Jacobian conjecture. This approach constructs a fractal structure on the semigroup of the (normalized) Keller mappings and outlines a new method of a possible attack on this open problem (in preparation). The construction uses the left composition operator and the injectivity problem is essential. In this paper we will completely solve the injectivity problems of the two composition operators for (normalized) Keller mappings. We will also solve the much easier surjectivity problem of these composition operators.


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## Body

## 1. Introduction

LetX be a topological space. A mapping F:X→X is called a local homeomorphism of X or an étale mapping of X if for any point x∈X there exists a neighborhood U of x such that the restriction of F to U, denoted by F|U, is an homeomorphism. The set of all the étale mappings of X, denoted by et(X), is a semigroup with a unit with the composition of mappings taken to be the binary operation. If G∈et(X), then the G-right composition operator on et(X) is defined by
(1)RG:et(X)⟶et(X)RG(F)=F∘G.
The G-left composition operator on et(X) is defined by
(2)LG:et(X)⟶et(X)LG(F)=G∘F.
We were interested in the injectivity of these two composition operators in two particular cases. The first is the case of entire functions C→C that are étale (and normalized). The second case is that of the polynomial mappings C2→C2 with determinant of their Jacobian matrix equal (identically) to 1 and whose Y-degrees equal their total degrees. For the first case we use the following.Definition 1.
Consider the following(3)elhC=f:C⟶C∣fisentire,∀z∈Cf′(z)≠0,f′(0)=1h∀z∈Cf′(z)≠0,f′(0)=1.
Thus we use in this case the symbol elh(C) instead of et(X). Then we have the following.Proposition 2 (see [1]).
Consider the following∀f∈elh(C), Rf is injective.Theorem 3 (see [1]).
Letf∈elh(C). Then Lf is not injective if and only if
(4)f(z)=1bebz+aforsomea∈C,b∈C×.
This settled the first case. It should be noted (see [1]) that the proof for the left composition operator is much more involved than the proof for the right composition operator (which follows directly from the Picard little theorem). It is in fact the second case that initiated our interest in the injectivity of the composition operators. It results from a new approach to study étale polynomial mappings C2→C2 and in particular the two-dimensional Jacobian conjecture [2–4]. This approach constructs a fractal structure on the semigroup of the (normalized) Keller mappings and outlines a new method of a possible attack on this open problem (in preparation). The construction uses the left composition operator and the injectivity problem is essential. In this paper we will completely solve the injectivity problems of the two composition operators for (normalized) Keller mappings. We will also solve the much easier surjectivity problem of these composition operators.


## 2. The Semigroupet(C2) of Normalized Keller Mappings and a Few Facts on Their Asymptotic Variety

LetF∈C[X,Y]2 be an étale mapping that satisfies the two normalizations(1)
det⁡JF≡1;(2)
deg⁡P=deg⁡YP and deg⁡Q=deg⁡YQ, where F(X,Y)=(P(X,Y),Q(X,Y))∈C[X,Y]2.The set of all such mappings F will be denoted by et(C2). This semigroup (with respect to composition of mappings) is the parallel of the semigroup elh(C) for entire functions. The 2-dimensional Jacobian conjecture can be rephrased in each of the following forms: (a)
et(C2)⊆Aut(C2);(b)
(et(C2),∘) is a group.For the next survey of results we refer to the following paper: [5]. We denote by A(F) the asymptotic variety of F, that is, the curve of all the asymptotic values of the mapping F. The canonical geometric basis of F will be denoted by R0(F). This basis consists of finitely many rational mappings of the following form: R(X,Y)=(X-α,XβY+X-αΦ(X)), where α∈Z+, β∈Z+∪{0}, Φ(X)∈C[X], and deg⁡Φ<α+β. Also the effective X powers in Xα+βY+Φ(X) have a gcd which equals 1. The cardinality of the geometric basis, |R0(F)|, equals the number of components of the affine algebraic curve A(F). ∀R∈R0(F) we have the double asymptotic identity F∘R=GR∈C[X,Y]2, where the polynomial mapping GR is called the R-dual of F. Each R∈R0(F) generates exactly one component of A(F). This component is normally parametrized by {GR(0,Y)∣Y∈C}. We will denote by HR(X,Y)=0 the implicit representation of this component in terms of the irreducible polynomial HR∈C[X,Y]. There exists a natural number γ(R)≥2 and a polynomial SR(X,Y)∈C[X,Y]. The affine curve SR(X,Y)=0 is called the R-phantom curve of F. The R-component of A(F), HR(X,Y)=0, is a polynomial curve, which is not isomorphic to A1, and hence in particular must be a singular irreducible curve. We have the relation HR(GR(X,Y))=Xγ(R)SR(X,Y). The exponent γ(R) satisfies the double inequality 2≤γ(R)≤β-α. In our case of the canonical rational mappings R∈R0(F), we have sing(R)={X=0}. The following is true:
(5)GR-1HRX,Y=0=GR-1GRsingR=sing(R)∪SRX,Y=0.
Thus the GR-preimage of the R-component of A(F) (which is the GR-image of sing(R)) is the union of two curves: the first is sing(R) and the second is the so-called R-phantom curve of F. Even if for a single R(X,Y) the R-phantom curve is empty, then JC(2) follows. Also if ∀R∈R0(F)sing(R)∩{SR(X,Y)=0}=∅, then F is a surjective mapping.Proposition 4.
IfF,G∈et(C2), then R0(G)⊆R0(F∘G), F(A(G))⊆A(F∘G).Proof.
R
∈
R
0
(
G
)
⇒
G
∘
R
∈
C
[
X
,
Y
]
2
⇒
F
∘
(
G
∘
R
)
∈
C
[
X
,
Y
]
2
⇒
(
F
∘
G
)
∘
R
∈
C
[
X
,
Y
]
2
⇒
R
∈
R
0
(
F
∘
G
). Next we have (a,b)∈F(A(G))⇒∃R∈R0(G)∃Y∈C such that (a,b)=F((G∘R)(0,Y))⇒∃R∈R0(F∘G)∃Y∈C such that (a,b)=((F∘G)∘R)(0,Y)⇒(a,b)∈A(F∘G).The proposition tells us that compositions of étale mappings do not decrease the geometric basis of the right factor and consequently do not decrease the left image of its asymptotic variety. We naturally ask under what conditions the geometric basis ofF∘G is actually larger than that of G? In other words, we would like to know when is it true that R0(G)⊂R0(F∘G)? This happens exactly when ∃R∈R0(F∘G)-R0(G). This means that (F∘G)∘R∈C[X,Y]2, G∘R∉C[X,Y]2. Let R(X,Y)=(X-α,XβY+X-αΦ(X)), G(X,Y)=(P(X,Y),Q(X,Y)). Then
(6)G∘RX,Y=PX-α,XβY+X-αΦX,hhhQ(X-α,XβY+X-αΦ(X))∈CX,Y2-CX,Y2.
We clearly have sing(G∘R)⊆sing(R) and so sing(G∘R)={X=0}. By F∘(G∘R)=(F∘G)∘R∈C[X,Y]2 we have G∘R∈R(F). This is not necessarily a member of the geometric basis of F. The canonical geometric basis of F, R0(F) contains finitely many rational mappings of the form S(X,Y)=(X-a,XbY+X-aΨ(X)). Since G∈et(C2), it follows that |C2-G(C2)|<∞ (a similar phenomenon as the Picard little theorem). If L is an asymptotic tract of F, then G-1(L) cannot be a bounded subset of C2. The reason is that if G-1(L)¯ is compact, then G(G-1(L)¯) is compact and since L⊆G(G-1(L))⊆G(G-1(L)¯) this would imply the contradiction that L is bounded (and hence cannot be an asymptotic tract). Hence G-1(L) has at least one component, say L1, that goes to infinity. This is because the number of components of G-1(L) is finite and G-1(L) is not bounded. So F∘G has a limit along L1 which equals the above asymptotic value of F. This proves the following generalization of the second part of Proposition 4, namely, the following proposition.Proposition 5.
IfF,G∈et(C2), then A(F)∪F(A(G))=A(F∘G).This proposition implies that ifA(F)⊂F(A(G)), then necessarily R0(G)⊂R0(F∘G) because, as shown in the proof of Proposition 4  ∀R∈R0(G), ((F∘G)∘R)(sing(R))⊆F(A(G)).Proposition 6.
LetF∈et(C2). If ∃G∈et(C2) such that R0(G)=R0(F∘G), then F(C2)=C2; that is, F is a surjective mapping.Proof.
SinceF∈et(C2) we have C2-F(C2)⊆A(F) because in this case the only points in the complement of the image of F are the finitely many Picard exceptional values of F which are asymptotic values of F. If, as the assumption says, R0(G)=R0(F∘G), then by Proposition 5 we must have A(F)⊆F(A(G))⊆F(C2) and so there are no Picard exceptional values of the mapping F.IfF∈et(C2) is not a surjective mapping, then the last proposition implies that ∀G∈et(C2); we must have R0(G)⊂R0(F∘G). In particular R0(F)⊂R0(F∘F). This is the choice G=F. If we choose G=F∘F, we get R0(F∘F)⊂R0(F∘(F∘F)). Now it is clear that by induction we get the infinite chain of strict inequalities:
(7)R0F⊂R0F∘F⊂R0F∘F∘F⊂⋯⊂R0(F∘n)⊂⋯,
where F∘n=F∘⋯∘Fn times fold composition. Since the cardinality of the geometric basis R0(G) is the number of components of the asymptotic variety A(G), it follows in this case that the asymptotic varieties of iterates of F are of increasing complexity in the sense that the number of components of the curve A(F∘(n+1)) is strictly larger than the number of components of the curve A(F∘n). By Proposition 5 we always have A(F)∪F(A(F))=A(F∘F). Hence A(F)∪F(A(F))∪(F∘F)(A(F))=A(F)∪F(A(F)∪F(A(F))) = A(F)∪F(A(F∘F))=A(F∘F∘F). By induction we get in general:
(8)A(F)∪⋃k=1n‍F∘k(A(F))=A(F∘(n+1)).


## 3. The Composition Operators onet(C2) Are Not Surjective, but the Right Composition Operator Is Injective

Proposition 7.
The mappingsRF, LF are not surjective if and only if F∉Aut(C2). In fact in this case we have RF(et(C2))⊂et(C2)-Aut(C2), LF(et(C2))⊂et(C2)-Aut(C2).Proof.
Consider the following:(9)R0(RF(G))=R0(G∘F)⊇R0(F)≠∅,ALFG=AF∘G⊇AF≠∅.Proposition 8.
R
F is injective.Proof.
R
F
(
G
)
=
R
F
(
H
)
⇒
G
∘
F
=
H
∘
F. Since F∈et(C2) we have |C2-F(C2)|<∞ and by the assumption G|F(C2)=H|F(C2). Hence G≡H.We naturally inquire if alsoLF is injective. So let us assume that LF(G)=LF(H). Then F∘G=F∘H. If we denote T=F∘G, then R0(G),R0(H)⊆R0(T) and also A(F)∪F(A(G))=A(F)∪F(A(H)). If A(F)=∅, then F∈Aut(C2) and so G=F-1(F∘G)=F-1(F∘H)=H. If, on the other hand, G≠H, then there are points (X,Y)∈C2 for which G(X,Y)≠H(X,Y). By our assumption, F(G(X,Y))=F(H(X,Y)), so F is not in Aut(C2) and it identifies different images of G and H of the same (X,Y). We ask the following question: suppose that F∘G=F∘H, G≠H, is there a point (X,Y)∈C2 for which G(X,Y)=H(X,Y)?Based on our experience with entire functions we tend to prove that the answer to the question is negative. Indeed this is the case and the proof is almost identical to the entire case; see [1]. Namely, if the answer is affirmative, then we have two types of points in C2: (U,V)∈C2 for which G(U,V)≠H(U,V) and the complimentary set, where both sets are nonempty. Let us denote by N the first subset of C2; that is,
(10)N=U,V∈C2∣GU,V≠HU,V.
The subset N of C2 is open in the strong topology because G, H are étale mappings and if G(U,V)≠H(U,V), then ∃O an open neighborhood of (U,V) in the strong topology such that G(O)∩H(O)=∅. So the complimentary subset of N is a closed nonempty subset of C2. Let (X,Y)∈∂Nc be a boundary point of Nc. Let (Un,Vn)∈N satisfy lim⁡(Un,Vn)=(X,Y). Then ∀n∈Z+, G(Un,Vn)≠H(Un,Vn), F(G(Un,Vn))=F(H(Un,Vn)), and G(X,Y)=H(X,Y). This implies that in any strong neighborhood of G(X,Y)=H(X,Y) there are different points, say G(Un,Vn)≠H(Un,Vn) for n∈Z+ large enough, so that F(G(Un,Vn))=F(H(Un,Vn)). Hence F is not injective in any strong neighborhood of G(X,Y)=H(X,Y). Thus F∉et(C2). This contradiction proves the following.Proposition 9.
Suppose thatLF is not injective for some F∈et(C2). Then ∀G≠H, G,H∈et(C2) such that LF(G)=LF(H) and ∀(X,Y)∈C2 we have G(X,Y)≠H(X,Y).Remark 10.
Proposition7 asserts that ∀F∈et(C2)-Aut(C2)∃G,H∈et(C2)-Aut(C2) such that ∀M∈et(C2) we have M∘F≠G and F∘M≠H.


## 4. The Size,dF, of the Generic Fiber of a Keller Mapping F∈et(C2)

We will need the generic size of a fiber of a mappingF=(P,Q)∈et(C2). If we denote deg⁡P(X,Y)=n and deg⁡Q(X,Y)=m, then ∀(a,b)∈C2 the F fiber over (a,b) is F-1(a,b)={(x,y)∈C2∣F(x,y)=F(a,b)}. It is well known that this set is a finite subset of C2 and, by the Bezout theorem, we have |{(x,y)∈C2∣F(x,y)=(a,b)}|=|F-1(a,b)|≤n·m. Moreover, there is a number that we will denote by dF such that generically in (a,b) we have |F-1(a,b)|=dF. This means that {(a,b)∈C2∣|F-1(a,b)|≠dF} is a closed and proper Zariski subset of C2. In fact ∀(a,b)∈C2,|F-1(a,b)|≠dF⇒|F-1(a,b)|<dF. Thus we have dF=max⁡{|F-1(a,b)|∣(a,b)∈C2}.Definition 11.
LetF∈et(C2). We will denote dF=max⁡{|F-1(a,b)|∣(a,b)∈C2}. We will call dF the geometrical degree of the étale mapping F.Proposition 12.
Consider the following∀F,G∈et(C2),dF∘G=dF·dG.This is a well-known result. We include one of its proofs for convenience.Proof.
∀
(
a
,
b
)
∈
C
2
,
(
F
∘
G
)
-
1
(
a
,
b
)
=
G
-
1
(
F
-
1
(
a
,
b
)
). But generically in a,b,  |F-1(a,b)|=dF and generically in (c,d), |G-1(c,d)|=dG.Definition 13.
An étale mappingF∈et(C2) is composite if ∃G,H∈et(C2)-Aut(C2) such that F=G∘H. An étale mapping A∈et(C2)-Aut(C2) is prime if it is not composite. This is equivalent to A=B∘C for some B,C∈et(C2)⇒B∈Aut(C2)∨C∈Aut(C2). The subset of et(C2) of all the prime mappings will be denoted by etp(C2). Thus the set of all the composite étale mappings is et(C2)-etp(C2).Proposition 14.
∀
F
∈
et
(
C
2
)
-
et
p
(
C
2
), dF is not a prime number. Equivalently, ∀F∈et(C2), dF is a prime number ⇒F∈etp(C2).Proof.
F
∈
et
(
C
2
)
-
e
t
p
(
C
2
)
⇒
∃
G
,
H
∈
et
(
C
2
)
-
Aut
(
C
2
) such that F=G∘H (by the definition) ⇒dF=dG·dH,dG,dH>1 (by Proposition 12 and the fact that dM=1⇔M∈Aut(C2))⇒dF is a composite integer.Theorem 15.
The following hold true.(1)
Ifet(C2)-Aut(C2)≠∅, then etp(C2)≠∅.(2)
∀F∈et(C2)∃k∈Z+∪{0}∃A0∈Aut(C2)∃P1,…,Pk∈etp(C2) such that F=A0∘P1∘⋯∘Pk.Proof.
Ifetp(C2)=∅, then et(C2)-Aut(C2) are all composite étale mappings. Let F∈et(C2)-Aut(C2), then ∃G1,G2′∈et(C2)-Aut(C2) such that F=G1∘G2′. So ∃G2,G3′∈et(C2)-Aut(C2) such that G2′=G2∘G3′. Hence F=G1∘G2∘G3′. Continuing this we get for any k∈Z+∃G1,…,Gk∈et(C2)-Aut(C2) such that F=G1∘⋯∘Gk and by Proposition 12  dF=∏j=1k‍dGj. But ∀1≤j≤k, dGj≥2 and so ∀k∈Z+, dF≥2k, a contradiction to dF<∞. Thus etp(C2)≠∅.
Now part 2 is standard, for ifF∈Aut(C2) we take A0=F and k=0. If F∈etp(C2) we take A0=id, k=1, and P1=F. If F∈et(C2)-etp(C2), then F=G∘H for some G,H∈et(C2)-Aut(C2). So by Proposition 12  dF=dG·dH and since dG,dH≥2 it follows that dG,dH<dF and we conclude the proof of part 2 using induction on the geometrical degree. Namely, G=P1∘⋯∘Pm, H=Pm+1∘⋯∘Pk for m≥1, k≥m+1, and some primes P1,…,Pk∈etp(C2).


## 5. The Metric Spaces(et(C2),ρD)

We will need a special kind of four (real) dimensional subsets ofR4. These will serve us to construct suitable metric structures on et(C2).Definition 16.
LetD be an open subset of C2 with respect to the strong topology that satisfies the following conditions. (1)
int⁡(D¯)=D (D has no “slits”).(2)
D¯ is a compact subset of C2 (in the strong topology).(3)
∀G1,G2∈et(C2), G1(D)=G2(D)⇔G1=G2.We define the following real valued function:
(11)ρD:et(C2)×et(C2)⟶R+∪{0},ρD(G1,G2)=thevolumeofG1(D)ΔG2(D).
Here we use the standard set-theoretic notation of the symmetric difference between two sets A and B; that is, AΔB=(A-B)∪(B-A).Remark 17.
It is not clear how to construct an open subsetD of C2 that will satisfy the three properties that are required in Definition 16. We will postpone for a while the demonstration that such open sets exist.Proposition 18.
ρ
D is a metric on et(C2).Proof.
(1)ρD(G1,G2)=0⇔ the volume of G1(D)ΔG2(D)=0⇔G1(D)=G2(D) (where the last equivalence follows by the fact that G1 and G2 are local homeomorphisms in the strong topology and because of condition (1) in Definition 16) ⇔G1=G2 (by condition (3) in Definition 16).
(2) ByG1(D)ΔG2(D)=G2(D)ΔG1(D), it follows that ρD(G1,G2)=ρD(G2,G1).
(3) Here we use a little technical set-theoretic containment. Namely, for any three setsA, B, and C we have
(12)AΔC⊆(AΔB)∪(BΔC).
This implies that G1(D)ΔG3(D)⊆(G1(D)ΔG2(D))∪(G2(D)ΔG3(D)) from which it follows that
(13)thevolumeofG1DΔG3D≤thevolumeofG1DΔG2D+thevolumeofG2DΔG3D.
Hence the triangle inequality ρD(G1,G3)≤ρD(G1,G2)+ρD(G2,G3) holds.So far we thought of the volume ofG1(D)ΔG2(D) as the volume of the open set which is the symmetric difference between the G1 image and the G2 image of the open set D. However, the mappings G1 and G2 are étale and in particular need not be injective. We will take into the volume computation the multiplicities of G1 and of G2. By Theorem 3 on page 39 of  [6] we have the following: given that F:Cn→Cn we define F~=(ReF1,Im⁡F1,…,ReFn,Im⁡Fn);R2n→R2n. Then det⁡JF~=|det⁡JF|2. Thus the Jacobian condition det⁡JF≡1 implies that det⁡JF~≡1. So the real mapping F~ preserves the usual volume form. In order to take into account the multiplicities of the étale mappings G1 and G2 when computing the volume of the symmetric difference G1(D)ΔG2(D) we had to do the following. For any G∈et(C2) instead of computing,
(14)⨌Ddet⁡JG~·dV=⨌DdV,
we compute
(15)⨌G~(D)dX1dX2dY1dY2whereX=X1+iX2,Y=Y1+iY2.
For every j=1,2,…,dG we denote by Dj the subset of D such that for each point of Dj there are exactly j points of D that are mapped by G to the same image of that point. In other words, Dj={α∈D∣|G~-1(G~(α))∩D|=j}. We assume that D is large enough so that ∀j=1,…,dG we have Dj≠∅. For our étale mappings it is well known that if j<dG, then dim⁡Dj<dim⁡D, so the volume of these Dj’s contribution equals 0. The dimension claim follows by the well-known fact that the size of a generic fiber |G-1(x)| equals dG and that dG is also the maximal size of any of the fibers of G. However, for the sake of treating more general families of mappings, we denote by vol(Dj) the volume of the set Dj. Then D has a partition into exactly j subsets of equal volume. The volume of each such a set is vol(Dj)/j and each such a set has exactly one of the j points in G~-1(G~(α))∩D for each α∈Dj. We note that vol(G~(Dj))=vol(Dj)/j by the Jacobian Condition. Thus the volume with the multiplicity of G~ taken into account is given by
(16)volG~D+∑j=2dG‍j-1·volDjj=volG~D+∑j=2dG‍(j-1)·vol(G~(Dj)).
We note that G~(D)=⋃j=1dG‍G~(Dj) is a partition, so vol(G~(D))=∑j=1dG‍vol(G~(Dj)). Hence we can express the desired volume by
(17)volG~D+∑j=2dG‍j-1·volG~Dj=∑j=1dG‍j·vol(G~(Dj)).
We note that this equals ∑j=1dG‍vol(Dj) and since D=⋃j=1dG‍Dj is a partition we have vol(D)=∑j=1dG‍vol(Dj). As expected, the volume computation that takes into account the multiplicity of G is in general larger than the geometric volume vol(G~(D)). The access can be expressed in several forms:
(18)volD-volG~D=∑j=2dG‍j-1·volG~Dj=∑j=2dG‍1-1jvol(Dj).
Coming back to the computation of the metric distance ρD(G1,G2)= the volume of G1(D)ΔG2(D) we compute the volume of  G1(D)-G2(D) with the multiplicity of  G1 while the volume of  G2(D)-G1(D) is computed with the multiplicity of G2.


## 6. Characteristic Sets of Families of Holomorphic Local HomeomorphismsC2→C2

In this section we prove the existence of setsD that satisfy the three properties that are required in Definition 16. The third property will turn out to be the tricky one.Definition 19.
LetΓ be a family of holomorphic local homeomorphisms F:C2→C2. A subset D⊆C2 is called a characteristic set of Γ if it satisfies the following condition: ∀F1,F2∈Γ,  F1(D)=F2(D)⇔F1=F2.We start by recalling the well-known rigidity property of holomorphic functions in one complex variable. Also known as the permanence principle, or the identity theorem. The identity theorem for analytic functions of one complex variable says that ifD⊆C is a domain (an open and a connected set) and if  E is a subset of D that has a nonisolated point and if f(z) is an analytic function defined on D and vanishing on E, then f(z)=0 for all z∈D.There is an identity theorem for analytic functions in several complex variables, but for more than one variable the above statement is false. One correct statement is as follows: “Letf(z) and g(z) be holomorphic functions in a domain D in Cn. If f(z)=g(z) for all z in a non-empty set δ in D, then f(z)=g(z) in D. Hence, analytic continuation of holomorphic functions in several complex variables can be performed as in the case of one complex variable. Contrary to the case of one complex variable, the zero set of a holomorphic function in a domain D⊆Cn, n≥2, contains no isolated points. Thus even if f(z)=g(z) in a set with accumulation points in D, it does not necessarily follow that f(z)=g(z) in D. For example, in C2 with variables z and w we can take f(z,w)=z and g(z,w)=z2” [7, Chapter 1, page 16].In spite of the above standard identity theorem forn≥2 complex variables that requires a thick set E (i.e., an open set) on which f(z)=g(z) one can do much better. Let us start with the following. Let F(Z,W) be an entire function of two complex variables Z and W. Let us define a subset E of C2 as follows. We take a convergent sequence {Zk}k=1∞ of different numbers. Thus lim⁡Zk=a and j≠k⇒Zj≠Zk. For each k, let {Wj(k)}j=1∞ be a convergent sequence of different numbers, such that their limit is lim⁡j→∞Wj(k)=Zk′. We define E={(Zk,Wj(k))∣j,k=1,2,3,…}. Now we have the following.Proposition 20.
IfF(Z,W) vanishes on E, that is, F(Zk,Wj(k))=0 for j,k=1,2,3,…, then F(Z,W)≡0 is the zero function.Remark 21.
We note thatE is a thin set, in fact a countable set. Even the closure E¯ is thin.Proof of Proposition20.
SinceF(Z,W) is an entire function, it can be represented as a convergent power series centered at (0,0) with an infinite radius of convergence. We can sum the terms in the order we please. Let us write F(Z,W) as a power series in W with coefficients that are entire functions in Z. Thus we have F(Z,W)=∑k=0∞‍ak(Z)Wk, where for each k=0,1,2,…, ak(Z) is an entire function in the variable Z. For a fixed l∈{1,2,3,…} we have by our assumptions the following: F(Zl,Wj(l))=0 for j=1,2,3,…. But lim⁡j→∞Wj(l)=Zl′ so that gl(W)=F(Zl,W) is an entire function of the single variable W, which vanishes on a convergent sequence {Wj(l)}j=1∞. By the identity theorem of one complex variable we deduce that gl(W)≡0, the zero function. Since gl(W)=∑k=0∞‍ak(Zl)Wk it follows that the Maclaurin coefficients ak(Zl), k=0,1,2,… vanish. Now, this is valid for each l, and lim⁡Zl=a converges. Since each ak(Z) is an entire function which vanishes on a convergent sequence {Zl}l=1∞ it follows, once again, by the identity theorem in one complex variable, that is, ak(Z)≡0, the zero function k=0,1,2,…. Hence we conclude that F(Z,W)=∑k=0∞‍ak(Z)Wk≡0.This type of elementary arguments that was used to construct a thin setE for identity purpose is not new. For example, see the following.Theorem 22 (see [8]).
LetD⊆C be a domain, and let E be a subset of D that has a nonisolated point. Let F(Z,W) be a function defined for Z,W∈D such that F(Z,W) is analytic in Z for each fixed W∈D and analytic in W for each fixed Z∈D. If F(Z,W)=0 whenever Z and W both belong to E, then F(Z,W)=0 for all Z,W∈D.Advancing along the lines of the construction of the thin set in Proposition20 we note that if {Zk}k=1∞ is a sequence of different numbers that converges to lim⁡Zk=a, and if for each k=1,2,3,… there is a straight line segment [αk,βk] of W’s such that two entire functions F(Z,W) and G(Z,W) agree on the union (a countable union) of the segments {Zk}×[αk,βk], that is, F(Zk,W)=G(Zk,W), ∀W∈[αk,βk], then F(Z,W)≡G(Z,W), ∀(Z,W)∈C2.We now will construct characteristic sets of familiesΓ of holomorphic local homeomorphisms F:C2→C2.Definition 23.
Letm be a natural number and α∈C2. An m-star at α is the union of m line segments, so that any pair intersects in α.Definition 24.
Letl be a line segment and let {αk} be a countable dense subset of  l. Let {nk} be a sequence of different natural numbers and ∀k, let Snk be an nk-star at αk such that one of the star’s segments lies on l, and such that ∀k1≠k2, S~nk1∩S~nk2=∅. Here we denoted S~=S-l. Moreover, we group the stars in bundles of, say 5, thus getting the sequence of star bundles:
(19){Sn1,Sn2,Sn3,Sn4,Sn5},{Sn6,…,Sn10},…,{Sn5j+1,…,Sn5j+5},…
and for each bundle of five we take the maximal length of its rays to be at most 1/10 the length of the maximal length of the previous bundle. We define
(20)l0{nk}=l∪⋃k=1∞‍Snk.
Let {Zk}k=1∞ be a sequence of different complex numbers that converges to lim⁡Zk=a. Let {{nj(k)}j=1∞}k=1∞ be a partition of the natural numbers, Z+. In fact all we need is the disjointness, that is, k1≠k2⇒{nj(k1)}j=1∞∩{nj(k2)}j=1∞=∅. Let us consider the starred segments
(21)l0njkj=1∞∣k=1,2,3,…
and define the following countable union of starred segments in C2:
(22)⋃k=1∞‍{Zk}×l0njkj=1∞,
where we assume that the lengths of the star rays were chosen to satisfy disjointness in C2; namely,
(23)k1≠k2⟹Zk1×l0njk1j=1∞∩Zk2×l0njk2j=1∞=∅.
We let
(24)E=⋃k=1∞‍Zk×l0njkj=1∞,
or if we need a closed (compact) set, the closure of this union.Proposition 25.
LetΓ be any family of entire holomorphic local homeomorphisms F:C2→C2. Then E is a characteristic set of  Γ.Proof.
LetF1,F2∈Γ satisfy F1(E)=F2(E). Then each starred line segment
(25){Zk}×l0njkj=1∞
must be mapped onto a curve
(26)F1Zk×l0njkj=1∞=F2Zk×l0njkj=1∞
and each nj(k)-star on l, Snj(k) is mapped onto a holomorphic nj(k)-star
(27)F1Zk×Snjk=F2Zk×Snjk.
This is because the valence sequences of the stars
(28)njkj=1∞k=1∞
are pairwise disjoint natural numbers, and F1, F2 are local homeomorphisms and hence preserve the star valencies nj(k). The centers of the holomorphic stars
(29){F1(αnj(k))}={F2(αnj(k))}
form a countable and a dense subset of the curves F1({Zk}×l)=F2({Zk}×l). By continuity this implies that the restrictions
(30)F1|{Zk}×l,F2|{Zk}×l
coincide. Since F1 and F2 are holomorphic, this implies by Proposition 20 (which is a variant of the identity theorem for entire functions C2→C2) that F1≡F2.Remark 26.
Proposition25 holds true for any rigid family of local homeomorphisms. Rigidity here means that
(31)F1|{Zk}×l=F2|{Zk}×l⟺F1≡F2.
So the proposition holds true for holomorphic mappings, for harmonic mappings, and in particular for et(C2).We recall that Definition16 required also two additional topological properties; namely, the open set D should satisfy int⁡(D¯)=D; D¯ is compact (all in the strong topology). These automatically exclude the set E that was constructed in Definition 24. However, we can modify this construction to get at least an open set.Proposition 27.
LetΓ be any family of holomorphic local homeomorphisms F:C2→C2. Let U be any open subset of 2 with a smooth boundary that contains the compact E. Then the open set U-E is a characteristic set of  Γ.Proof.
SinceE can not be mapped in the smooth ∂U by an holomorphic local homeomorphism, we have for any F1,F2∈Γ for which F1(U-E)=F2(U-E) that also F1(E)=F2(E). Now the result follows by Proposition 25.Remark 28.
We note that ifU¯ is a compact then U-E satisfies, at least, the requirement that U-E¯ is compact. However, the “no-slit” condition int⁡(U-E¯)=int⁡(U¯)≠U-E fails.Now that we gained some experience with the topological construction ofE we are going to make one more step and fix its shortcomings that were mentioned above. We need to construct a domain D of C2 which has the following three properties.(1)
int⁡(D¯)=D relative to the complex topology.(2)
D¯ is a compact subset of C2 relative to the strong topology.(3)
∀G1,G2∈et(C2), G1(D)=G2(D)⇔G1≡G2.The complex topology and the strong topology are the same. Our construction will be a modification of the construction of the domain that was constructed in Proposition 27. We start by modifying the notion of an m-star that was introduced in Definition 23.Definition 29.
Letm be a natural number and α∈Cn. A thick m-star at α is a union of 2m triangles, so that any pair intersects exactly at one vertex, and this vertex (i.e., common to all the 2m triangles) is α.Definition 30.
LetE be the construction of Definition 24 that uses thick m-stars.Proposition 31.
LetΓ be any family of holomorphic local homeomorphisms F:C2→C2. Then E is a characteristic set of  Γ.Proof.
The proof is the same word by word as that of Proposition25 where we replace k-star Sk by thick k-star Sk.We finally obtain our construction.Proposition 32.
LetΓ be any family of holomorphic local homeomorphisms F:C2→C2. Let B(0,R) be an open ball centered at 0 with a radius R large enough so that E⊂B(0,R) (where E is the set in Proposition 31). Then the domain D=B(0,R)-E is a characteristic set of Γ.Proof.
The proof is the same as that of Proposition27 where we replace k-star Sk by thick k-star Sk.


## 7. Injectivity of the Left Composition OperatorLF

We would like our natural mappings, the right mappingRF and the left mapping LF, to be say bi-Lipschitz with respect to the metric ρD (that reflects the fact that our mappings et(C2) satisfy the Jacobian condition). Considering first the right mapping RF, it would mean that given three étale mappings G1,G2,F∈et(C2) and a characteristic set D of et(C2) we need to compare the volume of G1(D)ΔG2(D) (multiplicities of G1 and of G2 are taken into account) with the volume of the RF deformed set (G1∘F)(D)Δ(G2∘F)(D). A short reflection shows that the two volumes are not comparable (in the sense of bi-Lipschitz). The situation is completely different when we replace the right mapping, RF by the left mapping, LF. For example, we have the following.Proposition 33.
∀
F
∈
Aut
(
C
2
) the mapping LF is an isometry of the metric space (et(C2),ρD).Proof.
For any two mappingsG1 and G2 in et(C2) we need to compare ρD(G1,G2) with ρD(F∘G1,F∘G2). We have (using our assumption on F)
(32)(F∘G1)(D)Δ(F∘G2)(D)=FG1(D)ΔG2(D).
Since F is also (globally) volume preserving we have
(33)thevolumeofFG1DΔG2D=thevolumeofG1(D)ΔG2(D).
This proves that ρD(G1,G2)=ρD(F∘G1,F∘G2).We now drop the restrictive assumption thatF∈Aut(C2). Thus we merely have F∈et(C2) and we still want to compare ρD(G1,G2) with ρD(F∘G1,F∘G2), for any pair G1,G2∈et(C2). We only know that F is a local diffeomorphism of C2 and (by the Jacobian condition) that it preserves (locally) the volume. In this case the geometrical degree of F, dF can be larger than 1. We have the identity dF=|F-1({(a,b)})| which holds generically (in the Zariski sense) in (a,b)∈C2. Hence the (complex) dimension of the set {(a,b)∈C2∣|F-1(a,b)|<dF} is at most 1. The Jacobian condition det⁡JF≡1 implies (as we noticed before) that F preserves volume taking into account the multiplicity. The multiplicity is a result of the possibility that F is not injective and hence the deformation of the characteristic set D by F convolves (i.e., might overlap at certain locations). However, this overlapping is bounded above by dF. So if A⊆C2 is a measurable subset of C2 and we compare the volume of A with the volume of its image F(A), then
(34)thevolumeofFA≤thevolumeofA≤dF·{thevolumeofF(A)}.
This can be rewritten as follows:
(35)1dF·thevolumeofA≤thevolumeofFA≤thevolumeofA.
This is the place to emphasize also the following conclusion (that follows by the generic identity dF=|F-1({(a,b)})|); namely,
(36)lim⁡A→C2thevolumeofF(A)thevolumeofA=1dF,
provided that the set A tends to cover the whole of the complex space C2 in an appropriate manner. To better understand why the quotient tends to the lower limit 1/dF rather than to any number in the interval [1/dF,1] (if at all) we recall that our mapping belongs to et(C2) and so is a polynomial étale mapping. So any point (a,b)∈C2 for which |F-1(a,b)|<dF is an asymptotic value of  F and hence belongs to the curve AF which is the asymptotic variety of  F. In other words the identity dF=|F-1(a,b)| is satisfied exactly on the semialgebraic set C2-AF which is the complement of an algebraic curve. We now state and prove the main result of this paper.Theorem 34.
LetF,G1,G2∈et(C2). Then one has the following. (i)
ρD(F∘G1,F∘G2)≤ρD(G1,G2).(ii)
Suppose thatD is a family of characteristic sets of et(C2) such that D→C2, then ∀ϵ>0 one has
(37)1dF-ϵ·ρD(G1,G2)≤ρD(F∘G1,F∘G2)forD being large enough.(iii)
Under the assumptions in (ii) one has(38)lim⁡D→C2ρD(F∘G1,F∘G2)ρD(G1,G2)=1dF.In particular, the left mapping LF:et(C2)→et(C2), LF(G)=F∘G, is a bi-Lipschitz self-mapping of the metric space (et(C2),ρD) with the constants 1/dF≤1.Proof.
(i)x∈(F∘G1)(D)Δ(F∘G2)(D)⇒∃y∈Gj(D), j=1or2 such that x=F(y) and x∉(F∘G3-j)(D). By x∉(F∘G3-j)(D) it follows that y∉G3-j(D) and so y∈G1(D)ΔG2(D) and x=F(y)∈F(G1(D)ΔG2(D)). Hence (F∘G1)(D)Δ(F∘G2)(D)⊆F(G1(D)ΔG2(D)), so vol((F∘G1)(D)Δ(F∘G2)(D))≤vol(F(G1(D)ΔG2(D)), and finally ρD(F∘G1,F∘G2)≤ρD(G1,G2).
(ii) and (iii). Here the proof is not just set theoretic. We will elaborate more in the remark that follows this proof. We recall thatF,G1,G2∈et(C2). This implies that ∀(α,β)∈C2 we have |F-1(α,β)|≤[C(X,Y):C(F)], the extension degree of F; see [3]. This is the so-called Fiber theorem for étale mappings. Moreover the image is cofinite; |C2-F(C2)|<∞ [3]. Also F has a finite set of exactly dF maximal domains {Ω1,…,ΩdF}. This means that F is injective on each maximal domain Ωj, and i≠j⇒Ωi∩Ωj=∅, and C2=⋃j=1dF‍Ωj¯ and the boundaries ∂Ωj are piecewise smooth (even piecewise analytic). For the theory of maximal domains of entire functions in one complex variable see [9], and for that theory for meromorphic functions in one complex variable see [10, 11]. Here we use only basic facts of the theory which are valid also for more than complex variable. If D is a family of characteristic sets of et(C2) such that D→C2, then by the above G1(D),G2(D)→C2-A, where A is a finite set, and if G1≢G2, then we have the identity
(39)FG1DΔG2D-F∘G1DΔF∘G2D=x=Fy=Fz∣y∈G1D-G2Dhhhhh∧z∈G2(D)-G1(D).
Recalling that (F∘G1)(D)Δ(F∘G2(D))⊆F(G1(D)ΔG2(D))) we write the last identity as follows:
(40)FG1DΔG2D=F∘G1DΔF∘G2D∪x=F(y)=F(z)∣y∈G1(D)-G2(D)hhhhh∧z∈G2(D)-G1(D).
Taking any two points y∈G1(D)-G2(D) and z∈G2(D)-G1(D) (as in the defining equation of the set on the right-hand side in the last identity), we note that there are i≠j, 1≤i,j≤dF such that y∈Ωi∧z∈Ωj (for F(y)=F(z)!). For D~ a large enough characteristic set of et(C2), we will have z∈G1(D~) and y∈G2(D~) and so y,z∈G1(D~)∩G2(D~) (since G1(D),G2(D)→C2-{afiniteset}). Hence F(G1(D~)ΔG2(D~))-(F∘G1)(D~)Δ(F∘G2)(D~) will not include the point x. We conclude that if y and z are F-equivalent (F(y)=F(z)), then x=F(y)=F(z) will not belong to F(G1(D)ΔG2(D))-(F∘G1)(D)Δ(F∘G2)(D) for large enough D. We obtain the following crude estimate:
(41)volx=Fy=Fz∣y∈G1(D)-G2(D)hhhh∧z∈G2(D)-G1(D)=ovolF∘G1DΔF∘G2D.
One can think of D as a large open ball centered at the origin of R4, D≈B(R) and with the radius R and look at the images of the two polynomial étale mappings (F∘G1)(B(R)) and (F∘G2)(B(R)) and compare the volume of (F∘G1)(B(R))Δ(F∘G2)(B(R)) which is of the order of magnitude R4d, where d depends on the algebraic degrees of F∘G1 and F∘G2, with the volume of the set in the left-hand side of the last equation. Similar estimates are used in the theory of covering surfaces by Ahlfors, see [12, chapter 5]. We conclude that
(42)lim⁡D→C2volFG1DΔG2DvolF∘G1DΔF∘G2D=1.
Hence
(43)lim⁡D→C2ρDF∘G1,F∘G2ρDG1,G2=lim⁡D→C2volF∘G1DΔF∘G2DvolG1DΔG2D=lim⁡D→C2volF∘G1DΔF∘G2DvolFG1DΔG2D·volFG1DΔG2DvolG1DΔG2D=1·1dF=1dF.Remark 35.
The facts we used in proving (ii) and (iii) for étale mappings are in fact true in any dimensionn, that is, in Cn. In dimension n=2 it turns out that the codimension of the image of the mapping is 0 and in fact the coimage is a finite set. Also the fibers are finite and have a uniform bound on their cardinality (one can get a less tight uniform bound by the Bezout theorem). Here are few well-known facts (which one can find in Hartshorne’s book on Algebraic Geometry, [13]). (1)
The following two conditions are equivalent.
(a)
The Jacobian condition: the determinantdet⁡JF is a nonzero constant.(b)
The mapF* is étale (in standard sense of algebraic geometry). In particular it is flat.Let F*:Y→X be étale. Let Xim:=F*(Y)⊆X. (2)
For every prime ideal℘⊆A (X=spec(A)), with residue field k(℘) the ring B⊗Ak(℘) is finite over k(℘) (Y=spec(B)).(3)
F* is a quasifinite mapping.(4)
The setXim is open in X.(5)
For every pointx∈X(C) the fiber (F*)-1(x) is a finite subset of Y(C).(6)
The ring homomorphismA→B is injective, and the induced field extension K→L is finite.(7)
There is a nonempty open subsetXfin⊆Xim such that on letting Yfin:=(F*)-1(Xfin)⊆Y, the map of schemes F*|Yfin:Yfin→Xfin is finite. For any point x∈Xfin(C) we have the equality dx=dF* the geometrical degree of F*.(8)
The dimension of the setZ:=X-Xim is at most n-2.(9)
IfXim=X-Z is affine, then Z=∅ and Xim=X.Let Xcl be the topological space which is the set X(C)≅Cn given the classical topology. Similarly for Ycl, the map of schemes F*:Y→X induces a map of topological spaces Fcl:Ycl→Xcl(Fcl=f*|Y(C)). (10)
The mapFcl:Ycl→Xcl is a local homeomorphism.An immediate conclusion from Theorem 34 is the following,Corollary 36.
∀
F
∈
et
(
C
2
) the left mapping LF:et(C2)→LF(et(C2)), LF(G)=F∘G is an injective mapping.


## 8. Extending the Notion of Geometrical Degree

In this section we will outline the fact that some of the notions and results that are related to geometrical degree of an étale mapping originate, in fact, in the more basic topological spaces (no algebraic or holomorphic structure is needed). We will skip most of the proofs (that are elementary).Definition 37.
LetX be a topological space. The semigroup of all the continuous mappings, F:X→X, will be denoted by C(X). Here, as usual, the binary operation is composition of mappings.Proposition 38.
(
1
) Let X be a topological space, F∈C(X), and F(X)¯=X. Then RF:C(X)→C(X) is injective.
(
2
) Let X be a topological space, F∈C(X), and RF:C(X)→C(X) is injective. Then for any G,H∈C(X), the property G|F(X)=H|F(X) implies that G≡H; that is, any G∈C(X) is determined by its restriction G|F(X).
(
3
) Let X be a topological space that has the following property: for any closed C⊆X and any point x∈X-C there exist two continuous mappings F,G:X→X such that F|C=G|C, but F(x)≠G(x).
LetF∈C(X) be such that RF:C(X)→C(X) is injective. Then F(X)¯=X.Remark 39.
Proposition8 follows.We are ready to discuss the notion of the geometrical degree,dF, of appropriate mappings in C(X).Lemma 40.
LetX be a topological space, F∈C(X) and n∈Z+∪{0}. If F is open, then the set Bn={x∈X∣|F-1(x)|≤n} is closed.Corollary 41.
LetX be a topological space, F∈C(X), F is open, and n∈Z+. Then one has the following. (1)
{x∈X∣|F-1(x)|≠n}=Bn-1∪(X-Bn), the union of a closed set and an open set.(2)
{x∈X∣|F-1(x)|=n}=Bn-Bn-1=Bn∩(X-Bn-1), the intersection of a closed set and an open set.(3)
IfdF=max⁡{|F-1(x)|∣x∈X}<∞ exists, then {x∈X∣|F-1(x)|≠dF} is a closed set.Remark 42.
B
n
⊆
B
n
+
1.Definition 43.
LetX be a topological space, F∈C(X), F is open, and the maximum dF=max⁡{|F-1(x)|∣x∈X}<∞ exists. Then we call dF the geometrical degree of F.Example 44.
IfX=C2 with the complex topology and F∈et(C2), then we know that dF exists. We also know that the set AF={x∈C2∣|F-1(x)|<dF} is a plane algebraic curve (possibly empty). Thus it is closed in C2. Moreover it is also small because dim⁡AF<2=dim⁡C2.We need one more property to hold for our mappings, namely, that the fiber size will generically bedF, that is, that the set of all x∈X for which dF=|F-1(x)| will be a large set measured in the topology of  X. This leads us to the following.Definition 45.
LetX be a topological space. We will denote by E(X) the set of all the mappings F:X→X that have the following properties. (1)
F∈C(X).(2)
F is open.(3)
The maximumdF=max⁡{|F-1(x)|∣x∈X}<∞ exists.(4)
X-BdF-1¯=X.Proposition 46.
LetX be an Hausdorff space. Then one has the following. (1)
E(X) is a semigroup with an identity (where the binary operation is composition of mappings). In fact Aut(X)⊆E(X).(2)
∀F,G∈E(X), dF∘G=dF·dG.Proof.
Checking thatE(X) is closed for composition: F,G∈C(X)⇒F∘G∈C(X). Also F,G open ⇒F∘G is open. If x∈X, then |F-1(x)|≤dF and ∀y∈F-1(x); we have |G-1(y)|≤dG. Since (F∘G)-1(x)=G-1(F-1(x)), it follows that dF∘G exists and, in fact, that dF∘G≤dF·dG. This gives the first three properties in the last definition. We need to check that X-BdF∘G-1¯=X. For that matter it will be convenient to denote SF={x∈X∣|F-1(x)<dF}. Let x∈X-SG=X-BdG-1. This set is open and dense in X. By our definitions G-1(x)={a1,a2,…,adG}, a finite set of exactly dG points. Since X is Hausdorff we can find dG open neighborhoods Va1,Va2,…,VadG of a1,a2,…,adG, respectively which are pairwise disjoint. The images G(Va1),G(Va2),…,G(VadG) are open (since G is open) neighborhoods of the point x=Ga1=Ga2=⋯=G(adg). Let us take the intersection
(44)V=⋂j=1dG‍G(Vaj).
Then V is an open neighborhood of x and G-1(V) consists of dG neighborhoods V1,V2,…,VdG of a1,a2,…,adG. We define Uj=Vj-SG, j=1,2,…,dG. Then since SG is closed and SGc is open and dense in X, the Uj’s are open and dense subsets of the Vj’s. Since G is continuous and G(Vj)=V, it follows that G(Uj) are open and dense in V. We note that each point y∈Uj is such that |F-1(y)|=dF because SF is disjoint of Uj. Thus the point x∈X-SG has a neighborhood V and dG open and dense subsets G(Uj) of V such that each point
(45)x′∈⋂j=1dG‍G(Uj)
(this set is still open and dense in V) is such that |G-1(x′)|=dG and each a∈G-1(x′) is such that |F-1(a)|=dF. Hence |(F∘G)-1(x′)|=dF·dG. This proves both that X-BdF∘G-1¯=X and that dF∘G=dF·dG.Corollary 47.
LetX be Hausdorff and F a semigroup of mappings F:X→X which are continuous and open and suppose that there is an absolute constant c∈Z+ such that ∀F∈F, dF≤c. Then the sets X-BdF-1, F∈F cannot be dense in X, unless ∀F∈F, dF=1.Corollary 48.
LetX be Hausdorff and F a family of mappings F:X→X which are continuous and open and satisfy X-BdF∘G-1¯=X, ∀F∈F (in particular ∀F∈F, dF<∞). Let S=F be the semigroup generated by F (composition of mappings is the binary operation). Then ∀F,G∈S, dF∘G=dF·dG.Definition 49.
LetX be an Hausdorff space. A mapping F∈E(X) is called a composite mapping if ∃G,H∈E(X)-Aut(X) such that F=G∘H. A mapping A∈E(X)-Aut(X) is a prime mapping if it is not composite. This is equivalent to the following: if A=B∘C for some B,C∈E(X), then B∈Aut(X)∨C∈Aut(X). The subset of E(X) of all the prime mappings will be denoted by Ep(X). Thus the set of all the composite mappings is E(X)-Ep(X).Proposition 50.
∀
F
∈
E
(
X
)
-
E
p
(
X
), dF is not a prime integer. Equivalently, ∀F∈E(X), dF is a prime integer ⇒F∈Ep(X).Theorem 51.
(
1
) If E(X)-Aut(X)≠∅, then Ep(X)≠∅.
(
2
)  ∀F∈E(X), ∃k∈Z+∪{0}, ∃A0∈Aut(X), ∃P1,…,Pk∈Ep(X) such that F=A0∘P1∘⋯∘Pk.Maybe few elementary examples are in place.Example 52.
In Definition45 we take X=C with the complex topology and (with an abuse of notation) take C[T]-C for E(X). Then Proposition 46(2), dF∘G=dF·dG, is the elementary fact from algebra that ∀P,Q∈C[T] we have deg⁡P∘Q=deg⁡P·deg⁡Q.A second example is given in Section4 of this paper.Example 53.
In Definition45 we take X=C2 with the complex topology, and E(X)=et(C2). Then the theory that was outlined in Propositions 12 and 14 and Theorem 15 is a special case of the above more general topological theory.



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