# On quadratic polynomial mappings from $\mathbb{C}^{3}$ to $\mathbb{C}^{2}$

M. Farnik  Jagiellonian University
Faculty of Mathematics and Computer Science
Łojasiewicza 6, 30-348 Kraków, Poland<EMAIL_ADDRESS>
###### Abstract.

We classify quadratic polynomial mappings from $\mathbb{C}^{3}$ to
$\mathbb{C}^{2}$ up to affine equivalence and topological equivalence. This is
a part of a larger project, we have already classified mappings from
$\mathbb{C}^{2}$ to $\mathbb{C}^{n}$ and inted to classify mappings from
$\mathbb{C}^{3}$ to $\mathbb{C}^{3}$.

###### Key words and phrases:

quadratic polynomial mappings, linear equivalence, topological equivalence

###### 2020 Mathematics Subject Classification:

14D06, 14Q20, 14R99, 51M99

The author was partially supported by the grant of Narodowe Centrum Nauki
number 2019/33/B/ST1/00755.

##  1\. Introduction

Let $\Omega_{\mathbb{K}^{k}}(d_{1},\ldots,d_{n})$ denote the space of
polynomial mappings $F=(f_{1},\ldots,f_{n}):\mathbb{K}^{k}\to\mathbb{K}^{n}$
where $\deg f_{i}\leq d_{i}$ for $1\leq i\leq n$. Let
$F,G\in\Omega_{\mathbb{K}^{k}}(d_{1},\ldots,d_{n})$. We say that $F$ is
_topologically_ (respectively _affinely_ or _linearly_) _equivalent_ to $G$ if
there are homeomorphisms (respectively affine or linear automorphisms)
$\Phi:\mathbb{K}^{k}\to\mathbb{K}^{k}$ and
$\Psi:\mathbb{K}^{n}\to\mathbb{K}^{n}$ such that $F=\Psi\circ G\circ\Phi$. In
the papers [3, 4] it was shown that for $\mathbb{K}$ equal $\mathbb{C}$ or
$\mathbb{R}$ and a fixed $n>0$ the space $\Omega_{\mathbb{K}^{2}}(2,\ldots,2)$
of quadratic $\mathbb{K}^{2}\to\mathbb{K}^{n}$ mappings splits into a finite
number of equivalence classes with respect to affine equivalence (hence also
with respect to topological equivalence). Moreover, the authors provided a
full classification of mappings in $\Omega_{\mathbb{K}^{2}}(2,\ldots,2)$. Here
we focus on classifying the mappings in $\Omega_{\mathbb{C}^{3}}(2,2)$. We
believe that obtaining the classification of $\Omega_{\mathbb{R}^{3}}(2,2)$
using similar methods is also possible but significantly more complicated than
in the complex case. In the future we intend to classify mappings in
$\Omega_{\mathbb{C}^{3}}(2,2,2)$, classifying quadratic mappings from
$\mathbb{C}^{3}$ to $\mathbb{C}^{2}$ is a natural first step towards achieving
this goal.

Unlike in the $\mathbb{K}^{2}\to\mathbb{K}^{n}$ case, there are infinitely
many equivalence classes of quadratic $\mathbb{C}^{3}\to\mathbb{C}^{2}$
mappings with respect to affine equivalence. This is quite obvious since
$\Omega_{\mathbb{C}^{3}}(2,2)$ is an affine space of dimension $20$ and the
groups of of affine automorphisms of $\mathbb{C}^{3}$ and $\mathbb{C}^{2}$
have dimensions $12$ and $6$, respectively. However,
$\Omega_{\mathbb{C}^{3}}(2,2)$ splits into a finite number of equivalence
classes with respect to topological equivalence. In fact there is a Zariski
open dense subset $U\subset\Omega_{\mathbb{C}^{3}}(2,2)$ such that every
mapping $F\in U$ has the same, generic, topological type. If a mapping $f$ has
a generic topological type then we say that $f$ is a _topologically generic_
mapping. The affine equivalence classes of topologically generic mappings are
parametrized by a two dimensional variety. There are also three topological
types with affine equivalence classes parametrized by one dimensional
varieties.

Now we present our main result – we give the complete classification of
quadratic polynomial mappings from $\mathbb{C}^{3}$ to $\mathbb{C}^{2}$ with
respect to affine equivalence and with respect to topological equivalence. The
notation and definitions are set up in Section 2. We select only one
representative for each family of topologically equivalent mappings having
infinitely many affine equivalence classes, they are $F_{1}$, $F_{2}$, $F_{4}$
and $F_{8}$. Items are enumerated according to the topological equivalence. We
use the same number and distinct letters to enumerate affine equivalence
classes of topologically equivalent mappings. There are $47$ classes for
topological equivalence. There are $4$ infinite families and $60$ classes for
affine equivalence.

We have the following cases:

  * (1)

(generic case) $F_{1}=(x^{2}+z^{2}+2y,y^{2}+z^{2}+2x+2z)$. The mapping $F_{1}$
is topologically equivalent to mappings $(x^{2}+z^{2}+y,y^{2}+z^{2}+\alpha
x+\beta z)$ for $\alpha\beta\neq 0$ and $H_{1}(\alpha,\beta)\neq 0$ (see
equation (4.9)). The critical set $C(F_{1})=V(xy-1,x(z+1)-z,yz-(z+1))$ is a
smooth curve of degree $3$ parametrized by
$\varphi:\mathbb{C}\setminus\\{0,-1\\}\ni
t\mapsto\left(t(t+1)^{-1},(t+1)t^{-1},t\right)\in C(F_{1})$. The mapping
$F_{1}$ has $6$ cusps at points $\varphi(t)$ for
$t^{3}(t+1)^{3}+t^{3}-(t+1)^{3}=0$ and $4$ nodes at points $\varphi(t)$ for
$(t+1)^{4}(t^{2}+1)^{2}+2t^{2}(t+1)^{2}(t^{2}-1)+t^{4}=0$.

  * (2)

$F_{2}=(x^{2}+z^{2}+y,y^{2}+z^{2}+4x+iz)$. The mapping $F_{2}$ is
topologically equivalent to mappings $(x^{2}+z^{2}+y,y^{2}+z^{2}+\alpha
x+\beta z)$ for $\alpha\beta\neq 0$ and $H_{1}(\alpha,\beta)=0$ (see equation
(4.9)) and $(\alpha^{2},\beta^{2})\notin\\{(1,4),(-4,-1),(-4^{-1},4^{-1})\\}$.
The critical set $C(F_{2})=V(xy-1,x(2z+i)-4z,4yz-(2z+i))$ is a smooth curve of
degree $3$ parametrized by $\varphi:\mathbb{C}\setminus\\{0,-1\\}\ni
t\mapsto\left(4t(2t+i)^{-1},(2t+i)(4t)^{-1},t\right)\in C(F_{2})$. The mapping
$F_{2}$ has $4$ cusps at points $\varphi(t)$ for $(t+1)^{4}+t^{3}+6t^{2}+t=0$,
a double cusp at $\varphi(1)$ and $3$ nodes at points $\varphi(t)$ for
$(t^{2}+1)((t+1)^{4}+2t^{3}+6t^{2}+2t)=0$.

  * (3)

$F_{3}=(x^{2}+z^{2}+y,y^{2}+z^{2}+x+2z)$. The critical set
$C(F_{3})=V(4xy-1,2x(z+1)-z,2yz-(z+1))$ is a smooth curve of degree $3$
parametrized by $\varphi:\mathbb{C}\setminus\\{0,-1\\}\ni
t\mapsto\left(t(2t+2)^{-1},(t+1)(2t)^{-1},t\right)\in C(F_{3})$. The mapping
$F_{3}$ has two cusps at points $\varphi(t)$ for $t^{2}+t-1=0$, two double
cusps at points $\varphi(t)$ for $t^{2}+t+2^{-1}=0$ and two nodes at points
$\varphi(t)$ for $4(t^{2}+t)^{2}=0$.

  * (4)

$F_{4}=(x^{2}+z^{2}+y,y^{2}+z^{2}+2x)$. The mapping $F_{4}$ is topologically
equivalent to mappings $(x^{2}+z^{2}+y,y^{2}+z^{2}+\alpha x)$ for
$\alpha\notin\\{-1,0,1\\}$. The critical set
$C(F_{4})=V(2xy-1,(x-1)z,(2y-1)z)$ is reducible and consists of the hyperbola
$C_{1}=V(2xy-1,z)$ and the line $C_{2}=V(x-1,2y-1)$. The components $C_{1}$
and $C_{2}$ intersect at $(1,2^{-1},0)$. The restriction $F_{4}|_{C_{1}}$ is
injective, the restriction $F_{4}|_{C_{2}}$ is generically $2:1$ and is
branched at $(1,2^{-1},0)$. The mapping $F_{4}$ has $3$ cusps at points
$(x,(2x)^{-1},0)$ for $4x^{3}-1=0$, the curves $F_{4}(C_{1})$ and
$F_{4}(C_{2})$ intersect at points $F_{4}(x,(2x)^{-1},0)$ for
$x\in\\{1,2^{-1},-2^{-1}\\}$.

  * (5)

$F_{5}=(x^{2}+z^{2}+2y,y^{2}+z^{2}+2x)$. The critical set
$C(F_{5})=V(xy-1,(x-1)z,(y-1)z)$ is reducible and consists of the hyperbola
$C_{1}=V(xy-1,z)$ and the line $C_{2}=V(x-1,y-1)$. The curves $C_{1}$ and
$C_{2}$ intersect at $(1,1,0)$. The restriction $F_{5}|_{C_{1}}$ is injective,
the restriction $F_{5}|_{C_{2}}$ is generically $2:1$ and is branched at
$(1,1,0)$. The mapping $F_{5}$ has two cusps at points
$(\varepsilon,\varepsilon^{2},0)$ for $\varepsilon^{3}=1$ and $\varepsilon\neq
1$, the curves $F_{5}(C_{1})$ and $F_{5}(C_{2})$ intersect at points
$F_{5}(1,1,0)$ and $F_{5}(-1,-1,0)$.

  * (6)

$F_{6}=(x^{2}+z^{2}+2y,y^{2}+z^{2})$. The critical set
$C(F_{6})=V(xy,xz,(y-1)z)$ is reducible and consists the $3$ lines:
$C_{1}=V(x,y-1)$, $C_{2}=V(x,z)$, $C_{3}=V(y,z)$. The line $C_{2}$ intersects
$C_{1}$ and $C_{3}$, the lines $C_{1}$ and $C_{3}$ do not intersect. The
restriction $F_{6}|_{C_{2}}$ is injective, the restrictions $F_{6}|_{C_{1}}$
and $F_{6}|_{C_{3}}$ are generically $2:1$ and are branched at $(0,1,0)$ and
$(0,0,0)$, respectively. The curves $F_{6}(C_{1})$, $F_{6}(C_{2})$ and
$F_{6}(C_{3})$ intersect pairwise at one point, the three points are distinct.

  * (7)

$F_{7}=(x^{2}+z^{2},y^{2}+z^{2})$. The critical set $C(F_{7})=V(xy,xz,yz)$ is
reducible and consists of the $3$ lines: $C_{1}=V(x,y)$, $C_{2}=V(x,z)$,
$C_{3}=V(y,z)$. The $3$ lines intersect at $(0,0,0)$. The restrictions
$F_{7}|_{C_{i}}$, $i=1,2,3$, are generically $2:1$ and are branched at
$(0,0,0)$.

  * (8)

$F_{8}=(x^{2}+z^{2}+2y,yz+x)$. The mapping $F_{8}$ is topologically equivalent
to mappings $(x^{2}+z^{2}+2y,yz+x+\alpha y)$ for $\alpha^{4}\neq-16$. The
critical set $C(F_{8})=V(xz-1,xy-z,y-z^{2})$ is a smooth curve of degree $3$
parametrized by $\varphi:\mathbb{C}\setminus\\{-1\\}\ni
t\mapsto\left(t^{-1},t^{2},t\right)\in C(F_{8})$. The mapping $F_{8}$ has $4$
cusps at points $\varphi(t)$ for $3t^{4}-1=0$ and $2$ nodes at points
$\varphi(t)$ for $t^{4}+1=0$.

  * (9)

$F_{9}=(x^{2}+z^{2}+2y,yz+x+Ay)$ for $A=\sqrt{2}(1+i)$. The critical set
$C(F_{9})=V(x(z+A)-1,xy-z,y-z(z+A))$ is a smooth curve of degree $3$
parametrized by $\varphi:\mathbb{C}\setminus\\{-1\\}\ni
t\mapsto\left((t+A)^{-1},t^{2}+At,t\right)\in C(F_{9})$. The mapping $F_{9}$
has $2$ cusps at points $\varphi(t)$ for $3t^{2}+7At+17i=0$, a double cusp at
$\varphi(-A/2)$ and a node at $\varphi(t)$ for $t^{2}+At-i=0$.

  * (10)

$F_{10}=(x^{2}+z^{2}+2y,yz+y)$. The critical set
$C(F_{10})=V(xy,x(z+1),y-z(z+1))$ is reducible and consists of the parabola
$C_{1}=V(x,y-z(z+1))$ and the line $C_{2}=V(y,z+1)$. The curves $C_{1}$ and
$C_{2}$ intersect at $(0,0,-1)$. The restriction $F_{10}|_{C_{1}}$ is
injective, the restriction $F_{10}|_{C_{2}}$ is generically $2:1$ and is
branched at $(0,0,-1)$. The mapping $F_{10}$ has a cusp at $(0,-2/9,-1/3)$,
the curves $F_{10}(C_{1})$ and $F_{10}(C_{2})$ intersect at points $(0,0)$ and
$(1,0)$.

  * (11)

$F_{11}=(x^{2}+z^{2},yz+x+y)$. The critical set
$C(F_{11})=V(x(z+1),xy-z,z(z+1))$ is reducible and consists of the hyperbola
$C_{1}=V(xy+1,z+1)$ and the line $C_{2}=V(x,z)$. The curves $C_{1}$ and
$C_{2}$ do not intersect. The restrictions $F_{11}|_{C_{1}}$ and
$F_{11}|_{C_{2}}$ are injective. The curves $F_{11}(C_{1})$ and
$F_{11}(C_{2})$ intersect at points $(0,i)$ and $(0,-i)$.

  * (12)

$F_{12}=(x^{2}+z^{2},yz+y)$. The critical set $C(F_{12})=V(x(z+1),xy,z(z+1))$
is reducible and consists of the $3$ lines: $C_{1}=V(x,z)$, $C_{2}=V(x,z+1)$,
$C_{3}=V(y,z+1)$. The line $C_{1}$ does not intersect $C_{2}$ and $C_{3}$, the
lines $C_{2}$ and $C_{3}$ intersect at $(0,0,-1)$. The restriction
$F_{12}|_{C_{1}}$ is injective, the restriction $F_{12}|_{C_{2}}$ is constant
and the restriction $F_{12}|_{C_{3}}$ is generically $2:1$ and branched at
$(0,0,-1)$.

  * (13)

$F_{13}=(x^{2}+z^{2}+2y,yz)$. The critical set $C(F_{13})=V(xy,xz,y-z^{2})$ is
reducible and consists of the parabola $C_{1}=V(x,y-z^{2})$ and the line
$C_{2}=V(y,z)$. The curves $C_{1}$ and $C_{2}$ intersect at $(0,0,0)$. The
restriction $F_{13}|_{C_{1}}$ is injective, the restriction $F_{13}|_{C_{2}}$
is generically $2:1$ and is branched at $(0,0,0)$. The curves $F_{13}(C_{1})$
and $F_{13}(C_{2})$ intersect only at $(0,0)$.

  * (14)

$F_{14}=(x^{2}+z^{2},yz+x)$. The critical set
$C(F_{14})=V(x(z+1),xy-z,z(z+1))$ is reducible and consists of the double line
$C_{1}=V(x^{2},xy-z)$ and the line $C_{2}=V(y,z)$. The lines $C_{1}$ and
$C_{2}$ intersect at $(0,0,0)$. The restriction $F_{14}|_{C_{1}}$ is constant
and the restriction $F_{14}|_{C_{2}}$ is injective.

  * (15)

$F_{15}=(x^{2}+z^{2},yz)$. The critical set $C(F_{15})=V(x(z+1),xy-z,z(z+1))$
is reducible and consists of the double line $C_{1}=V(x,z^{2})$ and the line
$C_{2}=V(y,z)$. The lines $C_{1}$ and $C_{2}$ intersect at $(0,0,0)$. The
restriction $F_{15}|_{C_{1}}$ is constant and the restriction
$F_{15}|_{C_{2}}$ is generically $2:1$ and is branched at $(0,0,0)$.

  * (16)

$F_{16}=(x^{2}+y^{2}+2z,z^{2}+2x)$. The critical set $C(F_{16})=V(xz-1,y)$ is
a hyperbola. The mapping $F_{16}$ has $3$ cusps at points
$(\varepsilon,0,\varepsilon^{2})$ for $\varepsilon^{3}=1$.

  * (17)

$F_{17}=(x^{2}+y^{2},z^{2}+2x)$. The critical set $C(F_{17})=V(xz,y)$ is
reducible and consists of the two lines: $C_{1}=V(x,y)$, $C_{2}=V(y,z)$. The
lines $C_{1}$ and $C_{2}$ intersect at $(0,0,0)$. The restriction
$F_{17}|_{C_{1}}$ is injective, the restriction $F_{17}|_{C_{2}}$ is
generically $2:1$ and is branched at $(0,0,0)$. The curves $F_{17}(C_{1})$ and
$F_{17}(C_{2})$ intersect only at $(0,0)$.

  * (18)

$F_{18}=(xy+z,z^{2}+2x)$. The critical set $C(F_{18})=V(x,yz-1)$ is a
hyperbola. The restriction $F_{18}|_{C(F_{18})}$ is injective. The image
$F_{18}(C(F_{18}))$ is the parabola $V(q-p^{2})$ without $(0,0)$.

  * (19)

$F_{19}=(xy,z^{2}+2x)$. The critical set $C(F_{19})=V(x,yz)$ is reducible and
consists of the two lines: $C_{1}=V(x,y)$, $C_{2}=V(x,z)$. The lines $C_{1}$
and $C_{2}$ intersect at $(0,0,0)$. The restriction $F_{19}|_{C_{1}}$ is
generically $2:1$ and is branched at $(0,0,0)$, the restriction
$F_{19}|_{C_{2}}$ is constant.

  * (20)

$F_{20}=(x^{2}+y^{2}+2z,z^{2})$. The critical set $C(F_{20})=V(xz,yz)$ is
reducible and consists of the line $C=V(x,y)$ and the plane $H=V(z)$. The
restriction $F_{20}|_{C}$ is injective and the image $F_{20}(H)$ is a line.

  * (21)

$F_{21}=(x^{2}+y^{2},z^{2})$. The critical set $C(F_{21})=V(xz,yz)$ is
reducible and consists of the line $C=V(x,y)$ and the plane $H=V(z)$. The
restriction $F_{21}|_{C}$ is generically $2:1$ and is branched at $(0,0,0)$,
the image $F_{21}(H)$ is a line.

  * (22)

$F_{22}=(x^{2}+2yz,y^{2}+2xy+2z)$. The critical set
$C(F_{22})=V(x-y^{2},z-y^{3}-y^{2})$ is a smooth curve of degree $3$
parametrized by $\varphi:\mathbb{C}\setminus\ni
t\mapsto\left(t^{2},t,t^{3}+t^{2}\right)\in C(F_{22})$. The mapping $F_{22}$
has two cusps at $\varphi(0)$ and $\varphi(-6^{-1})$ and a node at
$\varphi(t)$ for $8t^{2}+4t-1=0$.

  * (23)

$F_{23}=(x^{2}+2yz,y^{2}+2xy+3y/8+2z)$. The critical set
$C(F_{23})=V(x-y^{2},z-y^{3}-y^{2}-3y/8)$ is a smooth curve of degree $3$
parametrized by $\varphi:\mathbb{C}\setminus\ni
t\mapsto\left(t^{2},t,t^{3}+t^{2}+3t/8\right)\in C(F_{23})$. The mapping
$F_{23}$ has a double cusp at $\varphi(-4^{-1})$.

  * (24a)

$F_{24}=(x^{2}+2yz,y^{2}+2xy+2x)$. The critical set
$C(F_{24})=V(x(x+y)-(y+1)z,y(y+1),y(x+y))$ is reducible and consists of the
parabola $C_{1}=V(y,x^{2}-z)$ and the line $C_{2}=V(x-1,y+1)$. The curves
$C_{1}$ and $C_{2}$ do not intersect. The restrictions $F_{24}|_{C_{i}}$,
$i=1,2$, are injective. The curves $F_{24}(C_{1})$ and $F_{24}(C_{2})$
intersect at $(4^{-1},1)$. The mapping is topologically equivalent to
$(xy,(y+1)z)$.

  * (24b)

$F_{37}=(xy,yz+z)$. The critical set $C(F_{37})=V(x(y+1),y(y+1),yz)$ is
reducible and consists of the two lines: $C_{1}=V(x,y)$, $C_{2}=V(y+1,z)$. The
lines $C_{1}$ and $C_{2}$ do not intersect. The restrictions
$F_{37}|_{C_{i}}$, $i=1,2$, are injective. The images $F_{37}(C_{1})$ and
$F_{37}(C_{2})$ intersect at $(0,0)$.

  * (25)

$F_{25}=(x^{2}+2yz,y^{2}+2xy+2y)$. The critical set
$C(F_{25})=V(x(x+y+1)-yz,xy+y,y^{2})$ is reducible and consists of the line
$C_{1}=V(x,y)$ and the double line $C_{2}=V(x+1+y(z+1),y^{2})$. The lines
$C_{1}$ and $C_{2}$ do not intersect. The restrictions $F_{25}|_{C_{i}}$,
$i=1,2$, are constant.

  * (26)

$F_{26}=(x^{2}+2yz,y^{2}+2xy+2y)$. The critical set
$C(F_{26})=V(x^{2}-yz,xy,y^{2})$ is a triple line. The image
$F_{26}(C(F_{26}))$ is a point.

  * (27)

$F_{27}=(x^{2}+2yz,z^{2}+2y)$. The critical set $C(F_{27})=V(x,y-z^{2})$ is a
parabola. The mapping $F_{27}$ has a cusp at $(0,0,0)$.

  * (28a)

$F_{28}=(x^{2}+2yz,z^{2}+2x)$. The critical set $C(F_{28})=V(y,z)$ is a line.
The restriction $F_{28}|_{C(F_{28})}$ is injective. The mapping is
topologically equivalent to $(x,yz)$.

  * (28b)

$F_{31}=(x^{2}+2z,y^{2}+2z)$. The critical set $C(F_{31})=V(x,y)$ is a line.
The restriction $F_{31}|_{C(F_{31})}$ is injective. The mapping is
topologically equivalent to $(x,yz)$.

  * (28c)

$F_{40}=(xy,y^{2}+2z)$. The critical set $C(F_{40})=V(x,y)$ is a line. The
restriction $F_{40}|_{C(F_{40})}$ is injective. The mapping is topologically
equivalent to $(x,yz)$.

  * (28d)

$F_{46}=(x^{2}+yz,x)$. The critical set $C(F_{46})=V(y,z)$ is a line. The
restriction $F_{46}|_{C(F_{46})}$ is injective. The mapping is topologically
equivalent to $(x,yz)$.

  * (28e)

$F_{49}=(x^{2}+y^{2},z)$. The critical set $C(F_{49})=V(x,y)$ is a line. The
restriction $F_{49}|_{C(F_{49})}$ is injective. The mapping is topologically
equivalent to $(x,yz)$.

  * (29)

$F_{29}=(x^{2}+2yz,z^{2}+2z)$. The critical set $C(F_{29})=V(x(z+1),z(z+1))$
is reducible and consists of the line $C=V(x,z)$ and the plane $H=V(z+1)$. The
image $F_{29}(C)$ is a point and the image $F_{29}(H)$ is a line.

  * (30)

$F_{30}=(x^{2}+2yz,z^{2})$. The critical set $C(F_{30})=V(xz,z^{2})$ is the
plane $H=V(z)$ with the embedded double line $C=V(x,z^{2})$. The image
$F_{30}(C)$ is a point and the image $F_{30}(H)$ is a line.

  * (31a)

$F_{32}=(x^{2}+z,y^{2}+x)$ with empty critical set. The mapping is
topologically equivalent to $(x,y)$.

  * (31b)

$F_{41}=(xy+z,y^{2}+x)$ with empty critical set. The mapping is topologically
equivalent to $(x,y)$.

  * (31c)

$F_{50}=(x^{2}+y^{2}+z,x)$ with empty critical set. The mapping is
topologically equivalent to $(x,y)$.

  * (31d)

$F_{52}=(xy+z,x)$ with empty critical set. The mapping is topologically
equivalent to $(x,y)$.

  * (31e)

$F_{56}=(x^{2}+z,y)$ with empty critical set. The mapping is topologically
equivalent to $(x,y)$.

  * (31f)

$F_{58}=(x^{2}+y,x)$ with empty critical set. This is $f_{10}$ from [3]. The
mapping is topologically equivalent to $(x,y)$.

  * (31g)

$F_{62}=(x,y)$ with empty critical set. This is $f_{12}$ from [3].

  * (32a)

$F_{33}=(x^{2}+z,y^{2})$. The critical set $C(F_{33})=V(y)$ is a plane. The
image $F_{33}(C(F_{33}))$ is a line. The mapping is topologically equivalent
to $(x,y^{2})$.

  * (32b)

$F_{42}=(xy+z,y^{2})$. The critical set $C(F_{42})=V(y)$ is a plane. The image
$F_{42}(C(F_{42}))$ is a line. The mapping is topologically equivalent to
$(x,y^{2})$.

  * (32c)

$F_{51}=(x^{2}+y^{2},x)$. This is $f_{7}$ from [3]. The critical set
$C(F_{51})=V(y)$ is a plane. The image $F_{51}(C(F_{51}))$ is a parabola. The
mapping is topologically equivalent to $(x,y^{2})$.

  * (32d)

$F_{57}=(x^{2},y)$. This is $f_{9}$ from [3]. The critical set
$C(F_{57})=V(x)$ is a plane. The image $F_{57}(C(F_{57}))$ is a line.

  * (33)

$F_{34}=(x^{2}+2y,y^{2}+2x)$. This is $f_{1}$ from [3]. The critical set
$C(F_{34})=V(xy-1)$ is a product of a line and hyperbola. The image
$F_{34}(C(F_{34}))$ is a curve with three cusps.

  * (34)

$F_{35}=(x^{2}+2y,y^{2})$. This is $f_{3}$ from [3]. The critical set
$C(F_{35})=V(xy)$ is reducible and consists of two intersecting planes. The
image $F_{35}(C(F_{35}))$ consist of a parabola and a tangent line.

  * (35)

$F_{36}=(x^{2},y^{2})$. This is $f_{4}$ from [3]. The critical set
$C(F_{36})=V(xy)$ is reducible and consists of two intersecting planes. The
image $F_{36}(C(F_{36}))$ consist of two intersecting lines.

  * (36)

$F_{38}=(xy+z,yz)$. The critical set $C(F_{38})=V(xy-z,y^{2})$ is a double
line. The image $F_{38}(C(F_{38}))$ is a point.

  * (37)

$F_{39}=(xy,yz)$. The critical set $C(F_{39})=V(xy,y^{2},yz)$ is the plane
$H=V(y)$ with the embedded double point $C=V(x,y^{2},z)$. The image
$F_{39}(H)$ is a point.

  * (38)

$F_{43}=(xy,y^{2}+2x)$. This is $f_{2}$ from [3]. The critical set
$C(F_{43})=V(x-y^{2})$ is a product of a line and parabola. The image
$F_{43}(C(F_{43}))$ is a curve with one cusp.

  * (39)

$F_{44}=(xy,y^{2}+2y)$. This is $f_{5}$ from [3]. The critical set
$C(F_{44})=V(y^{2}+y)$ is reducible and consists of two planes. The image
$F_{44}(C(F_{44}))$ consists of a line and a point.

  * (40)

$F_{45}=(xy,y^{2})$. This is $f_{6}$ from [3]. The critical set
$C(F_{45})=V(y^{2})$ is a double plane planes. The image $F_{45}(C(F_{45}))$
is a point.

  * (41)

$F_{47}=(x^{2}+yz,y)$. The critical set $C(F_{47})=V(x,y)$ is a line. The
image $F_{47}(C(F_{47}))$ is a point.

  * (42)

$F_{48}=(x^{2}+y^{2}+z^{2},0)$ with critical equal $\mathbb{C}^{3}$.

  * (43)

$F_{53}=(xy,x)$. This is $f_{8}$ from [3]. The critical set $C(F_{53})=V(y)$
is a plane. The image $F_{53}(C(F_{53}))$ is a point.

  * (44a)

$F_{54}=(x^{2}+y^{2}+z,0)$ with critical set equal to $\mathbb{C}^{3}$. The
mapping is topologically equivalent to $(x,0)$.

  * (44b)

$F_{59}=(x^{2},x)$ with critical set equal to $\mathbb{C}^{3}$. This is
$f_{11}$ from [3]. The mapping is topologically equivalent to $(x,0)$.

  * (44c)

$F_{60}=(x^{2}+y,0)$ with critical set equal to $\mathbb{C}^{3}$. This is
$f_{14}$ from [3]. The mapping is topologically equivalent to $(x,0)$.

  * (44d)

$F_{63}=(x,0)$ with critical set equal to $\mathbb{C}^{3}$. This is $f_{16}$
from [3].

  * (45)

$F_{55}=(x^{2}+y^{2},0)$ with critical set equal to $\mathbb{C}^{3}$. This is
$f_{13}$ from [3].

  * (46)

$F_{61}=(x^{2},0)$ with critical set equal to $\mathbb{C}^{3}$. This is
$f_{15}$ from [3].

  * (47)

$F_{64}=(0,0)$ with critical set equal to $\mathbb{C}^{3}$. This is $f_{17}$
from [3].

We will show in Sections 3–6 that the classification above enumerates all
affine equivalence classes.

In most cases the topological equivalence or the lack thereof is quite
obvious. Observe that for any quadratic $\mathbb{C}^{3}\to\mathbb{C}^{2}$
mapping having at least one cusp, double cusp or node the topological type of
the mapping is uniquely determined by the number of those singularities. For
mappings without those singularities we look at the number and types of
irreducible components of the critical set (or the scheme associated with the
ideal generated by the minors of the Jacobian matrix), at the behavior of the
restriction of the mapping to each component of the critical set, at the image
of each component and at the intersections of those images. We will now give
brief arguments for the least obvious cases.

  * •

Both $F_{13}(C(F_{13}))$ and $F_{17}(C(F_{17}))$ have two components
intersecting only at $(0,0)$, however, one of the components of
$F_{13}(C(F_{13}))$ has a cusp at $(0:0)$ and both components of
$F_{17}(C(F_{17}))$ are smooth.

  * •

We compose $F_{24}$ with $(p/2-q^{2}/8,q-1)$ to obtain
$(y(z+h(2x,y)),y^{2}+2xy+2x-1)$. Then with $(x/2,y,z-h(x,y))$ to obtain
$(yz,(x+y-1)(y+1))$ and with $(z-y+1,y,x)$ to obtain $(xy,(y+1)z)$.

  * •

We compose $F_{28}$ with $(2x,y,2z)$ and $(p/4,q/4)$ to obtain
$(x^{2}+yz,z^{2}+x)$. Then with $(x-z^{2},y+2xz+z^{3},z)$ to obtain
$(x^{2}+yz,x)$ and with $(q,p-q^{2})$ to obtain $(x,yz)$.

  * •

We compose $F_{31}$ with $(p,p-q)$ to obtain $(x^{2}+2z,x^{2}-y^{2})$ and with
$(x,y,(z-x^{2})/2)$ to obtain $(z,x^{2}-y^{2})$. Finally compose with
$((y+z)/2,(y-z)/2,x)$ to obtain $(x,yz)$.

##  2\. Notation and definitions

Throughout the paper we will consider mappings
$F=(f,g):\mathbb{C}^{3}\rightarrow\mathbb{C}^{2}$. We take
$f=a_{1}x^{2}+a_{2}xy+a_{3}xz+a_{4}y^{2}+a_{5}yz+a_{6}z^{2}+a_{7}x+a_{8}y+a_{9}z+a_{10}$
and, similarly, $g=b_{1}x^{2}+\ldots+b_{10}$. Note that $a_{i}$ and $b_{i}$
denote coefficients at the appropriate monomials and often change. After
composing $F$ with an affine or linear automorphism the coefficients change
but we do not introduce new symbols. By $h_{x}=\frac{\partial h}{\partial x}$
we denote the partial differential of a function $h$ with respect to a
variable $x$. We denote $m_{xy}=f_{x}g_{y}-f_{y}g_{x}$ and similarly for
$m_{xz}$ and $m_{yz}$. We denote the critical set of $F$ by
$C(F)=V(m_{xy},m_{xz},m_{yz})$, we call the set of critical of $F$ the
discriminant of $F$ and denote it by by $\Delta(F)=F(C(F))$.

We denote by $\tilde{F}=(\tilde{f},\tilde{g})$ the mapping obtained by taking
the homogeneous parts of maximal degree of the components of a mapping
$F=(f,g)$. If both components of $F$ are of degree two then we will often
consider $\tilde{F}$ as a birational mapping
$(\tilde{f}:\tilde{g}):\mathbb{P}^{2}\rightarrow\mathbb{P}^{1}$. We take
$\tilde{m}_{xy}=\tilde{f}_{x}\tilde{g}_{y}-\tilde{f}_{y}\tilde{g}_{x}$, it is
the homogeneous part of maximal degree of $m_{xy}$.

We use $x$, $y$ and $z$ for the coordinates in the source $\mathbb{C}^{3}$ or
$\mathbb{P}^{2}$ and $p$, $q$ for the coordinates in the target
$\mathbb{C}^{2}$ or $\mathbb{P}^{1}.$ The letters $a_{i}$, $b_{i}$, $A$, $B$
will denote parameters or coordinates in parameter space.

If there is no danger of confusion we will omit the arguments of a mapping,
i.e., we will write
$(f_{1}(x_{1},\ldots,x_{k}),\ldots,f_{n}(x_{1},\ldots,x_{k}))$ instead of
$(x_{1},\ldots,x_{k})\mapsto$$(f_{1}(x_{1},\ldots,x_{k}),\ldots,f_{n}(x_{1},\ldots,x_{k}))$.

Let $F:\mathbb{C}^{3}\rightarrow\mathbb{C}^{2}$ be a mapping. We say that
$P\in C(F)$ is a _cusp_ of $F$ if $F^{-1}(F(P))\cap\mathbb{C}(F)=\\{P\\}$ and
there are neighborhoods $U_{1}$ of $P_{1}$ and $U_{2}$ of $F(P)$, open sets
$V_{1}\subset\mathbb{C}^{3}$ and $V_{2}\subset\mathbb{C}^{2}$ and
biholomorphic mappings $h_{1}:(U_{1},P)\rightarrow(V_{1},0)$,
$h_{2}(U_{2},F(P))\rightarrow(V_{2},0)$ such that $F|_{U_{1}}=h_{2}^{-1}\circ
F_{1}\circ h_{1}$ for $F_{1}(x,y,z)=(x,y^{3}+xy+z^{2})$. Note that if $P$ is a
cusp of $F$ then $P$ is a smooth point of $C(F)$ and a critical point of the
restriction $F|_{C(F)}$. Moreover the irreducible component of $\Delta(F)$
containing $F(P)$ is a curve with a cusp singularity at $F(P)$.

Similarly, we call $P\in C(F)$ a _double cusp_ if $F$ restricted to some
neighborhood of $P$ is biholomorphically equivalent to $(x,y^{4}+xy+z^{2})$.

We say that $P_{1},P_{2}\in C(F)$ are a _node_ of $F$ if
$F^{-1}(F(P_{1}))\cap\mathbb{C}(F)=\\{P_{1},P_{2}\\}$, the points $P_{1}$ and
$P_{2}$ belong to the same irreducible component of $C(F)$ and there are
neighborhoods $U_{1}$ of $P_{1}$, $U_{2}$ of $P_{2}$ and $U_{3}$ of
$F(P_{1})$, open sets $V_{1},V_{2}\subset\mathbb{C}^{3}$ and
$V_{3}\subset\mathbb{C}^{2}$ and biholomorphic mappings
$h_{1}:(U_{1},P_{1})\rightarrow(V_{1},0)$,
$h_{2}(U_{2},P_{2})\rightarrow(V_{2},0)$ and
$h_{3}(U_{3},F(P_{1}))\rightarrow(V_{3},0)$ such that
$F|_{U_{1}}=h_{3}^{-1}\circ F_{1}\circ h_{1}$ for
$F_{1}(x,y,z)=(x,y^{2}+z^{2})$ and $F|_{U_{2}}=h_{3}^{-1}\circ F_{2}\circ
h_{2}$ for $F_{1}(x,y,z)=(y^{2}+z^{2},x)$. Note that if $P_{1}$ and $P_{2}$
are a node of $F$ then $P_{1},P_{2}$ are smooth points of $C(F)$ and regular
points of the restriction $F|_{C(F)}$. Moreover the irreducible component of
$\Delta(F)$ containing $F(P_{1})$ is a curve with a node singularity at
$F(P_{1})$.

##  3\. Classification of homogeneous mappings

In this section we will consider $F=(f,g)$ such that $f$ and $g$ are
homogeneous. We will enumerate up to linear equivalence all possible cases for
$F$. First, we will consider the case when $f$ and $g$ are of degree $2$ and
one is not a multiple of the other, i.e., $F$ is not equivalent to $(f,0)$.
Later we will enumerate the cases with lower degrees.

We begin by taking care of some obvious special cases. First, consider the
case when $F$ is equivalent to $(f,z^{2})$. If $z$ divides $f$ then $F$ is
obviously equivalent to

(3.1) $\tilde{F}_{8}=(xy,y^{2}).$

If $z$ does not divide $f$ then $a_{1}$, $a_{2}$ or $a_{4}$ is nonzero. By
composing with $(x,y+ax,z)$ for generic $a$ we may assume that $a_{1}\neq 0$.
By composing with $(a_{1}^{-1/2}x,y,z)$ we obtain
$(x^{2}+a_{2}xy+a_{3}xz+a_{4}y^{2}+a_{5}yz+a_{6}z^{2},z^{2})$. Composing with
$(x-a_{2}/2y-a_{3}/2z,y,z)$ we obtain
$(x^{2}+a_{4}y^{2}+a_{5}yz+a_{6}z^{2},z^{2})$. If $a_{4}\neq 0$ then we
compose with $(x,a_{4}^{-1/2}y-\sqrt{a_{4}}a_{5}/2z,z)$ to obtain
$(x^{2}+y^{2}+a_{6}z^{2},z^{2})$ and with $(p-a_{6}q,q)$ to obtain

(3.2) $\tilde{F}_{3}=(x^{2}+y^{2},z^{2}).$

Now assume $a_{4}=0$. If $a_{5}\neq 0$ then composing with
$(x,a_{5}^{-1}(2y-a_{6}z),z)$ we obtain

(3.3) $\tilde{F}_{5}=(x^{2}+2yz,z^{2}).$

If $a_{5}=0$ then we compose with $(x,z,y)$ and $(p-a_{6}q,q)$ to obtain

(3.4) $\tilde{F}_{6}=(x^{2},y^{2}).$

Now we adopt a geometric approach. We will consider $F$ as a birational
$\mathbb{P}^{2}\rightarrow\mathbb{P}^{1}$ mapping. It is easy to see that for
generic $F$ the sets $V(m_{xy})$ and $V(m_{xz})$ are quadrics, their
intersection consists of $4$ points. The curve $V(m_{yz})$ omits the point
$V(f_{x},g_{x})$ and contains the other three. Thus, we will begin with the
case when $C(F)$ has three points, then follow with the case of double point
and a singular point and other nongeneric cases.

Recall, that we have

$F=(a_{1}x^{2}+a_{2}xy+a_{3}xz+a_{4}y^{2}+a_{5}yz+a_{6}z^{2},b_{1}x^{2}+b_{2}xy+b_{3}xz+b_{4}y^{2}+b_{5}yz+b_{6}z^{2}).$

We compute

$m_{xy}=(2a_{1}x+a_{2}y+a_{3}z)(b_{2}x+2b_{4}y+b_{5}z)-(a_{2}x+2a_{4}y+a_{5}z)(2b_{1}x+b_{2}y+b_{3}z),$
$m_{xz}=(2a_{1}x+a_{2}y+a_{3}z)(b_{3}x+b_{5}y+2b_{6}z)-(a_{3}x+a_{5}y+2a_{6}z)(2b_{1}x+b_{2}y+b_{3}z),$
$m_{yz}=(a_{2}x+2a_{4}y+a_{5}z)(b_{3}x+b_{5}y+2b_{6}z)-(a_{3}x+a_{5}y+2a_{6}z)(b_{2}x+2b_{4}y+b_{5}z).$

Assume that $C(F)$ has $3$ distinct points. If the points are collinear, then
by Bézout’s Theorem each of the three quadrics defining $C(F)$ contains the
line passing through the $3$ points. Consequently, $C(F)$ contains a line,
which is a special case that we will consider later. Assume that the points
are not collinear. By composing $F$ with a linear automorphism of
$\mathbb{P}^{2}$ we may assume that the points of $C(F)$ are $(1:0:0)$,
$(0:1:0)$ and $(0:0:1)$.

We will show that if $F$ is well defined at two distinct points of $C(F)$ and
has the same value for both of them then we can reduce to the already solved
case $F=(f,z^{2})$ or $F=(f,0)$. Indeed, we may assume that
$(a_{1}:b_{1})=F(1:0:0)=(1:0)$ and $(a_{4}:b_{4})=F(0:1:0)=(1:0)$, which means
that $a_{1},a_{4}\neq 0$ and $b_{1}=b_{4}=0$. Moreover,
$0=m_{xy}(1:0:0)=2a_{1}b_{2}$, $0=m_{xz}(1:0:0)=2a_{1}b_{3}$ and
$0=m_{yz}(0:1:0)=2a_{4}b_{5}$, so $b_{2}=b_{3}=b_{5}=0$. Thus, from now on we
may assume that $F$ is injective on all points of $C(F)$ on which it is well
defined.

Now we look in how many points of $C(F)$ the birational mapping $F$ is well
defined. First consider the case when $F$ is well defined at $3$ points of
$C(F)$. By composing with a linear automorphism of $\mathbb{P}^{1}$ we may
assume that the corresponding points of $F(C(F))$ are $(1:0)$, $(0:1)$ and
$(1:1)$.

From $(1:0)=F(1:0:0)=(a_{1}:b_{1})$ we obtain $a_{1}\neq 0$ and $b_{1}=0$.
Furthermore, $(0:1)=F(0:1:0)=(a_{4}:b_{4})$ and $(1:1)=F(0:0:1)=(a_{6}:b_{6})$
yields $a_{4}=0$, $b_{4}\neq 0$ and $a_{6}=b_{6}\neq 0$. Replacing $(x,y,z)$
with $(x/\sqrt{a_{1}},y/\sqrt{b_{4}},x/\sqrt{a_{6}})$ we may assume that
$a_{1}=b_{4}=a_{6}=b_{6}=1$.

We have $0=m_{xy}(1:0:0)=2b_{2}$ and $0=m_{xy}(0:1:0)=2a_{2}$. Next,
$0=m_{xz}(1:0:0)=2b_{3}$ and $0=m_{xz}(0:0:1)=2a_{3}-2b_{3}$. Finally,
$0=m_{yz}(0:1:0)=-2a_{5}$ and $0=m_{yz}(0:0:1)=2a_{5}-2b_{5}$.

Thus we obtain

(3.5) $\tilde{F}_{1}=(x^{2}+z^{2},y^{2}+z^{2}).$

Now consider the case when $F$ is well defined at $2$ points of $C(F)$ and not
well defined at the third. As before, $F(1:0:0)=(1:0)$ and $F(0:1:0)=(0:1)$
yield $a_{1}=b_{4}=1$ and $a_{4}=b_{1}=0$. Furthermore, $F(0:0:1)=(0:0)$ means
that $a_{6}=b_{6}=0$. Next, we have $m_{xy}(1:0:0)=2b_{2}=0$,
$m_{xy}(0:1:0)=2a_{2}=0$, $m_{xz}(1:0:0)=2b_{3}=0$ and
$m_{yz}(0:1:0)=2a_{5}=0$. We arrive at $F=(x^{2}+a_{3}xz,y^{2}+b_{5}yz)$. We
have $m_{xy}(0:0:1)=a_{3}b_{5}=0$. So at least one of $f$ and $g$ is a square.

Now consider the case when $F$ is well defined at one point of $C(F)$ and not
well defined at two other. As before, $F(1:0:0)=(1:0)$ yields $a_{1}=1$ and
$b_{1}=0$. Furthermore, $F(0:1:0)=F(0:0:1)=(0:0)$ means that
$a_{4}=a_{6}=b_{4}=b_{6}=0$. Next, we have $m_{xy}(1:0:0)=2b_{2}=0$ and
$m_{xz}(1:0:0)=2b_{3}=0$. If $b_{5}=0$ then $F=(f,0)$. So we can assume
$b_{5}\neq 0$ and by dividing $g$ by $b_{5}$ obtain $b_{5}=1$. We have
$m_{xz}(0:1:0)=a_{2}=0$ and $m_{xy}(0:0:1)=a_{3}=0$. We arrive at
$F=(x^{2}+a_{5}yz,yz)$, composing with $(q,p-a_{5}q)$ we obtain
$F=(yz,x^{2})$, so $g$ is a square.

Now, if $F$ is not well defined at $3$ points of $C(F)$ then we obtain
$F=(a_{2}xy+a_{3}xz+a_{5}yz,b_{2}xy+b_{3}xz+b_{5}yz)$. The conditions
$m_{xy}(0:0:1)=m_{xz}(0:1:0)=m_{yz}(1:0:0)=0$ mean that the vectors
$[a_{2},a_{3},a_{5}]$ and $[b_{2},b_{3},b_{5}]$ are proportional, hence $F$ is
equivalent to $(f,0)$.

Now, we assume that $C(F)$ has two points, one of which is double. Note that
if the tangent to the double point passes through the smooth point, then by
Bézout’s Theorem $C(F)$ must contain the line through these points. We assume
that this is not the case. So we may assume that $(1:0:0)$ and $(0:1:0)$ are
critical points and that the line $V(y)$ (but not the whole plane) is tangent
to $C(F)$ at $(1:0:0)$. From $m_{xy}(1:0:0)=m_{xz}(1:0:0)=m_{yz}(1:0:0)=0$ we
obtain that the vectors $[a_{1},a_{2},a_{3}]$ and $[b_{1},b_{2},b_{3}]$ are
proportional. Thus we may assume that $b_{1}=b_{2}=b_{3}=0$. Since $V(y)$ is
tangent to $C(F)$ at $(1:0:0)$ we have $\partial m_{xy}/\partial
z(1:0:0)=2a_{1}b_{5}=0$ and $\partial m_{xz}/\partial z(1:0:0)=4a_{1}b_{6}=0$.
If we had $a_{1}\neq 0$, then $b_{5}=b_{6}=0$, so $g=b_{4}y^{2}$ would be a
square. So $a_{1}=0$. W have $m_{xy}(0:1:0)=2a_{2}b_{4}=0$ and
$m_{xz}(0:1:0)=a_{2}b_{5}=0$. Again, $a_{2}\neq 0$ would imply $b_{4}=b_{5}=0$
and $g=b_{6}z^{2}$, so $a_{2}=0$. We have $\partial m_{xy}/\partial
y(1:0:0)=\partial m_{xz}/\partial z(1:0:0)=0$, since $V(y)$ is tangent but the
whole plane not, we must have $\partial m_{yz}/\partial
z(1:0:0)=2a_{3}b_{4}\neq 0$. By multiplying $f$ and $g$ we may assume
$a_{3}=b_{4}=1$. Next, $\partial m_{yz}/\partial z(1:0:0)=-b_{5}=0$. As
remarked before, $b_{5}=0$ implies $b_{6}\neq 0$. We have
$F=(xz+a_{4}y^{2}+a_{5}yz+a_{6}z^{2},y^{2}+b_{6}z^{2})$, composing with
$(p-a_{4}q,q)$ and $(x-a_{5}y-(a_{6}b_{4}^{-1}-a_{4})z,y,b_{6}^{-1/2}z)$ we
obtain $F=(xz,y^{2}+z^{2})$. Permuting the variables we write it down as

(3.6) $\tilde{F}_{2}=(x^{2}+z^{2},yz).$

Now assume that $C(F)$ is a triple point. We may assume that it is $(1:0:0)$
and that $V(y)$ is tangent to $C(F)$ at $(1:0:0)$. We may pick two points at
which $F$ is well defined and by composing with linear transformations obtain
$F(0:1:0)=(0:1)$ and $F(1:1:0)=(1:0)$. We obtain $a_{4}=0$, $b_{4}\neq 0$ and
$a_{1}+a_{2}\neq 0$. As with the double point, from $(1:0:0)\in C(F)$ we
obtain that the vectors $[a_{1},a_{2},a_{3}]$ and $[b_{1},b_{2},b_{3}]$ are
proportional. Thus we may assume that $b_{1}=b_{2}=b_{3}=0$. From the tangency
of $V(y)$ at $(1:0:0)$ w obtain $a_{1}=0$, consequently $a_{2}\neq 0$.
Multiplying $f$ and $g$ by constants we obtain $a_{2}=1$ and $b_{4}=1$. Thus,
we have $F=(xy+a_{3}xz+a_{5}yz+a_{6}z^{2},y^{2}+b_{5}yz+b_{6}z^{2})$.
Composing with $(x-a_{5}z,y-b_{5}/2z,z)$ we obtain
$F=(xy+a_{3}xz+a_{6}z^{2},y^{2}+b_{6}z^{2})$. Since $V(y)$ is not a component
of $C(F)$ we have $b_{6}\neq 0$, dividing $z$ by $i\sqrt{b_{6}}$ we obtain
$F=(xy+a_{3}xz+a_{6}z^{2},y^{2}-z^{2})$. Now
$C(F)=V(y^{2}+a_{3}yz,yz+a_{3}z^{2},xz+a_{3}xy+2a_{6}yz)$, so if
$a_{3}^{2}\neq 1$ then $C(F)$ contains $(1:0:0)$ and
$(2a_{3}a_{6}/(1-a_{3}^{2}):-a_{3}:1)$ which is a contradiction. If $a_{3}=-1$
then we multiply $z$ by $-1$ and obtain $F=(xy+xz+a_{6}z^{2},y^{2}-z^{2})$. If
$a_{6}=0$ then $y+z$ divides $f$ and $g$, consequently $V(y+z)\in C(F)$ which
is a contradiction. So $a_{6}\neq 0$ and composing with $(a_{6}x,y,z)$ and
$(a_{6}^{-1}p,q)$ we obtain $F=(xy+xz+z^{2},y^{2}-z^{2})$. This is already a
good form, however we prefer to compose it with $(-2z,-y-z,x)$ and obtain

(3.7) $\tilde{F}_{4}=(x^{2}+2yz,y^{2}+2xy).$

Now we pass to the case when $C(F)$ does contain the line $V(y)$, but is not
the whole plane. If $C(F)$ contains a point outside $V(y)$, then it contains
three noncollinear points. According to the analysis above it means that $F$
is equivalent to $(f,z^{2})$, in particular to $\tilde{F}_{3}$ if $C(F)$ is a
line and a point or $\tilde{F}_{6}$ if $C(F)$ is two lines. It remains to
examine the case when $C(F)$ is precisely the line $V(y)$ (as a set). First
assume that $F$ is well defined at some point of $C(F)$. We may additionally
pick a point outside $C(F)$ for which $F$ is well defined. By composing linear
transformations we may assume that $F(1:0:0)=(1:0)$ and $F(0:1:0)=(0:1)$. It
follows that $a_{1},b_{4}\neq 0$ and $a_{4}=b_{1}=0$ and we may assume that
$a_{1}=b_{4}=1$. We have $C(F)=V(y)$, so $y$ must divide $m_{xy}$, $m_{xz}$
and $m_{yz}$. The resulting equations yield $b_{2}=b_{3}=b_{5}=b_{6}=0$, so
$g=y^{2}$. The only mapping with the required geometry is $\tilde{F}_{5}$.

Now assume that $F$ is not well defined along $V(y)$, i.e., $f=yl$ and $g=yk$
for linear forms $l$ and $k$. If $y$, $l$ and $k$ are linearly dependent then
$F$ is equivalent to $\tilde{F}_{8}$. If $y$, $l$ and $k$ are linearly
independent then $F$ is equivalent to

(3.8) $\tilde{F}_{7}=(xy,yz).$

We sum up the results obtained above in the following corollary:

######  Corollary 3.1.

Let $F=(f,g):\mathbb{C}^{3}\rightarrow\mathbb{C}^{2}$ be a mapping with
components homogeneous of degree $2$. If $C(F)\neq\mathbb{C}^{3}$ then the
linear equivalence class of $F$ is determined by the critical set
$\tilde{C}(F)$ of the corresponding $\mathbb{P}^{2}\rightarrow\mathbb{P}^{1}$
mapping. We have the following possibilities:

  1. (1)

$\tilde{F}_{1}=(x^{2}+z^{2},y^{2}+z^{2})$, if $\tilde{C}(F)$ is three
noncollinear points.

  2. (2)

$\tilde{F}_{2}=(x^{2}+z^{2},yz)$, if $\tilde{C}(F)$ is a double and a smooth
point.

  3. (3)

$\tilde{F}_{3}=(x^{2}+y^{2},z^{2})$, if $\tilde{C}(F)$ is a line and a point.

  4. (4)

$\tilde{F}_{4}=(x^{2}+2yz,y^{2}+2xy)$, if $\tilde{C}(F)$ is a triple point.

  5. (5)

$\tilde{F}_{5}=(x^{2}+2yz,z^{2})$, if $\tilde{C}(F)$ is a line with a
nonreduced point.

  6. (6)

$\tilde{F}_{6}=(x^{2},y^{2})$, if $\tilde{C}(F)$ is two lines.

  7. (7)

$\tilde{F}_{7}=(xy,yz)$, if $\tilde{C}(F)$ is a line.

  8. (8)

$\tilde{F}_{8}=(xy,y^{2})$, if $\tilde{C}(F)$ is a double line.

Next, we enumerate $F=(f,g)$ such that $f$ is of degree $2$ and $g$ is of a
lower degree. It is a classic result that the quadratic form $f$ is equivalent
to $x^{2}+y^{2}+z^{2}$, $x^{2}+y^{2}$ or $x^{2}$.

First consider the case where $f=x^{2}+y^{2}+z^{2}$ and $g$ is a linear form.
Then $V(f,g)$ is either two points or a double point in $\mathbb{P}^{2}$. It
is easy to see that in the former case $F$ is equivalent to

(3.9) $\tilde{F}_{9}=(x^{2}+yz,x)$

and in the latter case $F$ is equivalent to

(3.10) $\tilde{F}_{10}=(x^{2}+yz,y).$

If $g$ is zero then we simply have

(3.11) $\tilde{F}_{11}=(x^{2}+y^{2}+z^{2},0).$

Next, if $f=x^{2}+y^{2}$ and $g$ is a linear form then $V(f)$ is two lines and
$V(g)$ a line in $\mathbb{P}^{2}$. The line $V(g)$ can meet the lines of
$V(f)$ in distinct points or in their intersection point or it can be one of
the lines in $V(f)$. Correspondingly we obtain:

(3.12)
$\tilde{F}_{12}=(x^{2}+y^{2},z),\quad\tilde{F}_{13}=(x^{2}+y^{2},x),\quad\tilde{F}_{14}=(xy,x),\quad\tilde{F}_{15}=(x^{2}+y^{2},0).$

If $f=x^{2}$ then a linear form $g$ can either divide $f$ or not. We obtain:

(3.13)
$\tilde{F}_{16}=(x^{2},y),\quad\tilde{F}_{17}=(x^{2},x),\quad\tilde{F}_{18}=(x^{2},0).$

Finally, we are left with the simple cases when neither $f$ nor $g$ is of
degree $2$. We have:

(3.14)
$\tilde{F}_{19}=(x,y),\quad\tilde{F}_{20}=(x,0),\quad\tilde{F}_{21}=(0,0).$

We sum up the results obtained above in the following corollary:

######  Corollary 3.2.

Let $F=(f,g):\mathbb{C}^{3}\rightarrow\mathbb{C}^{2}$ be a mapping with $f$
homogeneous of degree $2$ and $g$ homogeneous of degree less than $2$. Then
$F$ is linearly equivalent to one of the following:

  1. (1)

$\tilde{F}_{9}=(x^{2}+yz,x)$, $\tilde{F}_{10}=(x^{2}+yz,y)$,
$\tilde{F}_{12}=(x^{2}+y^{2},z)$, if $C(F)$ is a line.

  2. (2)

$\tilde{F}_{13}=(x^{2}+y^{2},x)$, $\tilde{F}_{14}=(xy,x)$,
$\tilde{F}_{16}=(x^{2},y)$, if $C(F)$ is a plane.

  3. (3)

$\tilde{F}_{11}=(x^{2}+y^{2}+z^{2},0)$, $\tilde{F}_{15}=(x^{2}+y^{2},0)$,
$\tilde{F}_{17}=(x^{2},x)$, $\tilde{F}_{18}=(x^{2},0)$,
$\tilde{F}_{20}=(x,0)$, $\tilde{F}_{21}=(0,0)$, if $C(F)$ is $\mathbb{C}^{3}$.

  4. (4)

$\tilde{F}_{19}=(x,y)$, if $C(F)$ is empty.

##  4\. Generic behavior at infinity

In this section we enumerate the linear and topological equivalence classes of
mappings $F$ having general behavior at infinity, i.e., with
$\tilde{F}=\tilde{F}_{1}$. There are infinitely many linear equivalence
classes, the families of topologically equivalent mappings have up to $2$
parameters.

We have

(4.1)
$F=(x^{2}+z^{2}+a_{7}x+a_{8}y+a_{9}z+a_{10},y^{2}+z^{2}+b_{7}x+b_{8}y+b_{9}z+b_{10}).$

By composing $F$ with the translation $(x-a_{7}/2,y-b_{8}/2,z-a_{9}/2)$ we may
assume that

$F=(x^{2}+z^{2}+a_{8}y+a_{10},y^{2}+z^{2}+b_{7}x+b_{9}z+b_{10}).$

Next, we use the translation $(p-a_{10},q-b_{10})$ on $\mathbb{C}^{2}$ and
obtain

(4.2) $F=(x^{2}+z^{2}+a_{8}y,y^{2}+z^{2}+b_{7}x+b_{9}z).$

Now we have two cases: the generic case when $a_{8}\neq 0$ and the nongeneric
case when $a_{8}=0$. We will investigate the nongeneric case later and assume
now that $a_{8}\neq 0$. Then we can multiply $x$, $y$ and $z$ by $a_{8}$ and
divide $f$ and $g$ by $a_{8}^{2}$ obtaining

(4.3) $F=(x^{2}+z^{2}+y,y^{2}+z^{2}+\alpha x+\beta
z),\quad\alpha,\beta\in\mathbb{C}.$

Note that the space $\Omega_{\mathbb{C}^{3}}(2,2)$ has dimension $20$, whereas
the spaces of affine automorphisms on $\mathbb{C}^{3}$ and $\mathbb{C}^{2}$
have dimensions $12$ and $6$, respectively. Thus $20-12-6=2$ is the smallest
number of parameters we can obtain by simplifying a generic member of
$\Omega_{\mathbb{C}^{3}}(2,2)$ with affine automorphisms.

Again, we will first investigate the generic case when $\alpha,\beta\neq 0$
and leave the nongeneric case $\alpha\beta=0$ for later.

Now let us take $A=\alpha^{-2}$ and $B=\beta^{2}\alpha^{-2}$, and compose $F$
with $\left(\alpha x/2,y/2,\beta z/2\right)$ and
$(4\alpha^{-2}p,4\alpha^{-2}q)$. We obtain

(4.4) $F=(x^{2}+Bz^{2}+2Ay,Ay^{2}+Bz^{2}+2x+2Bz),\quad A,B\in\mathbb{C}^{*}.$

Observe that $C(F)=V(xy-1,x(z+1)-z,yz-(z+1))$ is a smooth curve of degree $3$.
It can be easily parametrized by

(4.5) $\varphi:\mathbb{C}\setminus\\{0,-1\\}\ni
t\mapsto\left(\frac{t}{t+1},\frac{t+1}{t},t\right)\in C(F).$

The critical set of $F\circ\varphi$ is equal to the set
$V((f\circ\varphi)^{\prime},(g\circ\varphi)^{\prime})$ which is equal to the
set of zeroes of the derivative of $f\circ\varphi$ and set of zeroes of the
polynomial

(4.6) $H_{c}=Bt^{3}(t+1)^{3}+t^{3}-A(t+1)^{3}.$

The computation of the selfintersections of $\Delta(F)$ requires some tedious
calculations, so we use the computational algebra system Magma [1] to compute
that the set $\\{t\in\mathbb{C}\setminus\\{0,-1\\}:\exists
t_{1}\in\mathbb{C}\setminus\\{0,-1\\}:t\neq t_{1},\
F\circ\varphi(t)=F\circ\varphi(t_{1})\\}$ is contained in the set of zeroes of
the polynomial

(4.7) $H_{n}=(t+1)^{4}(Bt^{2}+A)^{2}+2t^{2}(t+1)^{2}(Bt^{2}-A)+t^{4}.$

For generic values of $A$ and $B$ the polynomials $H_{c}$ and $H_{n}$ have
neither multiple roots nor common roots. In that case $\Delta(F)$ has $6$
cusps at $F\circ\varphi(V(H_{c}))$ and $4$ nodes at $F\circ\varphi(V(H_{n}))$
and no other singularities. We use Magma to compute relevant resultants and
determine that $H_{c}$ has a multiple root if and only if $H_{n}$ has a
multiple root if and only if $H_{c}$ and $H_{n}$ have a common root if and
only if $(A,B)$ is a zero of

(4.8) $H_{0}=(A+B)^{4}+(A-1)^{4}+(B+1)^{4}-A^{4}-B^{4}-1+124AB(A-B+1).$

Note that the resultant of $H_{c}$ and $H_{c}^{\prime}$ is actually equal to
$-729A^{2}B^{3}H_{0}$, however $AB\neq 0$. Similarly, for $H_{n}$ and
$H_{n}^{\prime}$ we obtain $16777216A^{8}B^{10}H_{0}$ and for $H_{c}$ and
$H_{n}$ we obtain $A^{4}B^{4}H_{0}^{2}$.

Now, substituting $(A,B)$ for $\left(\alpha^{-2},\beta^{2}\alpha^{-2}\right)$
in equation (4.8) and multiplying the equation by $\alpha^{8}$ we obtain

(4.9)
$H_{1}=(\alpha^{2}+\beta^{2})^{4}+(\alpha^{2}-1)^{4}+(\beta^{2}+1)^{4}-\alpha^{8}-\beta^{8}-1+124\alpha^{2}\beta^{2}(\alpha^{2}-\beta^{2}+1).$

Observe that any $F=(x^{2}+z^{2}+y,y^{2}+z^{2}+\alpha x+\beta z)$ for
$\alpha\beta\neq 0$ is affinely equivalent to
$(x^{2}+Bz^{2}+2Ay,Ay^{2}+Bz^{2}+2x+2Bz)$ with some $AB\neq 0$. Moreover
$H_{1}(\alpha,\beta)\neq 0$ if and only if $H_{0}(A,B)\neq 0$. We can now
prove the following theorem:

######  Theorem 4.1.

Let $F=(x^{2}+z^{2}+y,y^{2}+z^{2}+\alpha x+\beta z)$ for $\alpha\beta\neq 0$
and $H_{1}(\alpha,\beta)\neq 0$. Then $F$ is topologically equivalent to
$F_{1}=(x^{2}+z^{2}+y,y^{2}+z^{2}+x+z)$.

###### Proof.

Step 1. Scheme of the proof.

By the argument above it is sufficient to show that the family of mappings
$F_{A,B}=(x^{2}+Bz^{2}+2Ay,Ay^{2}+Bz^{2}+2x+2Bz)$, for $ABH_{0}(A,B)\neq 0$,
is topologically trivial. Let $V=\mathbb{C}^{2}\setminus V(ABH_{0}(A,B))$.
Since $V$ is connected it is enough to show local topological triviality. Let
$\mathcal{F}=(F_{A,B},\operatorname{id}_{V}):\mathbb{C}^{3}\times
V\rightarrow\mathbb{C}^{2}\times V$.

We will loosely follow the methods developed in Section 3 of [2]. We pass to
the real structure to construct and integrate real vector fields. To keep the
notation brief we will use the complex coordinates when possible. We identify
a complex point $(x_{1},\ldots,x_{k})$ with the real point
$(\operatorname{Re}x_{1},\operatorname{Im}x_{1},\ldots,\operatorname{Re}x_{k},\operatorname{Im}x_{k})$
and a complex mapping $(f_{1},\ldots,f_{k})$ with a real mapping
$(\operatorname{Re}f_{1},\operatorname{Im}f_{1},\ldots,\operatorname{Re}f_{k},\operatorname{Im}f_{k})$.

Fix some $(A_{0},B_{0})\in V$ and let $K$ be a small closed ball centered at
$(A_{0},B_{0})$ and contained in $V$. Take $(A,B)\in K\setminus(A_{0},B_{0})$
and let $\partial_{0}$ be the constant vector field on $K$ with value
$(A-A_{0},B-B_{0})/||(A-A_{0},B-B_{0})||$. Let
$\gamma:[0,\varepsilon]\rightarrow K$ be the unique integral curve of
$\partial_{0}$ with $\gamma(0)=(A_{0},B_{0})$. Obviously $\gamma(t_{1})=(A,B)$
for $t_{1}=||(A-A_{0},B-B_{0})||$. Let $\pi$ be the projection from
$\mathbb{C}^{2}\times K$ to $K$. We will construct vector fields $w_{0}$ on
$\mathbb{C}^{2}\times K$ and $v_{0}$ on $\mathbb{C}^{3}\times K$ such that
$d\pi(w_{0})=\partial_{0}$ and $d\mathcal{F}(v_{0})=w_{0}$. Note that we treat
here $\pi$ as a restriction of projection
$\mathbb{R}^{8}\rightarrow\mathbb{R}^{4}$ and $\mathcal{F}$ as a
$\mathbb{R}^{10}\rightarrow\mathbb{R}^{8}$ mapping. If we construct the vector
fields correctly, then for a point $(p_{0},q_{0})\in\mathbb{C}^{2}$ there will
be $\gamma_{1}$ – the unique integral curve of $w_{0}$ such that
$\gamma_{1}(0)=(p_{0},q_{0},A_{0},B_{0})$. Then,
$\gamma_{1}(t_{1})=(p_{1},q_{1},A,B)$ for some
$(p_{1},q_{1})\in\mathbb{C}^{2}$. We define a homeomorphism
$\Psi_{A,B}:\mathbb{C}^{2}\rightarrow\mathbb{C}^{2}$ by setting
$\Psi_{A,B}(p_{0},q_{0})=(p_{1},q_{1})$. Similarly, we define a homeomorphism
$\Phi_{A,B}:\mathbb{C}^{3}\rightarrow\mathbb{C}^{3}$ by setting
$\Phi_{A,B}(x_{0},y_{0},z_{0})=(x_{1},y_{1},z_{1})$, where
$(x_{1},y_{1},z_{1},A,B)=\gamma_{2}(t_{1})$ and $\gamma_{2}$ is the unique
integral curve of $v_{0}$ with
$\gamma_{2}(0)=(x_{0},y_{0},z_{0},A_{0},B_{0})$. By construction we have
$F_{A,B}\circ\Phi_{A,B}=\Psi_{A,B}\circ F_{A_{0},B_{0}}$.

There are two conditions that we will need $v_{0}$ and $w_{0}$ to satisfy.
Firstly, the vector fields will be rugose, this implies that $\gamma_{1}$ and
$\gamma_{2}$ are unique and $\Phi_{A,B}$ and $\Psi_{A,B}$ are homeomorphisms
(see the proof of [2, Lemma 3.1]). Secondly, we require that $||w_{0}(P)||\leq
C(||P||+1)$ for all $P\in\mathbb{C}^{2}\times K$. This condition allows to use
the Gronwall Lemma to ensure that $\gamma_{1}$ does not escape to infinity and
is well defined for $t_{1}$ (see [2, Claim 3.1]). Similarly, we need
$||v_{0}(P)||\leq C(||P||+1)$ for certain points that we will expose later.

Step 2. Stratification and rugosity.

We define the stratification

$\mathcal{S}=\\{Y_{1},Y_{2},Y_{3}\\}=\\{\mathbb{C}^{2}\times
V\setminus\Delta(\mathcal{F}),\Delta(\mathcal{F})\setminus\operatorname{Sing}(\Delta(\mathcal{F})),\operatorname{Sing}(\Delta(\mathcal{F}))\\}.$

We refer the reader to [2, Subsection 2.1] for definition and properties of
Whitney stratifications. Observe that $\mathcal{S}$ is a Whitney
stratification. Moreover, $(p,q,A,B)\in\Delta(\mathcal{F})$ if and only if
$(p,q)\in\Delta(F_{A,B})$ and
$(p,q,A,B)\in\operatorname{Sing}(\Delta(\mathcal{F}))$ if and only if
$(p,q)\in\operatorname{Sing}(\Delta(F_{A,B}))$. See [5, Lemma 3.5] for the
proof of an analogous fact for $\mathbb{C}^{2}\rightarrow\mathbb{C}^{2}$
mappings, the proof in our case is very similar.

We also define a Whitney stratification on $\mathbb{C}^{3}\times V$:

$\mathcal{S}^{\prime}=\\{X_{1},X_{2},X_{3},X_{4},X_{5}\\}=\\{\mathcal{F}^{-1}(Y_{1}),\mathcal{F}^{-1}(Y_{2})\cap
C(\mathcal{F}),\mathcal{F}^{-1}(Y_{3})\cap C(\mathcal{F})$
$\mathcal{F}^{-1}(Y_{2})\setminus
C(\mathcal{F}),\mathcal{F}^{-1}(Y_{3})\setminus C(\mathcal{F})\\}.$

Note that $(x,y,z,A,B)\in C(\mathcal{F})$ if and only if $(x,y,z)\in
C(F_{A,B})$ and $(x,y,z,A,B)\in X_{3}$ if and only if $(x,y,z)\in C(F_{A,B})$
and $F_{A,B}(x,y,z)\in\operatorname{Sing}(\Delta(F_{A,B}))$.

Let us now recall the definition of a rugose vector field. Let
$X\subset\mathbb{C}^{n}$ be a variety with stratification $\\{X_{i}\\}$, let
$\phi:X\rightarrow\mathbb{R}$ be a real function. We say that $\phi$ is a
_rugose function_ if the following conditions are fulfilled:

  * •

The restriction $\phi|_{X_{i}}$ to any stratum $X_{i}$ is a smooth function.

  * •

For any stratum $X_{i}$ and for any $x\in X_{i}$ there exists a neighborhood
$U$ of $x$ in $\mathbb{C}^{n}$ and a constant $C>0$ such that for any $y\in
X\cap U$ and any $x_{1}\in X_{i}\cap U$ we have $|\phi(y)-\phi(x_{1})|\leq
C||y-x_{1}||$.

A _rugose map_ is a map whose components are rugose functions. A vector field
$v$ on $X$ is called a rugose vector field if $v$ is a rugose map and $v(x)$
is tangent to the stratum containing $x$ for any $x\in X$.

The stratification $\mathcal{S}$ induces a Whitney stratification on
$\mathbb{C}^{2}\times K$ consisting of the restricted strata. We will
construct the vector field $w_{0}$ so that it will be rugose with respect to
the stratification induced by $\mathcal{S}$. For an open subset
$U\subset\mathbb{C}^{2}\times V$ by a rugose vector field we will mean a
vector field rugose with respect to the stratification $\\{Y_{i}\cap U\\}$.
For a biholomorphic mapping $\phi:U\mapsto U^{\prime}$ we obtain the Whitney
stratification $\\{\phi(Y_{i}\cap U)\\}$ of $U^{\prime}$ which we use to
define rugose vector fields on $U^{\prime}$. In this setting, if we construct
a rugose vector field on $U^{\prime}$ then by lifting it to $U$ via $\phi$ we
obtain a rugose vector field on $U$. Similarly, the vector field $v_{0}$ will
be rugose with respect to the stratification induced by
$\mathcal{S}^{\prime}$.

Step 3. Constructing the vector fields.

We will construct the vector fields $w_{0}$ and $v_{0}$ in several parts and
glue them together using a smooth Urysohn’s Lemma and smooth partition of
unity.

Let $Q_{1}=(0,0,0,A_{0},B_{0})$ and $Q_{2}=(0,0,A_{0},B_{0})$.

We begin with $X_{3}$ and $Y_{3}$. Take a point
$P_{1}=(x_{1},y_{1},z_{1},A_{0},B_{0})\in X_{3}$, then $(x_{1},y_{1},z_{1})$
is either a cusp or a node of $F_{A_{0},B_{0}}$. First assume that it is a
cusp, so $F_{A_{0},B_{0}}|_{C(F_{A_{0},B_{0}})}$ is not an immersion at
$(x_{1},y_{1},z_{1})$. We can express the cusp singularity in its normal form,
i.e., we can find neighborhoods $U_{1}$ of $P_{1}$ and $U_{2}$ of
$\mathcal{F}(P_{1})$, open sets $V_{1}\subset\mathbb{C}^{5}$, and
$V_{2}\subset\mathbb{C}^{4}$ and biholomorphic mappings
$h_{1}:(U_{1},P_{1})\rightarrow(V_{1},Q_{1})$,
$h_{2}(U_{2},\mathcal{F}(P_{1}))\rightarrow(V_{2},Q_{2})$ such that
$\mathcal{F}|_{U_{1}}=h_{2}^{-1}\circ\mathcal{F}_{1}\circ h_{1}$ for
$\mathcal{F}_{1}(x,y,z,A,B)=(x,y^{3}+xy+z^{2},A,B)$. Moreover, the mappings
$h_{1}$, $h_{2}$ are identity on the $A$ and $B$ coordinates. Thus we obtain
the following commutative diagram:

${(U_{1},P_{1})}$${(U_{2},\mathcal{F}(P_{1}))}$${(\mathbb{C}^{2},(A_{0},B_{0}))}$${(V_{1},Q_{1})}$${(V_{2},Q_{2})}$${(\mathbb{C}^{2},(A_{0},B_{0}))}$$\scriptstyle{\mathcal{F}}$$\scriptstyle{h_{1}}$$\scriptstyle{\pi}$$\scriptstyle{h_{2}}$$\scriptstyle{\operatorname{id}}$$\scriptstyle{\mathcal{F}_{1}}$$\scriptstyle{\pi}$

The construction above was executed over the complex numbers, but now we pass
to the real structure with all mappings being analytic. We lift $\partial_{0}$
trivially (by putting zeroes on first $4$ or $6$ coordinates) to
$V_{2}\cap\mathbb{C}^{2}\times K$ and $V_{1}\cap\mathbb{C}^{3}\times K$. Then
we lift via $h_{1}$ to obtain the vector field $v_{1}$ on
$U_{1}\cap\mathbb{C}^{3}\times K$ and via $h_{2}$ to obtain $w_{1}$ on
$U_{2}\cap\mathbb{C}^{2}\times K$. Note that $v_{1}$ and $w_{1}$ are rugose,
since they are obtained by lifting a constant vector field. Moreover, by
taking $K$ small enough we can assume that $U_{1}$ contains the connected
component of $X_{3}\cap\mathbb{C}^{3}\times K$ containing $P_{1}$ and $U_{2}$
contains the connected component of $Y_{3}\cap\mathbb{C}^{2}\times K$
containing $\mathcal{F}(P_{1})$.

Now, if $P_{3}=(x_{3},y_{3},z_{3},A_{0},B_{0})\in X_{3}$ is not a cusp of
$F_{A_{0},B_{0}}$, then it is a node of $F_{A_{0},B_{0}}$. This means that
there is a unique $P_{4}\in X_{3}$ such that
$\mathcal{F}(P_{3})=\mathcal{F}(P_{4})$. Again, we will use the normal form of
the node multisingularity to lift $\partial_{0}$. We have biholomorphic
mappings $h_{3}:(U_{3},P_{3})\rightarrow(V_{3},Q_{1})$,
$h_{4}:(U_{4},P_{4})\rightarrow(V_{4},Q_{1})$,
$h_{5}:(U_{5},\mathcal{F}(P_{3}))\rightarrow(V_{5},Q_{2})$ such that
$\mathcal{F}|_{U_{3}}=h_{5}^{-1}\circ\mathcal{F}_{3}\circ h_{3}$ for
$\mathcal{F}_{3}(x,y,z,A,B)=(x,y^{2}+z^{2},A,B)$ and
$\mathcal{F}|_{U_{4}}=h_{5}^{-1}\circ\mathcal{F}_{4}\circ h_{4}$ for
$\mathcal{F}_{4}(x,y,z,A,B)=(y^{2}+z^{2},x,A,B)$.

Since $F_{A_{0},B_{0}}$ has only $6$ cusps and $4$ nodes we will shrink $K$
only finitely many times. By shrinking the obtained neighborhoods connected
components of $X_{3}\cap\mathbb{C}^{3}\times K$ we can assume that they are
disjoint. So we can glue the various parts together to obtain the vector field
$v_{1}$ on a set $U_{6}$ open in $\mathbb{C}^{3}\times K$ and containing
$X_{3}\cap\mathbb{C}^{3}\times K$. Similarly, we obtain the vector field
$w_{1}$ on a set $U_{7}$. Since $Y_{3}\cap\mathbb{C}^{2}\times K$ is compact
we may assume that there is a constant $C$ such that $||w_{1}(P)||\leq C$ for
all $P\in U_{7}$. We may need to shrink $U_{6}$ and $U_{7}$ further to
maintain the condition $\mathcal{F}(U_{6})=U_{7}$.

Now consider $X_{2}$ and $Y_{2}$. We lift $\partial_{0}$ to $Y_{2}$ using the
horizontal lift, i.e., for $P\in Y_{2}$ we set $w_{2}(P)$ to be the unique
vector in $T_{P}Y_{2}$ which lifts $\partial_{0}$ and is orthogonal to $\ker
d_{P}(\pi|_{Y_{2}})$. We need to verify that there is a constant $C$ such that
for all $P\in Y_{2}\cap\mathbb{C}^{2}\times K$ we have $||w_{2}(P)||\leq
C(||x||+1)$, we will do that in Step 4. Since $\mathcal{F}|_{X_{2}}$ is an
isomorphism on $Y_{2}$ there is only one way that we can lift the vector field
$w_{2}$ on $Y_{2}$ to $v_{2}$ on $X_{2}$.

Now we would like to extend $v_{2}$ and $w_{2}$ to rugose vector fields on
neighborhoods of $X_{2}$ and $Y_{2}$, respectively, and combine them with
$v_{1}$ and $w_{1}$. For any $P\in X_{2}$ there are open sets $U_{P,1}$,
$U_{P,2}$, $U_{P,3}$, $U_{P,4}$ and biholomorphic mappings
$h_{P,1}:U_{P,1}\rightarrow U_{P,2}$ and $h_{P,2}:U_{P,3}\rightarrow U_{P,4}$
such that $\mathcal{F}|_{U_{P,1}}=h_{P,2}^{-1}\circ\mathcal{F}_{3}\circ
h_{P,1}$ and $h_{P,1}$ and $h_{P,2}$ are identities in the $A$, $B$
coordinates. Note that $h_{P,1}(X_{2})$ is given in
$U_{P,2}\subset\mathbb{C}^{3}\times\mathbb{C}^{2}$ by equations $y=z=0$. Now
we pass to the real structure and lift $v_{2}$ to $v_{2}^{\prime}$ on
$h_{P,1}(X_{2})$. Since $v_{2}(Q)\in T_{Q}X_{2}$ and $v_{2}$ is a lift of
$\partial_{0}$ we must have
$v_{2}^{\prime}=(\alpha,\beta,0,0,0,0,\partial_{0})$, where $\alpha$ and
$\beta$ are smooth functions in variables $\operatorname{Re}x$,
$\operatorname{Im}x$, $\operatorname{Re}A$, $\operatorname{Im}A$,
$\operatorname{Re}B$, $\operatorname{Im}B$. Let
$\pi_{1}:\mathbb{R}^{6}\times\mathbb{R}^{4}\rightarrow\mathbb{R}^{2}\times\mathbb{R}^{4}$
be the projection on first two and last four coordinates. We extend
$v_{2}^{\prime}$ to $U_{P,2}$ by composing it with $\pi_{1}$. Obviously,
$v_{2}^{\prime}\circ\pi_{1}$ is a rugose vector field, we lift it to a rugose
vector field $v_{P,3}$ on $U_{P,1}$ via $h_{P,1}$. In the same way we extend
$w_{2}$ to a rugose vector field $w_{P,3}$ on $U_{P,3}$. Note that thanks to
the construction $v_{3}$ is a lift of $w_{3}$ via $\mathcal{F}$. Since
$||w_{2}(Q)||\leq C(||Q||+1)$ for any $Q\in Y_{2}\cap U_{P,3}$, we can shrink
$U_{P,1}$ and $U_{P,3}$ so that we retain $U_{P,1}=\mathcal{F}^{-1}U_{P,3}$
and have $||w_{P,3}\mathcal{F}(Q)||\leq 2C(||\mathcal{F}(Q)||+1)$ for any
$Q\in U_{P,3}$.

Take a smooth partition of unity $1=\sum_{i}\varphi_{i}$ on a neighborhood
$U_{8}$ of $Y_{2}\cap\mathbb{C}^{2}\times K$ such that for all $i$ there is a
point $P_{i}$ such that $\varphi_{i}$ is zero outside $U_{P_{i},3}$. We define
$w_{3}=\sum_{i}\varphi_{i}\cdot w_{P_{i},3}$ and
$v_{3}=\sum_{i}(\varphi_{i}\circ\mathcal{F})\cdot v_{P_{i},3}$. Note that for
each $P\in X_{2}\cap\mathbb{C}^{3}\times K$ the value
$\varphi_{i}(\mathcal{F}(P))$ is nonzero for finitely many $i$, so $v_{3}$ is
well defined on the neighborhood $\bigcap\\{U_{P_{i},1}:\
\varphi_{i}(\mathcal{F}(P))\neq 0\\}$ of $P$. Thus $v_{3}$ is well defined on
a neighborhood $U_{9}$ of $X_{2}\cap\mathbb{C}^{3}\times K$ such that
$\mathcal{F}(U_{9})=U_{8}$.

Now, take a smooth function $\varphi_{1}:U_{7}\cup U_{8}\rightarrow[0,1]$ such
that $\varphi_{1}$ is equal $1$ on some neighborhood $U_{10}$ of
$Y_{3}\cap\mathbb{C}^{2}\times K$ and is equal $0$ outside the neighborhood
$U_{7}$ on which $w_{1}$ is defined. We define
$w_{4}=\varphi_{1}w_{1}+(1-\varphi_{1})w_{3}$ and
$v_{4}=(\varphi_{1}\circ\mathcal{F})v_{1}+(1-(\varphi_{1}\circ\mathcal{F}))v_{3}$.
By construction $w_{4}$ is rugose on the neighborhood $U_{8}\cup U_{10}$ of
$\Delta(\mathcal{F})\cap\mathbb{C}^{2}\times K$ and $v_{4}$ is rugose on the
neighborhood $U_{11}=(U_{6}\cap\mathcal{F}^{-1}(U_{10}))\cup U_{9}$ of
$C(\mathcal{F})\cap\mathbb{C}^{3}\times K$.

Now consider $Y_{3}$. Let $w_{5}$ be the horizontal lift of $\partial_{0}$ to
$Y_{3}$, note that this is a constant vector field
$(0,\ldots,0,\partial_{0})$. Take a smooth function
$\varphi_{2}:\mathbb{C}^{2}\times K\rightarrow[0,1]$ such that $\varphi_{2}$
is equal to $1$ on some neighborhood $U_{12}$ of
$\Delta(\mathcal{F})\cap\mathbb{C}^{2}\times K$ and $0$ outside $U_{8}\cup
U_{10}$. We define $w_{0}=\varphi_{1}w_{4}+(1-\varphi_{1})w_{5}$. Note that by
construction $w_{0}$ is rugose and $||w_{0}(P)||\leq 2C(||P||+1)$ for any
$P\in\mathbb{C}^{2}\times K$.

We define $v_{5}$ to be the horizontal lift of $w_{0}$ to
$(\mathbb{C}^{3}\setminus\mathbb{C}(\mathcal{F}))\times K=(X_{1}\cup X_{4}\cup
X_{5})\cap\mathbb{C}^{3}\times K$ via $\mathcal{F}$. Note that
$\mathcal{F}(X_{1})=Y_{1}$, $\mathcal{F}(X_{4})=Y_{2}$,
$\mathcal{F}(X_{5})=Y_{3}$ so rugosity of $w_{0}$ implies rugosity of $v_{5}$.
Take a smooth function $\varphi_{3}:\mathbb{C}^{3}\times K\rightarrow[0,1]$
such that $\varphi_{3}$ is equal to $1$ on some neighborhood of
$\mathbb{C}(\mathcal{F})\cap\mathbb{C}^{3}\times K$ and $0$ outside $U_{11}$.
We define $v_{0}=\varphi_{3}v_{4}+(1-\varphi_{3})v_{5}$. Note that by
construction $v_{0}$ is rugose.

Step 4. A bound for $||w_{2}(P)||$.

Since $Y_{2}\cap\mathbb{C}^{2}\times K$ is not bounded we must check for
critical points at infinity and ensure that we construct vector fields with no
trajectories going to infinity. Here we will rely heavily on the methods from
[2, Section 3]. We lifted $\partial_{0}$ to $Y_{2}$ using the horizontal lift
and need to verify that there is a constant $C$ such that for all $P\in
Y_{2}\cap\mathbb{C}^{2}\times K$ we have $||w_{2}(P)||\leq C(||x||+1)$. By [2,
Lemma 3.2] it is enough to show that
$(||x||+1)\nu(d_{P}(\pi|_{Y_{2}}))\geq\frac{1}{C}$, where $\nu$ is the Rabier
function. We can find a suitable constant if and only if $\pi|_{Y_{2}}$ does
not have any asymptotic critical values, i.e., there is no sequence of
$P_{n}\in Y_{2}$ such that $P_{n}\rightarrow\infty$, $\pi(P_{n})$ is
convergent and $||P_{n}||\nu(d_{P_{n}}(\pi|_{Y_{2}}))\rightarrow 0$. Due to
[6, Corollary 2.3] we can replace $\nu(d_{P_{n}}(\pi|_{Y_{2}}))$ with
$g^{\prime}(d_{P_{n}}(\pi),T_{P_{n}}Y_{2})$, which is much easier to compute.
We refer the reader to [6, Section 2] for the construction of the $g^{\prime}$
function, we will only compute its value in our setting.

Take $P\in Y_{2}$, to compute the value of $g^{\prime}(d_{P}(\pi),T_{P}Y_{2})$
we will need the equation of the hyperplane $T_{P}Y_{2}$ in
$T_{P}\mathbb{C}^{4}$, we represent the linear equation as a row vector of
coefficients. Note $\mathcal{F}|_{X_{2}}$ is an isomorphism of $X_{2}$ and
$Y_{2}$ since by definition $X_{2}$ consists of points at which
$\mathcal{F}|_{C(\mathcal{F})}$ is an immersion and one to one. So there is a
unique $P_{0}=(x,y,z,A,B)\in X_{2}$ such that $P=\mathcal{F}(P_{0})$. We have

$d_{P_{0}}\mathcal{F}=\left[\begin{matrix}2x&2A&2Bz&2y&z^{2}\\\
2&2Ay&2B(z+1)&y^{2}&z^{2}+2z\\\ 0&0&0&1&0\\\ 0&0&0&0&1\end{matrix}\right].$

Since $T_{P}Y_{2}$ is the image of $d_{P_{0}}\mathcal{F}$ it is generated by
the columns of the matrix representation of $d_{P_{0}}\mathcal{F}$. It is easy
to see that the linear equation which is satisfied by the columns of
$d_{P_{0}}\mathcal{F}$ has coefficients $[1,-x,-y,z-x]$ (we need to use the
fact $xy=1$ and $x(z+1)=z$). To compute $g^{\prime}(d_{P}(\pi),T_{P}Y_{2})$ we
append the row with the coefficients of the equation of $T_{P}Y_{2}$ to the
matrix of $d_{P}(\pi)$. We obtain the matrix:

$M=\left[\begin{matrix}0&0&1&0\\\ 0&0&0&1\\\ 1&-x&-y&z-x\end{matrix}\right].$

By [6, Definition 2.5]

$g^{\prime}(d_{P}(\pi),T_{P}Y_{2})=\max_{I}\left\\{\min_{J\subset I,\
j=1,2}\frac{|M_{I}|}{|M_{J}(j)|}\right\\}.$

where $M_{I}$ are the $3\times 3$ minors of $M$ with columns indexed by $I$
and $M_{J}(j)$ are the $2\times 2$ minors of $M$ with columns indexed by $J$
and without the $j$-th row. Thus, we have:

$g^{\prime}(d_{P}(\pi),T_{P}Y_{2})=\max\left\\{\min\left\\{1,\frac{1}{|z-x|},\frac{1}{|y|}\right\\},\min\left\\{1,\frac{|x|}{|z-x|},\frac{|x|}{|y|}\right\\}\right\\}.$

To compute the asymptotic critical values we take a sequence
$P_{n}=\mathcal{F}(x_{n},y_{n},z_{n},A_{n},B_{n})$. Since $\pi(P_{n})$ must be
convergent, $(A_{n},B_{n})$ converges to some $(A^{\prime},B^{\prime})$. The
curve $\Delta(F_{A^{\prime},B^{\prime}})$ has three points at infinity
($(1:0:0)$, $(0:1:0)$ and $(1:1:0)$), since $P_{n}\rightarrow\infty$ by taking
a subsequence we may assume that $F_{A_{n},B_{n}}(x_{n},y_{n},z_{n})$ tends to
one of these points. That means that $(x_{n},y_{n},z_{n})$ tends to one of the
three points at infinity of the curve $C(F_{A^{\prime},B^{\prime}})$, which
are ($(1:0:0:0)$, $(0:1:0:0)$ and $(0:0:1:0)$). This means that
$g^{\prime}(d_{P_{n}}(\pi),T_{P_{n}}Y_{2})$ is equal to approximately
$\min\\{|x_{n}|^{-1},|y_{n}|^{-1},|z_{n}|^{-1}\\}$. At the same time
$||P_{n}||$ is equal to approximately
$\max\\{|x|^{2},|Ay^{2}|,\sqrt{2}|Bz^{2}|\\}$. It follows that
$||P_{n}||\nu(d_{P_{n}}(\pi|_{Y_{2}}))$ does not converge to $0$, so
$\pi|_{Y_{2}}$ does not have any asymptotic critical values and there is a
constant $C$ such that $||w_{2}(P)||\leq C(||x||+1)$ for all $P\in
Y_{2}\cap\mathbb{C}^{2}\times K$.

Step 5. A bound for $||v_{0}(P)||$.

To conclude the proof we need to verify that the curve $\gamma_{2}$ which we
use in Step 1 is well defined, i.e., it does not go to infinity. By
construction $\mathcal{F}(\gamma_{2}(t))=\gamma_{1}(t)$ and by rugosity of
$v_{0}$ the curve $\gamma_{2}$ is contained in one of the strata. Consider
$(x_{0},y_{0},z_{0},A_{0},B_{0})\in X_{3}$, let $X_{3}^{\prime}$ be the
connected component of $X_{3}$ containing $(x_{0},y_{0},z_{0},A_{0},B_{0})$.
Since $\mathcal{F}|_{X_{3}^{\prime}}$ is an injection we have
$\gamma_{2}(t)=\mathcal{F}|_{X_{3}^{\prime}}^{-1}(\gamma_{1}(t))$. Similarly,
if $(x_{0},y_{0},z_{0},A_{0},B_{0})\in X_{2}$ then
$\gamma_{2}(t)=\mathcal{F}|_{X_{2}}^{-1}(\gamma_{1}(t))$. Thus we may assume
that $(x_{0},y_{0},z_{0},A_{0},B_{0})\notin C(\mathcal{F})$. To use Gronwall
Lemma we need to verify that $||v_{0}(P)||\leq C(||P||+1)$ for all
$P\in\mathcal{F}^{-1}(\gamma_{1}([0,t_{1}]))$. Suppose this is not true. Then
we have a sequence of points $P_{n}\rightarrow\infty$ such that
$||P_{n}||/||v_{0}(P_{n})||\rightarrow 0$. By compactness of
$\gamma_{1}([0,t_{1}])$ we may pass to a subsequence and assume that
$\mathcal{F}(P_{n})\rightarrow(p_{1},q_{1},A_{1},B_{1})$. Note that the
surfaces given by $x^{2}+B_{1}z^{2}+2A_{1}y=p_{1}$ and
$A_{1}y^{2}+B_{1}z^{2}+2x+2B_{1}z=q_{1}$ intersect at infinity at points given
by $x^{2}+B_{1}z^{2}=A_{1}y^{2}+B_{1}z^{2}=0$. Thus we may assume that
$P_{n}\rightarrow(a_{1}b_{1}:b_{1}:a_{1}i:0)$ for some $a_{1}$, $b_{1}$ such
that $a_{1}^{2}=A_{1}$ and $b_{1}^{2}=B_{1}$.

Take a point $P_{0}=(x,y,z,A,B)\in\mathcal{F}^{-1}(\gamma_{1}([0,t_{1}]))$.
For $||P_{0}||$ big enough $v_{0}(P_{0})$ is the orthogonal lift of
$w_{0}(\mathcal{F}(P_{0}))$. The three nonzero minors of the matrix associated
to $d_{P_{0}}\mathcal{F}$ are $m_{1}=4AB(z+1-yz)$, $m_{2}=4B(xz+x-z)$ and
$m_{3}=4A(xy-1)$ and the kernel of $d_{P_{0}}\mathcal{F}$ is generated by
$m=[m_{1},-m_{2},m_{3},0,0]$. Now we pass to the real structure and consider
the point
$P_{0}=(\operatorname{Re}x,\operatorname{Im}x,\ldots,\operatorname{Re}B,\operatorname{Im}B)\in\mathbb{R}^{10}$.
We will add subscripts $\mathbb{R}$ or $\mathbb{C}$ to highlight which
structure we consider. The matrix $d_{P_{0}}\mathcal{F}_{\mathbb{R}}$ can be
obtained from $d_{P_{0}}\mathcal{F}_{\mathbb{C}}$ by replacing each complex
entry $a+bi$ with the real block

$\left[\begin{matrix}a&-b\\\ b&a\end{matrix}\right].$

The kernel of $d_{P_{0}}\mathcal{F}_{\mathbb{R}}$ is generated by the two
columns of the real matrix corresponding $m$ as one column matrix, i.e.,
$[\operatorname{Re}m_{1},\operatorname{Im}m_{1},-\operatorname{Re}m_{2},-\operatorname{Im}m_{2},\operatorname{Re}m_{3},\operatorname{Im}m_{3},0,0,0,0]$
and
$[-\operatorname{Im}m_{1},\operatorname{Re}m_{1},\operatorname{Im}m_{2},-\operatorname{Re}m_{2},-\operatorname{Im}m_{3},\operatorname{Re}m_{3},0,0,0,0]$.
Now consider the matrix $Mm(P_{0})$ obtained by appending $\overline{m}$ as
the last row of $d_{P_{0}}\mathcal{F}_{\mathbb{C}}$:

$Mm=Mm(P_{0})=\left[\begin{matrix}2x&2A&2Bz&2y&z^{2}\\\
2&2Ay&2B(z+1)&y^{2}&z^{2}+2z\\\ 0&0&0&1&0\\\ 0&0&0&0&1\\\
\overline{m}_{1}&-\overline{m}_{2}&\overline{m}_{3}&0&0\end{matrix}\right].$

Observe that a vector $v\in\mathbb{R}^{10}$ is orthogonal to $\ker
d_{P_{0}}\mathcal{F}_{\mathbb{R}}$ if and only if $Mm_{\mathbb{R}}v$ has
zeroes on the last two coordinates. For a vector $w\in\mathbb{R}^{8}$ let
$w^{\prime}\in\mathbb{R}^{10}$ denote the vector obtained from $w$ by
appending two zeroes at the end. Then $v=Mm_{\mathbb{R}}^{-1}w^{\prime}$ is a
vector orthogonal to $\ker d_{P_{0}}\mathcal{F}_{\mathbb{R}}$ satisfying
$d_{P_{0}}\mathcal{F}_{\mathbb{R}}v=w$, so $v$ is the orthogonal lift of $w$.
Thus $v_{0}(P_{0})=Mm_{\mathbb{R}}^{-1}w_{0}(\mathcal{F}(P_{0}))^{\prime}$.
Since $w_{0}(\gamma_{1}([0,t_{1}]))$ is compact we have $||v_{0}(P_{0})||\leq
C\max\\{||Mm_{\mathbb{R}}^{-1}e_{i}||,i=1,\ldots,8\\}$ for some constant $C$.

In order to verify that $||P_{n}||/||v_{0}(P_{n})||$ does not tend to $0$ we
may approximate $P_{n}$ with $(a_{1}b_{1}k_{n},b_{1}k_{n},a_{1}ik_{n})$ for
some $k_{n}\in\mathbb{R}$ and $k_{n}\rightarrow\infty$. Obviously
$||P_{n}||=\Theta(k_{n})$, i.e., there is a constant $C_{1}>0$ such that
$C_{1}^{-1}k_{n}\leq||P_{n}||\leq C_{1}k_{n}$ for $n$ big enough. We have

$Mm(P_{n})_{\mathbb{C}}\approx\left[\begin{matrix}2a_{1}b_{1}k_{n}&2a_{1}^{2}&2a_{1}b_{1}^{2}ik_{n}&2b_{1}k_{n}&-a_{1}^{2}k_{n}^{2}\\\
2&2a_{1}^{2}b_{1}k_{n}&2a_{1}b_{1}^{2}ik_{n}&b_{1}^{2}k_{n}^{2}&-a_{1}^{2}k_{n}^{2}\\\
0&0&0&1&0\\\ 0&0&0&0&1\\\
4\overline{a}_{1}^{3}\overline{b}_{1}^{3}ik_{n}^{2}&4\overline{a}_{1}^{2}\overline{b}_{1}^{3}ik_{n}^{2}&4\overline{a}_{1}^{3}\overline{b}_{1}^{2}k_{n}^{2}&0&0\end{matrix}\right].$

Observe that $\det Mm(P_{n})_{\mathbb{C}}\approx
16a_{1}^{3}b_{1}^{2}\overline{a}_{1}^{2}\overline{b}_{1}^{3}ik_{n}^{4}$,
moreover, the norm of every $4\times 4$ minor of $Mm(P_{n})_{\mathbb{C}}$ is
in $O(k_{n}^{5})$. It follows that the norm of every entry of
$Mm(P_{n})^{-1}_{\mathbb{C}}$, and consequently of every entry of
$Mm(P_{n})^{-1}_{\mathbb{R}}$, is in $O(k_{n})$. Consequently
$||Mm(P_{n})_{\mathbb{R}}^{-1}e_{i}||\leq C_{2}k_{n}$ for some constant
$C_{2}$ and $n$ big enough. Thus, $||v_{0}(P_{n})||\leq C_{1}CC_{2}||P_{n}||$
for $n$ big enough. ∎

Now we will consider the case when $H_{0}(A,B)=0$ (see equation (4.8)), i.e.,
when $F$ has nongeneric singularities. In our computations we still assume
that $AB\neq 0$. We use Magma [1] to compute the Gröbner basis of the ideal
$(H_{c},\partial H_{c}/\partial t)$ with respect to lexicographical order with
$t$ as the first variable (see equation (4.6)). The only element of the basis
without the variable $t$ is $A\cdot H_{0}(A,B)$, which confirms that $H_{c}$
does not have multiple roots if and only if $H_{0}(A,B)\neq 0$. There are two
elements with leading term $t$. They are

$H_{c1}=tAB\cdot(B-4)(B-1/4)(B+1)+\text{poly}(A,B)$

and

$H_{c2}=tA\cdot(15A-28B^{3}+99B^{2}+52B-15)+\text{poly}(A,B).$

The leading coefficient with respect to $t$ of at least one of $H_{c1}$ and
$H_{c2}$ is nonzero if and only if $H_{c}$ has one double root and no other
multiple roots. This occurs when $(A,B)\notin\\{(1,4),(-1/4,1/4),(-4,-1)\\}$.
Setting $H_{c1}=0$ or $H_{c2}=0$ allows us to compute the double root. We will
denote it by $t_{0}$ but we will not write it down explicitly since the
formula is quite long and does not provide any additional insight.

We also compute the Gröbner bases of the ideals $(H_{n},\partial
H_{n}/\partial t)$ and $(H_{c},H_{n})$ and conclude that if
$(A,B)\notin\\{(1,4),(-1/4,1/4),(-4,-1)\\}$ then $t_{0}$ is a double root of
$H_{n}$ and $H_{n}$ does not have other multiple roots or common roots with
$H_{c}$. We can now prove the following theorem:

######  Theorem 4.2.

Let $F=(x^{2}+z^{2}+y,y^{2}+z^{2}+\alpha x+\beta z)$ for $\alpha\beta\neq 0$,
$H_{1}(\alpha,\beta)=0$ and
$(\alpha^{2},\beta^{2})\notin\\{(1,4),(-4,-1),(-1/4,1/4)\\}$. Then $F$ is
topologically equivalent to $F_{2}=(x^{2}+z^{2}+y,y^{2}+z^{2}+4x+iz)$.

###### Proof.

We proceed similarly as in Theorem 4.1. Using linear equivalence we transform
$F$ to $F_{A,B}=(x^{2}+Bz^{2}+2Ay,Ay^{2}+Bz^{2}+2x+2Bz)$ (see equation (4.4)).
We consider $V=V(H_{0}(A,B))\setminus
V(AB)\setminus\\{(1,4),(-1/4,1/4),(-4,-1)\\}$. Recall that
$(A,B)=(\alpha^{-2},\beta^{2}\alpha^{-2})$, so $\alpha$, $\beta$ satisfy the
assumptions if and only if $(A,B)\in V$. Again, $V$ is connected so it is
enough to show local topological triviality of $F_{A,B}$. Since the proof is
very similar to the proof of Theorem 4.1 we will not repeat it, we will only
highlight the differences.

Fix some $(A_{0},B_{0})\in V$ and let $K$ be a small closed ball centered at
$(A_{0},B_{0})$ such that $K\subset\mathbb{C}^{2}\setminus
V(AB)\setminus\\{(1,4),(-1/4,1/4),(-4,-1)\\}$ and $K\cap V(H_{0}(A,B))$ is
connected. Take $(A,B)\in K\cap V(H_{0}(A,B))\setminus(A_{0},B_{0})$. Now take
a smooth real arc $\gamma:[0,1]\rightarrow K\cap V(H_{0}(A,B))$ such that
$\gamma(0)=(A_{0},B_{0})$ and $\gamma(1)=(A,B)$. We define the vector field
$\partial_{0}=d\gamma([1])$ on $\gamma$, so $\gamma$ is the integral curve of
$\partial_{0}$.

As in Theorem 4.1 we lift $\partial_{0}$ to vector fields $w_{0}$ on
$\mathbb{C}^{2}\times\gamma$ and $v_{0}$ on $\mathbb{C}^{3}\times\gamma$. We
use integral curves of $w_{0}$ and $v_{0}$ to construct the required
homeomorphisms.

The mappings $F_{A,B}$ have now $4$ cusps and $3$ nodes, they also have a
double cusp – a point $P$ such that the germ $(F_{A,B}|_{C(F_{A,B})},P)$ is
biholomorphic to the germ of $x\mapsto(x^{3},0)$ at $0$. The normal form of a
double cusp is $(x,y,z)\mapsto(x,y^{4}+xy+z^{2})$. We refer the reader to [7]
for more information on classification of singularities, the double cusp is
enumerated as type $5$ on the list in [7, Proposition 1.3]. ∎

The three remaining cases are covered by the next theorem:

######  Theorem 4.3.

Let $F=(x^{2}+z^{2}+y,y^{2}+z^{2}+\alpha x+\beta z)$ for
$(\alpha^{2},\beta^{2})\in\\{(1,4),(-4,-1),(-1/4,1/4)\\}$. Then $F$ is
affinely equivalent to $F_{3}=(x^{2}+z^{2}+y,y^{2}+z^{2}+x+2z)$.

###### Proof.

Using linear equivalence we transform $F$ to
$F_{A,B}=(x^{2}+Bz^{2}+2Ay,Ay^{2}+Bz^{2}+2x+2Bz)$ for
$(A,B)\in\\{(1,4),(-1/4,1/4),(-4,-1)\\}$ (see equation (4.4)).

Observe that

$F_{1,4}=(4p+2,4p-4q+3)\circ F_{-1/4,1/4}\circ(-z,-y+1,-x)$

and

$F_{-1/4,1/4}=(-q/4+1/4,-p/4-1/4)\circ F_{-4,-1}\circ(y,x,-z-1).$

This concludes the proof. Note that to find the affine automorphism
$(-z,-y+1,-x)$ we computed the two cusps and two double cusps of
$F_{-1/4,1/4}$ and $F_{1,4}$ and then took an affine mapping which transformed
one quadruple into the other quadruple. Note that by the cusp of $F_{A,B}$ we
mean a point $P\in C(F_{A,B})$ such that $F_{A,B}(P)$ is a cusp of
$\Delta(F_{A,B})$. The cusps (double cusps) of $F_{A,B}$ are the points
$\varphi(t)$ (see equation (4.5)) where $t$ are the simple (double) roots of
$H_{c}$. ∎

Theorems 4.1, 4.2 and 4.3 classify mappings $F$ in the form of equation (4.3)
for $\alpha\beta\neq 0$. Let us now consider the case $\alpha\neq 0$,
$\beta=0$, i.e., $F=(x^{2}+z^{2}+y,y^{2}+z^{2}+\alpha x)$. We take
$A=\alpha^{-2}$, substitute $(x,y,z)$ for $\left(\alpha x/2,y/2,\alpha
z/2\right)$ and multiply both components of $F$ by $4\alpha^{-2}$. We obtain

(4.10) $F_{A}=(x^{2}+z^{2}+2Ay,Ay^{2}+z^{2}+2x),\quad A\in\mathbb{C}^{*}.$

Observe that $C(F_{A})$ is given by the ideal $(xy-1,(x-1)z,(y-1)z)$, so it is
reducible and consists of the hyperbola $C_{1}=V(xy-1,z)$ and the line
$C_{2}=V(x-1,y-1)$. The components $C_{1}$ and $C_{2}$ intersect at $(1,1,0)$.
The mapping $F_{A}|_{C_{2}}$ is a double cover of the line $V(p-q-A+1)$
branched at $(1,1,0)$. We parametrize $C_{1}$ by taking
$\varphi_{1}:\mathbb{C}^{*}\ni t\mapsto(t,1/t,0)\in C_{1}$ and obtain
$C(F_{A}\circ\varphi_{1})=V(t^{3}-A)$. To compute the intersections of
$F_{A}(C_{1})$ and $F_{A}(C_{2})$ we determine $t\in\mathbb{C}^{*}$ such that
$F_{A}\circ\varphi_{1}(t)$ lies on the line $V(p-q-A+1)$. We obtain
$(t-1)(t^{2}-A)=0$. We use Magma to verify that $F_{A}(C_{1})$ does not have
any selfintersections.

######  Theorem 4.4.

Let $F=(x^{2}+z^{2}+y,y^{2}+z^{2}+\alpha x)$ for $\alpha^{2}\notin\\{0,1\\}$.
Then $F$ is topologically equivalent to
$F_{4}=(x^{2}+z^{2}+y,y^{2}+z^{2}+2x)$.

###### Proof.

We begin by transforming $F$ into $F_{A}$ from equation (4.10) with
$A\in\mathbb{C}\setminus\\{0,1\\}$. Next, we proceed as in Theorem 4.1. We
will not repeat the proof, we will only focus on the important differences,
which are the singularities that occur and stratifications that are used.
Obviously, here we work over $V=\mathbb{C}\setminus\\{0,1\\}$ and take
$\mathcal{F}=(F_{A},\operatorname{id}_{V})$. Similarly as in Theorem 4.1, to
define the stratification $\mathcal{S}$ we take $Y_{1}=\mathbb{C}^{2}\times
V\setminus\Delta(\mathcal{F})$, however, we split
$\Delta(\mathcal{F})\setminus\operatorname{Sing}(\Delta(\mathcal{F}))$ into
two connected components. We take $Y_{2a}=\\{(p,q,A):\ (p,q)\in
F_{A}(C_{1})\\}$ and $Y_{2b}=\\{(p,q,A):\ (p,q)\in F_{A}(C_{2})\\}$, where
$C_{1}$ and $C_{2}$ are the two connected components of $C(F_{A})$ described
above. We take $Y_{3}=\operatorname{Sing}(\Delta(\mathcal{F}))$. Note that
$Y_{3}\cap\mathbb{C}^{2}\times\\{A\\}$ consists of $F_{A}(1,1,0)$, three cusps
of $F_{A}(C_{1})$ and two intersections of $F_{A}(C_{1})$ and $F_{A}(C_{2})$
other than $F_{A}(1,1,0)$. As in Theorem 4.1, we define the stratification
$\mathcal{S}^{\prime}$ of $\mathbb{C}^{3}\times V$ by taking
$\mathcal{F}^{-1}(Y)\cap C(\mathcal{F})$ and $\mathcal{F}^{-1}(Y)\setminus
C(\mathcal{F})$ for all strata $Y\in\mathcal{S}$. Now we exhibit the normal
forms of singularities of $F_{A}$, which are essential in defining the vector
fields. For $X_{2a}$ we have folds, as for $X_{2}$ in Theorem 4.1, with normal
form $(x,y,z)\mapsto(x,y^{2}+z^{2})$. For $X_{2b}$ the mapping
$F_{A}|_{C_{2}}$ is $2:1$ and we have a multisingularity with normal form
$(x_{1},y_{1},z_{1})\mapsto(x_{1},y_{1}^{2}+z_{1}^{2})$ and
$(x_{2},y_{2},z_{2})\mapsto(x_{2},y_{2}^{2}+z_{2}^{2})$. For $P\in
F_{A}(C_{1})\cap F_{A}(C_{2})\setminus\\{F_{A}(1,1,0)\\}$ we have three points
in $C(F_{A})\cap F_{A}^{-1}(P)$. For the first two the map germs are the same
as for $X_{2b}$ and for the third we have
$(x_{3},y_{3},z_{3})\mapsto(y_{3}^{2}+z_{3}^{2},x_{3})$. The cusps have the
normal form $(x,y,z)\mapsto(x,y^{3}+xy+z^{2})$. The singularity at $(1,1,0)$
has normal form $(x,y,z)\mapsto(x,xy^{2}+y^{4}+z^{2})$. This statement is not
obvious and we have not found a suitable reference in the literature. The
singularity is not finitely $\mathcal{A}$-determined (since
$(x,xy^{2}+y^{4}+y^{2k+1}+z^{2})$ is not equivalent and has the same $2k$-jet)
and therefore is absent in classifications with respect to biholomorphic
equivalence. Thus, we will provide a short proof.

First, we subtract second component from the first component and use
translations to move the singularity to the origin in source and target. We
obtain $F=(x^{2}-Ay^{2},Ay^{2}+Ay+2x+z^{2})$. As earlier, let $C_{1}$ be the
component of $C(F)$ such that $F|_{C_{1}}$ is generically $1:1$ and $C_{2}$ be
the component of $C(F)$ such that $F|_{C_{2}}$ is generically $2:1$. Note that
$F(C_{1})=V(p)$. We will introduce coordinates $p_{1}$ and $q_{1}$ so that
$F(C_{1})=V(p_{1})$ and $F(C_{1})=V(p_{1}-q_{1}^{2})$. We use Magma to compute
the equation of $F(C_{2})$, it is

(4.11) $0=4(A-1)^{3}p+(A-1)^{2}q^{2}+c_{1}(A)pq+c_{2}(A)p^{2}+\text{h.o.t.}$

We set $q_{1}=(A-1)q+c_{1}(A)/2(A-1)\cdot p+\text{h.o.t.}$ in such way that
the higher order terms in the definition of $q_{1}$ cancel all higher order
terms in equation (4.11) divisible by $q$. Thus we obtain
$0=4(A-1)^{3}p+p^{2}h_{1}(p,A)+q_{1}^{2}$ as the equation of $F(C_{2})$. We
take $p_{1}=-(A-1)^{3}p-p^{2}h_{1}(p,A)/4$ and obtain
$F(C_{1})=V(4p_{1}-q_{1}^{2})$. In the $p_{1}$, $q_{1}$ coordinates we have

$F=((1-A)^{3}(x^{2}-Ay^{2})+h_{2}(x^{2},y^{2},A),(A-1)(2x+2Ay+z^{2})+h_{3}(x,y,z^{2},A)),$

where $h_{2}$ and $h_{3}$ are power series of order $2$ with respect to all
variables except the last. Take $z_{1}=\sqrt{A-1}z+h_{4}(x,y,z,A)$ so that
$h_{4}$ does not have the $z$ term and $2\sqrt{A-1}zh_{4}+h_{4}^{2}$ coincides
with $h_{3}$ in all terms divisible by $z$. The second component of $F$ is
equal to $2(A-1)(x+Ay)+h_{5}(x,y,A)+z_{1}^{2}$, where $h_{5}$ is of order $2$
with respect to $x$, $y$ variables. Take $x_{1}=(A-1)(x+Ay)+h_{5}(x,y,A)/2$,
the first component of $F$ is equal to
$(1-A)x_{1}^{2}+2A(A-1)^{2}x_{1}y-A(A-1)^{4}y^{2}+h_{6}(x_{1},y,A)$, where
$h_{6}$ is of order $3$ with respect to $x_{1}$, $y$ variables. Take
$y_{1}=\sqrt{-A}(A-1)^{2}y-\sqrt{-A}x_{1}+h_{7}(x_{1},y,A)$ so that $h_{7}$ is
of order $2$ with respect to $x_{1}$, $y$ and $y_{1}^{2}$ coincides with
$2A(A-1)^{2}x_{1}y-A(A-1)^{4}y^{2}+h_{6}(x_{1},y,A)$ in all terms divisible by
$y$. We have

$F=(x_{1}^{2}+y_{1}^{2}+h_{8}(x_{1},A),2x_{1}+z_{1}^{2}).$

We have $C(F)=V(y_{1},(2x_{1}+\partial h_{8}/\partial x_{1})z_{1})$, in
particular $C_{2}=V(y_{1},z_{1})$. Thus
$4(x_{1}^{2}+h_{8}(x_{1},A))-(2x_{1})^{2}=0$, so $h_{8}=0$. To obtain the
desired normal form it is now enough to take $x_{2}=2x_{1}+z_{1}^{2}$,
$y_{2}=z_{1}/\sqrt{2}$, $z_{2}=y_{1}$, $p_{2}=q_{1}$ and
$q_{2}=p_{1}-q_{1}^{2}/4$. ∎

As a consequence of passing from equation (4.3) to equation (4.10) we obtain
the following:

######  Corollary 4.5.

Let $F=(x^{2}+z^{2}+y,y^{2}+z^{2}+\alpha x)$ for $\alpha^{2}=1$. Then $F$ is
affinely equivalent to $F_{5}=(x^{2}+z^{2}+2y,y^{2}+z^{2}+2x)$.

Next we will examine the case $\alpha=0$ in equation (4.3). We have
$F=(x^{2}+z^{2}+y,y^{2}+z^{2}+\beta z)$, $\beta\in\mathbb{C}$. Observe that

(4.12) $\left(-p+\frac{1}{2},q-p+\frac{1}{4}\right)\circ
F\circ\left(iz,-y+\frac{1}{2},ix\right)=(x^{2}+z^{2}+y,y^{2}+z^{2}+i\beta x).$

As a consequence of equation (4.12), Theorem 4.4 and Corollary 4.5 we obtain
the following:

######  Corollary 4.6.

If $F=(x^{2}+z^{2}+y,y^{2}+z^{2}+\beta z)$, then $F$ is affinely equivalent to
$F=(x^{2}+z^{2}+y,y^{2}+z^{2}+i\beta x)$. Consequently $F$ is

  1. (1)

topologically equivalent to $F_{4}=(x^{2}+z^{2}+y,y^{2}+z^{2}+2x)$ for
$\beta^{2}\notin\\{0,-1\\}$,

  2. (2)

affinely equivalent to $F_{5}=(x^{2}+z^{2}+2y,y^{2}+z^{2}+2x)$ for
$\beta^{2}=-1$.

The mapping $F_{6}=(x^{2}+z^{2}+y,y^{2}+z^{2})$ is not topologically
equivalent to any of the mappings considered earlier since it is the only one
with $C(F)$ being a union of three lines.

Now we consider the case $a_{8}=0$ in equation (4.2). We have

(4.13) $F=(x^{2}+z^{2},y^{2}+z^{2}+\alpha x+\beta
z),\quad\alpha,\beta\in\mathbb{C}$

For $\alpha\neq 0$ we have

(4.14)
$\left(\frac{q}{\alpha^{2}}+\frac{\beta^{2}}{4\alpha^{2}},\frac{p}{\alpha^{2}}-\frac{\beta^{2}}{4\alpha^{2}}\right)\circ
F\circ\left(\alpha y,\alpha x,\alpha
z-\frac{\beta}{2}\right)=\left(x^{2}+z^{2}+y,y^{2}+z^{2}-\frac{\beta}{\alpha}x\right).$

For $\alpha=0$ and $\beta\neq 0$ we have

(4.15) $\left(\frac{p-q}{\beta^{2}},\frac{p}{\beta^{2}}\right)\circ
F\circ\left(\beta z,i\beta x,-\beta
y\right)=\left(x^{2}+z^{2}+y,y^{2}+z^{2}\right).$

As a consequence of equation (4.14) and equation (4.15) we obtain the
following:

######  Corollary 4.7.

Let $F=(x^{2}+z^{2},y^{2}+z^{2}+\alpha x+\beta z)$. If $\alpha\neq 0$ then $F$
is affinely equivalent to $F=(x^{2}+z^{2}+y,y^{2}+z^{2}-\beta/\alpha x)$.
Consequently $F$ is

  1. (1)

topologically equivalent to $F_{4}=(x^{2}+z^{2}+y,y^{2}+z^{2}+2x)$ for
$(\beta/\alpha)^{2}\notin\\{0,1\\}$,

  2. (2)

affinely equivalent to $F_{5}=(x^{2}+z^{2}+2y,y^{2}+z^{2}+2x)$ for
$\beta^{2}=\alpha^{2}$.

  3. (3)

affinely equivalent to $F_{6}=(x^{2}+z^{2}+y,y^{2}+z^{2})$ for $\beta=0$.

Moreover, if $\alpha=0$ and $\beta\neq 0$ then $F$ is affinely equivalent to
$F=(x^{2}+z^{2}+y,y^{2}+z^{2})$.

The mapping $F_{7}=(x^{2}+z^{2},y^{2}+z^{2})$ is not topologically equivalent
to any of the mappings considered earlier since it is the only one with $C(F)$
being a union of three lines meeting in a point.

This concludes the enumeration of equivalence classes for mappings $F$ such
that $C(F)$ has $3$ points at infinity.

##  5\. Double critical point at infinity

In this section we enumerate the linear and topological equivalence classes of
mappings $F$ such that $\tilde{F}=\tilde{F}_{2}$, i.e., having a double a
smooth point at infinity. There are infinitely many linear equivalence
classes, there is one one-parameter family of topologically equivalent
mappings.

We have

(5.1)
$F=(x^{2}+z^{2}+a_{7}x+a_{8}y+a_{9}z+a_{10},yz+b_{7}x+b_{8}y+b_{9}z+b_{10}).$

By composing with translations we can assume that
$a_{7}=a_{9}=b_{9}=a_{10}=b_{10}=0$ and obtain

(5.2) $F=(x^{2}+z^{2}+a_{8}y,yz+b_{7}x+b_{8}y).$

There are several cases depending on whether the constants in equation (5.2)
are nonzero. First, we consider the case $a_{8}b_{7}\neq 0$. We compose $F$
with $(\sqrt{2^{-1}a_{8}b_{7}}x,b_{7}y,\sqrt{2^{-1}a_{8}b_{7}}z)$ and
$(2a_{8}^{-1}b_{7}^{-1}p,\sqrt{2a_{8}^{-1}b_{7}^{-3}}q)$ and obtain

(5.3) $F_{A}=(x^{2}+z^{2}+2y,yz+x+Ay),\quad A\in\mathbb{C}.$

Observe that $C(F_{A})=V(x(z+A)-1,xy-z,y-z^{2}-Az)$. Note that $C(F_{A})$ is a
smooth curve of degree $3$ and $\overline{C}(F_{A})$ has two points at
infinity: a smooth point $(1:0:0:0)$ and a singular point $(0:1:0:0)$.
Moreover, $C(F_{A})$ can be easily parametrized by

(5.4) $\varphi:\mathbb{C}\setminus\\{-1\\}\ni
t\mapsto\left(\frac{1}{t+A},t^{2}+At,t\right)\in C(F_{A}).$

The critical set of of $F\circ\varphi$ is equal to the set of zeroes of the
derivative of $f\circ\varphi$, which is equal to the set of zeroes of the
polynomial

(5.5) $H_{c1}=(3t+A)(t+A)^{3}-1.$

The selfintersections of $\Delta(F)$ are the points $F_{A}\circ\varphi(t)$,
where $t$ are zeroes of the polynomial

(5.6) $H_{n1}=t^{2}(t+A)^{2}+1.$

It follows that $F_{A}$ has $4$ cusps and $2$ nodes if and only if $H_{c1}$
and $H_{n1}$ have no double or common roots. This happens when $A^{4}\neq-16$.

We can now prove the following theorem:

######  Theorem 5.1.

Let $F_{A}=(x^{2}+z^{2}+2y,yz+x+Ay)$ for $A^{4}\neq-16$. Then $F_{A}$ is
topologically equivalent to $F_{8}=(x^{2}+z^{2}+2y,yz+x)$.

###### Proof.

Again, the proof is very similar to the proof of Theorem 4.1, we will not
repeat the proof but just highlight the differences. The only thing that needs
reexamination are the computations in Steps $4$ and $5$ of the original proof.

In Step $4$ we need to compute $g^{\prime}(d_{P}(\pi),T_{P}Y_{2})$. We compute
$d_{P_{0}}\mathcal{F}$ for $P_{0}=(x,y,z,A)\in X_{2}\cap\mathcal{F}^{-1}(P)$
and obtain

$d_{P_{0}}\mathcal{F}=\left[\begin{matrix}2x&2&2z&0\\\ 1&z+A&y&y\\\
0&0&0&1\end{matrix}\right].$

The linear equation which is satisfied by the columns $d_{P_{0}}\mathcal{F}$
has coefficients $[1,-2x,2z]$, so we obtain the matrix

$M=\left[\begin{matrix}0&0&1\\\ 1&-2x&2z\end{matrix}\right].$

By definition

$g^{\prime}(d_{P}(\pi),T_{P}Y_{2})=\max\left\\{\min\left\\{1,\frac{1}{|2z|}\right\\},\min\left\\{1,\frac{|x|}{|z|}\right\\}\right\\}.$

Thus for $P_{n}=\mathcal{F}(x_{n},y_{n},z_{n},A_{n})$ if
$\nu(d_{P_{n}}(\pi|_{Y_{2}}))\rightarrow 0$ then $|z_{n}|\rightarrow\infty$.
Since $y_{n}=z_{n}^{2}+A_{n}z_{n}$ the norm $||P_{n}||$ is equal to
approximately $|y_{n}z_{n}|$, so we cannot have
$||P_{n}||\nu(d_{P_{n}}(\pi|_{Y_{2}}))\rightarrow 0$. This had to be shown in
this step.

In Step $5$ we need to examine the matrix $Mm(P_{n})_{\mathbb{C}}$ for
$P_{n}=(x_{n},y_{n},z_{n},A_{n})\in\mathbb{C}^{3}\times\mathbb{C}$ such that
$||P_{n}||\rightarrow\infty$ but $\mathcal{F}(P_{n})$ converges to some
$(p,q,A)\in
Y_{1}\cap\mathbb{C}^{2}\times\mathbb{C}\setminus\\{\sqrt[4]{-16}\\}$. Note
that any fiber of $F_{A}$ has $3$ points at infinity: $(1:0:\pm i:0)$ and
$(0:1:0:0)$. First, consider the case $(x_{n}:y_{n}:z_{n})\rightarrow(1:0:\pm
i)$. Since $y_{n}(z_{n}+A_{n})+x_{n}\rightarrow q$ we must have
$y_{n}\rightarrow\pm i$. So

$Mm(P_{n})_{\mathbb{C}}\approx\left[\begin{matrix}2x_{n}&2&\pm 2ix_{n}&0\\\
1&\pm ix_{n}&\pm i&\pm i\\\ 0&0&0&1\\\ 2\overline{x_{n}}^{2}&-2\overline{(A\pm
iq)}&\pm 2\overline{ix_{n}^{2}}&0\end{matrix}\right].$

Observe that $\det Mm(P_{n})_{\mathbb{C}}\approx 8|x_{n}|^{4}$, moreover, the
norm of every $3\times 3$ minor of $Mm(P_{n})_{\mathbb{C}}$ is in
$O(|x_{n}|^{4})$. It follows that the norm of every entry of
$Mm(P_{n})^{-1}_{\mathbb{C}}$, and consequently of every entry of
$Mm(P_{n})^{-1}_{\mathbb{R}}$, is in $O(1)$.

Next, consider the case $(x_{n}:y_{n}:z_{n})\rightarrow(0:1:0)$. Since
$y_{n}(z_{n}+A_{n})+x_{n}\rightarrow q$ we must have $z_{n}\rightarrow-A$ and
since $x_{n}^{2}+z_{n}^{2}+2y_{n}\rightarrow p$ we must have
$x_{n}\approx\pm\sqrt{-2y_{n}}$. So

$Mm(P_{n})_{\mathbb{C}}\approx\left[\begin{matrix}\pm
2\sqrt{-2y_{n}}&2&-2A&0\\\ 1&o(1)&y_{n}&y_{n}\\\ 0&0&0&1\\\
2\overline{y_{n}}&\mp
2\overline{\sqrt{-2y_{n}}y_{n}}&o(\sqrt{|y_{n}|})&0\end{matrix}\right].$

Observe that $\det Mm(P_{n})_{\mathbb{C}}\approx-8|y_{n}^{3}|$, moreover, the
norm of every $3\times 3$ minor of $Mm(P_{n})_{\mathbb{C}}$ is in
$O(|y_{n}|^{3})$. It follows that the norm of every entry of
$Mm(P_{n})^{-1}_{\mathbb{C}}$, and consequently of every entry of
$Mm(P_{n})^{-1}_{\mathbb{R}}$, is in $O(1)$. Since $||P_{n}||\approx|y_{n}|$
we obtain the desired inequality. ∎

Observe that for $A\neq 0$ we have

$\left(\frac{1}{A^{2}}p,\frac{1}{A}q\right)\circ
F_{A}\circ(Ax,y,Az)=(x^{2}+z^{2}+2A^{-2}y,yz+x+y).$

Moreover,

$(x^{2}+z^{2}+2A^{-2}y,yz+x+y)=(p,-q)\circ(x^{2}+z^{2}-2A^{-2}y,yz+x+y)\circ(-x,-y,z).$

Thus, if $A^{4}=A_{1}^{4}$ then $F_{A}$ is linearly equivalent to $F_{A_{1}}$.
Consequently, if $A^{4}=-16$ then $F_{A}$ is linearly equivalent to

(5.7) $F_{9}=(x^{2}+z^{2}+2y,yz+x+\sqrt{2}(1+i)y).$

We return to equation (5.2) and consider the case $a_{8}b_{7}=0$.

For $b_{7}=0$ and $a_{8}b_{8}\neq 0$ we obtain

$F_{10}=(x^{2}+z^{2}+y,yz+y)=(b_{8}^{-2}p,a_{8}b_{8}^{-3}q)\circ(x^{2}+z^{2}+a_{8}y,yz+b_{8}y)\circ(b_{8}x,a_{8}^{-1}b_{8}^{2}y,b_{8}z).$

For $b_{7}=0$ and $a_{8}b_{8}=0$ we obtain

$F_{12}=(x^{2}+z^{2},yz+y),\quad F_{13}=(x^{2}+z^{2}+y,yz),\quad
F_{15}=(x^{2}+z^{2},yz).$

We are left with $b_{7}\neq 0$ and $a_{8}=0$. Depending on $b_{8}$ being zero
or not we obtain:

$F_{11}=(x^{2}+z^{2},yz+x+y),\quad F_{14}=(x^{2}+z^{2},yz+x).$

##  6\. Other behavior at infinity

In this section we consider mappings $F$ with nongeneric behavior at infinity.
We enumerate the linear and topological equivalence classes of mappings $F$
such that $\tilde{F}\neq\tilde{F}_{1}$ and $\tilde{F}\neq\tilde{F}_{2}$.

First, take $F$ such that $\tilde{F}=\tilde{F}_{3}$. We have

(6.1)
$F=(x^{2}+y^{2}+a_{7}x+a_{8}y+a_{9}z+a_{10},z^{2}+b_{7}x+b_{8}y+b_{9}z+b_{10}).$

By composing with translations we can assume that
$a_{7}=a_{8}=b_{9}=a_{10}=b_{10}=0$ and obtain

(6.2) $F=(x^{2}+y^{2}+a_{9}z,z^{2}+b_{7}x+b_{8}y).$

Now we have three cases regarding the linear form $b_{7}x+b_{8}y$: it is
nonzero and does not divide $x^{2}+y^{2}$, it does divide $x^{2}+y^{2}$ or it
is zero. After linearly changing the $x$ and $y$ coordinates we obtain one of
the following forms:

(6.3)
$(x^{2}+y^{2}+a_{9}z,z^{2}+x),\quad(xy+a_{9}z,z^{2}+x),\quad(x^{2}+y^{2}+a_{9}z,z^{2}).$

If $a_{9}\neq 0$ then by composing with
$(a_{9}^{2/3}x,a_{9}^{2/3}y,a_{9}^{1/3}z)$ and $(a_{9}^{-4/3}p,a_{9}^{-2/3}q)$
we obtain $a_{9}=1$. Thus we obtain the following corollary:

######  Corollary 6.1.

Let $F$ such that $\tilde{F}=\tilde{F}_{3}$. Then $F$ is linearly equivalent
to one of the following:

  1. (1)

$F_{16}=(x^{2}+y^{2}+z,z^{2}+x)$,

  2. (2)

$F_{17}=(x^{2}+y^{2},z^{2}+x)$,

  3. (3)

$F_{18}=(xy+z,z^{2}+x)$,

  4. (4)

$F_{19}=(xy,z^{2}+x)$,

  5. (5)

$F_{20}=(x^{2}+y^{2}+z,z^{2})$,

  6. (6)

$F_{21}=(x^{2}+y^{2},z^{2})$.

Next, take $F$ such that $\tilde{F}=\tilde{F}_{4}$. We have

(6.4)
$F=(x^{2}+2yz+a_{7}x+a_{8}y+a_{9}z+a_{10},y^{2}+2xy+b_{7}x+b_{8}y+b_{9}z+b_{10}).$

By composing with translations we can assume that
$a_{7}=a_{8}=a_{9}=a_{10}=b_{10}=0$ and obtain

(6.5) $F=(x^{2}+2yz,y^{2}+2xy+2b_{7}x+2b_{8}y+2b_{9}z).$

Now assume that $b_{9}\neq 0$. Composing with $(b_{9}x,b_{9}y,b_{9}z)$ and
$(b_{9}^{-2}p,b_{9}^{-2}q)$ we obtain

(6.6) $F_{A,B}=(x^{2}+2yz,y^{2}+2xy+2Ax+2By+2z).$

Note that $C(F_{A,B})=V(x-y^{2}-Ay,z-xy-y^{2}-By)$, it is a smooth curve of
degree $3$ that can be easily parametrized by

(6.7) $\varphi:\mathbb{C}\ni
t\mapsto\left(t^{2}+At,t,t^{3}+(A+1)t^{2}+Bt\right)\in C(F_{A,B}).$

The critical set of of $F_{A,B}\circ\varphi$ is equal to the set of zeroes of
the derivative of $g\circ\varphi$, which is equal to the set of zeroes of the
polynomial

(6.8) $H_{c2}=6t^{2}+(6A+3)t+(A^{2}+2B).$

Observe that $H_{c2}$ has two distinct roots if and only if
$4A^{2}+12A-16B+3\neq 0$. So let us assume that $H_{c2}$ has two distinct
roots. We denote the discriminant and one of the roots of $H_{c2}$ by
$T=(4/3\cdot A^{2}+4A-16/3\cdot B+1)^{1/2}$ and $S=1/4\cdot(-2A-1+T)$. Take

$R_{1}=T^{2}x+\frac{T^{2}-T}{2}y+S^{2}+AS,$ $R_{2}=Ty+S,$
$R_{3}=-(S+A)T^{2}x-\frac{1}{8}T(2AT-T+1)y+T^{3}z+\frac{1}{8}S(2A-T^{2}+T).$

We have

(6.9)
$F_{22}=F_{0,0}=\left(\frac{p-Sq}{T^{4}}-a_{10},\frac{q}{T^{3}}-b_{10}\right)\circ
F_{A,B}\circ(R_{1},R_{2},R_{3}).$

The constants $a_{10}$ and $b_{10}$ in equation (6.9) can be computed in an
elementary way. We do not include them because they are complicated and
provide no insight.

Now assume that $H_{c2}$ has a double root, i.e., $16B=4A^{2}+12A+3$. Take

$R_{4}=-\frac{A}{4}(2x+y)+z-\frac{1}{32}(4A^{2}+3A).$

We have

(6.10) $F_{23}=F_{0,3/16}=\left(p+\frac{A}{2}q-a_{10},q-b_{10}\right)\circ
F_{A,B}\circ\left(x-\frac{A^{2}}{4},y-\frac{A}{2},R_{4}\right)$

where $a_{10}=-1/32\cdot(6A^{4}+8A^{3}+3A^{2})$ and
$b_{10}=-1/8\cdot(4A^{3}+6A^{2}+3A)$.

Now we return to equation (6.5) and assume that $b_{9}=0$. If $b_{7}\neq 0$
then we compose $F$ with
$(b_{t}^{-2}p+b_{7}^{-3}b_{8}q+b_{7}^{-2}b_{8}^{2},b_{7}^{-2}q+2b_{7}^{-1}b_{8})$
and $(b_{7}x-b_{8},b_{7}y,-b_{8}/2\cdot(2x+y)+b_{7}z)$ and obtain

(6.11) $F_{24}=(x^{2}+2yz,y^{2}+2xy+2x).$

If $b_{7}=0$ and $b_{8}\neq 0$ then we compose $F$ with
$(b_{8}^{-2}p,b_{8}^{-2}q)$ and $(b_{8}x,b_{8}y,b_{8}z)$ and obtain

(6.12) $F_{25}=(x^{2}+2yz,y^{2}+2xy+2y).$

Finally, if both $b_{7}=0$ and $b_{8}=0$ then we have

(6.13) $F_{26}=(x^{2}+2yz,y^{2}+2xy).$

We gather all the cases that occur in equation (6.5) in the following
corollary:

######  Corollary 6.2.

Let $F$ such that $\tilde{F}=\tilde{F}_{4}$. Then $F$ is linearly equivalent
to one of the following:

  1. (1)

$F_{22}=(x^{2}+2yz,y^{2}+2xy+2z)$,

  2. (2)

$F_{23}=(x^{2}+2yz,y^{2}+2xy+3/8\cdot y+2z)$,

  3. (3)

$F_{24}=(x^{2}+2yz,y^{2}+2xy+2x)$,

  4. (4)

$F_{25}=(x^{2}+2yz,y^{2}+2xy+2y)$,

  5. (5)

$F_{26}=(x^{2}+2yz,y^{2}+2xy)$.

Next, take $F$ such that $\tilde{F}=\tilde{F}_{5}$. We have

(6.14)
$F=(x^{2}+2yz+a_{7}x+a_{8}y+a_{9}z+a_{10},z^{2}+b_{7}x+b_{8}y+b_{9}z+b_{10}).$

By composing with translations we can assume that
$a_{7}=a_{8}=a_{9}=a_{10}=b_{10}=0$ and obtain

(6.15) $F=(x^{2}+2yz,z^{2}+2b_{7}x+2b_{8}y+2b_{9}z).$

Now assume that $b_{8}\neq 0$. Compose $F$ with $(b_{8}^{-2}p,b_{8}^{-2}q)$
and $(b_{8}x,b_{8}y,b_{8}z)$ to obtain
$F=(x^{2}+2yz,z^{2}+2b_{7}x+2y+2b_{9}z)$. Denote
$T=1/3\cdot(b_{7}^{2}+2b_{9})$ and compose $F$ with $(p+Tq+2T^{3},q+3T^{2})$
and $(x+b_{7}z-b_{7}T,-b_{7}x+y-(2T-b_{9})z+T(T-b_{9}),z-T)$ to obtain

(6.16) $F_{27}=(x^{2}+2yz,z^{2}+2y).$

Now return to equation (6.15) and assume that $b_{8}=0$. If $b_{7}\neq 0$ then
compose $F$ with $(b_{7}^{-2}p,b_{7}^{-2}q)$ and
$(b_{7}x-b_{9}z,b_{9}x+b_{7}y-1/2\cdot b_{7}^{-1}b_{9}^{2}z,b_{7}z)$ to obtain

(6.17) $F_{28}=(x^{2}+2yz,z^{2}+2x).$

Now assume that $b_{8}=b_{7}=0$ in equation (6.15). If $b_{9}\neq 0$ then
composing $F$ with $(b_{9}^{-2}p,b_{9}^{-2}q)$ and $(b_{9}x,b_{9}y,b_{9}z)$ we
obtain

(6.18) $F_{29}=(x^{2}+2yz,z^{2}+2z).$

Otherwise we have

(6.19) $F_{30}=(x^{2}+2yz,z^{2}).$

We gather all the cases that occur in equation (6.15) in the following
corollary:

######  Corollary 6.3.

Let $F$ such that $\tilde{F}=\tilde{F}_{5}$. Then $F$ is linearly equivalent
to one of the following:

  1. (1)

$F_{27}=(x^{2}+2yz,z^{2}+2y)$,

  2. (2)

$F_{28}=(x^{2}+2yz,z^{2}+2x)$,

  3. (3)

$F_{29}=(x^{2}+2yz,z^{2}+2z)$,

  4. (4)

$F_{30}=(x^{2}+2yz,z^{2})$.

Next, take $F$ such that $\tilde{F}=\tilde{F}_{6}$. We have

(6.20)
$F=(x^{2}+a_{7}x+a_{8}y+a_{9}z+a_{10},y^{2}+b_{7}x+b_{8}y+b_{9}z+b_{10}).$

First assume that $a_{9}b_{9}\neq 0$. Composing $F$ with
$(x,y,-b_{7}b_{9}^{-1}x-a_{8}a_{9}^{-1}y+z)$ we obtain
$F=(x^{2}+a_{7}x+a_{9}z+a_{10},y^{2}+b_{8}y+b_{9}z+b_{10})$. Next, we compose
$F$ with $(x-a_{7}/2,y-b_{8}/2,z)$ to obtain
$F=(x^{2}+a_{9}z+a_{10},y^{2}+b_{9}z+b_{10})$. And finally we compose with
$(a_{9}^{-1}(p-a_{10}),b_{9}^{-1}(q-b_{10}))$ and
$(a_{9}^{1/2}x,b_{9}^{1/2}y,2z)$ to obtain

(6.21) $F_{31}=(x^{2}+2z,y^{2}+2z).$

Now, assume that in equation (6.20) exactly one of $a_{9}$ and $b_{9}$ is
nonzero. By symmetry we may assume $a_{9}\neq 0$ and $b_{9}=0$. Composing with
$(x,y-b_{8}/2,a_{9}^{-1}(-a_{7}x-a_{8}y+2z))$ we obtain
$F=(x^{2}+2z+a_{10},y^{2}+b_{7}x+b_{10})$. Obviously, depending on whether
$b_{7}\neq 0$ or $b_{7}=0$ we obtain

(6.22) $F_{32}=(x^{2}+2z,y^{2}+2x)$

or

(6.23) $F_{33}=(x^{2}+2z,y^{2}).$

The last case to consider in equation (6.20) is $a_{9}=b_{9}=0$. In this case
$F$ does not depend on the $z$ variable and we may use the list in [5] to
obtain that $F$ is affinely equivalent to one of the following:

(6.24) $F_{34}=(x^{2}+2y,y^{2}+2x),\quad F_{35}=(x^{2}+2y,y^{2}),\quad
F_{36}=(x^{2},y^{2}).$

We gather all the cases that occur in equation (6.20) in the following
corollary:

######  Corollary 6.4.

Let $F$ such that $\tilde{F}=\tilde{F}_{6}$. Then $F$ is linearly equivalent
to one of the following:

  1. (1)

$F_{31}=(x^{2}+2z,y^{2}+2z)$,

  2. (2)

$F_{32}=(x^{2}+2z,y^{2}+2x)$,

  3. (3)

$F_{33}=(x^{2}+2z,y^{2})$,

  4. (4)

$F_{34}=(x^{2}+2y,y^{2}+2x)$,

  5. (5)

$F_{35}=(x^{2}+2y,y^{2})$,

  6. (6)

$F_{36}=(x^{2},y^{2})$.

Next, take $F$ such that $\tilde{F}=\tilde{F}_{7}$. We have

(6.25) $F=(xy+a_{7}x+a_{8}y+a_{9}z+a_{10},yz+b_{7}x+b_{8}y+b_{9}z+b_{10}).$

By composing with translations we can assume that
$a_{7}=a_{8}=b_{8}=a_{10}=b_{10}=0$ and obtain

(6.26) $F=(xy+a_{9}z,yz+b_{7}x+b_{9}z).$

First assume that $a_{9}b_{7}\neq 0$. By composing with
$(a_{9}^{-1}b_{7}^{-1/2}p,a_{9}^{-1/2}b_{7}^{-1}q)$ and
$(\sqrt{a_{9}}x,\sqrt{a_{9}b_{7}}y,\sqrt{b_{7}}z)$ we obtain
$F=(xy+z,yz+x+b_{9}z)$. Observe that $C(F)=V(y^{2}+b_{9}y-1,x-yz)$. If
$b_{9}^{2}+4\neq 0$ then $C(F)$ consists of two lines intersecting at
infinity. If $b_{9}^{2}+4=0$ then $C(F)$ is a double line. Let us assume that
$b_{9}^{2}+4\neq 0$, i.e., $y^{2}+b_{9}y-1=0$ has two distinct roots. We
denote the discriminant and roots by $T=\sqrt{b_{9}^{2}+4}$,
$S_{1}=(-b_{9}+T)/2$ and $S_{2}=(-b_{9}+T)/2$. Let

(6.27) $F_{37}=(xy,yz+z).$

We have

(6.28) $F_{37}=(S_{2}p+q,S1p+q)\circ
F\circ((-x+z)T^{-2},Ty+S_{1},(S_{1}x-S_{2}z)T^{-2}).$

Now consider $b_{9}^{2}+4=0$. If $b_{9}=-2i$ then we compose $F$ with $(p,-q)$
and $(-x,-y,z)$ and obtain $b_{9}=2i$, i.e., $F=(xy+z,yz+x+2iz)$. We compose
$F$ with $(p,-ip+q)$ and $(x,y-i,ix+z)$ and obtain

(6.29) $F_{38}=(xy+z,yz).$

Now return to equation (6.26) and assume $a_{9}=0$ and $b_{7}\neq 0$. If
$b_{9}=0$, i.e., $F=(xy,yz+b_{7}x)$, then we compose $F$ with
$(-b_{7}p+q,b_{7}p)$ and $(b_{7}^{-1}z,y,x+z)$ and obtain $F_{38}$. If
$b_{9}\neq 0$, i.e., $F=(xy,yz+b_{7}x+b_{9}z)$ with $b_{7}b_{9}\neq 0$, then
we compose $F$ with $(p,b_{7}p+b_{9}q)$ and
$(b_{9}^{-1}x,b_{9}y,b_{9}^{-2}(-b_{7}x+z))$ and obtain $F_{37}$.

Now return to equation (6.26) and assume $a_{9}\neq 0$ and $b_{7}=0$. If
$b_{9}=0$, i.e., $F=(xy+a_{9}z,yz)$, then we compose $F$ with
$(a_{9}^{-1}p,q)$ and $(a_{9}x,y,z)$ and obtain $F_{38}$. If $b_{9}\neq 0$,
i.e., $F=(xy+a_{9}z,yz+b_{9}z)$ with $a_{9}b_{9}\neq 0$, then we compose $F$
with $(q,-b_{9}p+a_{9}q)$ and $(b_{9}^{-2}(-a_{9}x+z),-b_{9}y,-b_{9}^{-1}x)$
and obtain $F_{37}$.

Finally, we are left with the case $a_{9}=b_{7}=0$ in equation (6.26). If
$b_{9}\neq 0$ then we compose $F$ with $(b_{9}^{-1}p,b_{9}^{-1}q)$ and
$(x,b_{9}y,z)$ and obtain $F_{37}$. If $b_{9}=0$ then we have

(6.30) $F_{39}=(xy,yz).$

We gather all the cases that occur in equation (6.26) in the following
corollary:

######  Corollary 6.5.

Let $F$ such that $\tilde{F}=\tilde{F}_{7}$. Then $F$ is linearly equivalent
to one of the following:

  1. (1)

$F_{37}=(xy,yz+z)$,

  2. (2)

$F_{38}=(xy+z,yz)$,

  3. (3)

$F_{39}=(xy,yz)$.

Next, take $F$ such that $\tilde{F}=\tilde{F}_{8}$. We have

(6.31) $F=(xy+a_{7}x+a_{8}y+a_{9}z+a_{10},y^{2}+b_{7}x+b_{8}y+b_{9}z+b_{10}).$

First assume that $b_{9}\neq 0$. Composing $F$ with
$(x-a_{8},y,b_{9}^{-1}(-b_{7}x-b_{8}y+z))$ we obtain
$F=(xy+a_{7}x+a_{9}z+a_{10},y^{2}+z+b_{10})$. Next, we compose $F$ with
$(p-a_{9}(q-a_{7}^{2}-b_{10})-a_{10},q-a_{7}^{2}-b_{10})$ and
$(x+a_{9}y-2a_{7}a_{9},y-a_{7},2a_{7}y+2z)$ to obtain:

(6.32) $F_{40}=(xy,y^{2}+2z).$

Now return to equation (6.31) and assume $b_{9}=0$ and $a_{9}\neq 0$. We
compose $F$ with $(x,y-b_{8}/2,a_{9}^{-1}((-a_{7}+b_{8}/2)x-a_{8}y+z))$ and
obtain $F=(xy+z+a_{10},y^{2}+b_{7}x+b_{10})$. If $b_{7}\neq 0$ then by
composing with $(b_{7}(p-a_{10}),q-b_{10})$ and $(b_{7}^{-1}x,y,b_{7}^{-1}z)$
we obtain

(6.33) $F_{41}=(xy+z,y^{2}+x).$

If $b_{7}=0$ then we obtain

(6.34) $F_{42}=(xy+z,y^{2}).$

Finally, if $b_{9}=a_{9}=0$ in equation (6.31) then $F$ does not depend on the
$z$ variable and we may use the list in [5] to obtain that $F$ is affinely
equivalent to one of the following:

(6.35) $F_{43}=(xy,y^{2}+2x),\quad F_{44}=(xy,y^{2}+2y),\quad
F_{45}=(xy,y^{2}).$

We gather all the cases that occur in equation (6.31) in the following
corollary:

######  Corollary 6.6.

Let $F$ such that $\tilde{F}=\tilde{F}_{8}$. Then $F$ is linearly equivalent
to one of the following:

  1. (1)

$F_{40}=(xy,y^{2}+2z)$,

  2. (2)

$F_{41}=(xy+z,y^{2}+x)$,

  3. (3)

$F_{42}=(xy+z,y^{2})$,

  4. (4)

$F_{43}=(xy,y^{2}+2x)$,

  5. (5)

$F_{44}=(xy,y^{2}+2y)$,

  6. (6)

$F_{45}=(xy,y^{2})$.

This concludes the classification of all $F$ which are not linearly equivalent
to a mapping with a component of degree lower than $2$. We continue the
classification with $F$ such that $\tilde{F}=\tilde{F}_{9}$. We have

(6.36) $F=(x^{2}+yz+a_{7}x+a_{8}y+a_{9}z+a_{10},x+b_{10}).$

By composing $F$ with translations we can assume that
$a_{7}=a_{8}=a_{9}=a_{10}=b_{10}=0$ and obtain

(6.37) $F_{46}=(x^{2}+yz,x).$

Similarly, for $\tilde{F}=\tilde{F}_{10}$ and $\tilde{F}=\tilde{F}_{11}$ we
obtain

(6.38) $F_{47}=(x^{2}+yz,y),\quad F_{48}=(x^{2}+y^{2}+z^{2},0).$

Next, take $F$ such that $\tilde{F}=\tilde{F}_{12}$. We have

(6.39) $F=(x^{2}+y^{2}+a_{7}x+a_{8}y+a_{9}z+a_{10},z+b_{10}).$

We compose $F$ with $(x-a_{7}/2,y-a_{8}/2,z-b_{10})$ an obtain
$F=(x^{2}+y^{2}+a_{9}z+a_{10},z)$. Next we compose $F$ with
$(p-a_{9}q-a_{10},q)$ and obtain

(6.40) $F_{49}=(x^{2}+y^{2},z).$

For $F$ such that $\tilde{F}=\tilde{F}_{13}$ we have

(6.41) $F=(x^{2}+y^{2}+a_{7}x+a_{8}y+a_{9}z+a_{10},x+b_{10}).$

By composing $F$ with translations we can assume that
$a_{7}=a_{8}=a_{10}=b_{10}=0$ and obtain $F=(x^{2}+y^{2}+a_{9}z,x)$. Depending
on whether $a_{9}=0$ we obtain one of the following two:

(6.42) $F_{50}=(x^{2}+y^{2}+z,x),\quad F_{51}=(x^{2}+y^{2},x).$

Similarly, for $\tilde{F}=\tilde{F}_{14}$ we obtain

(6.43) $F_{52}=(xy+z,x),\quad F_{53}=(xy,x)$

and for $\tilde{F}=\tilde{F}_{15}$ we obtain

(6.44) $F_{54}=(x^{2}+y^{2}+z,0),\quad F_{55}=(x^{2}+y^{2},0).$

For $F$ such that $\tilde{F}=\tilde{F}_{16}$ we have

(6.45) $F=(x^{2}+a_{7}x+a_{8}y+a_{9}z+a_{10},y+b_{10}).$

By composing $F$ with translations we can assume that $a_{7}=a_{10}=b_{10}=0$
and obtain $F=(x^{2}+a_{8}y+a_{9}z,y)$. Composing with $(p-a_{8}q,q)$ we
obtain $a_{8}=0$ and depending on whether $a_{9}=0$ we obtain one of the
following two:

(6.46) $F_{56}=(x^{2}+z,y),\quad F_{57}=(x^{2},y).$

For $F$ such that $\tilde{F}=\tilde{F}_{17}$ we have

(6.47) $F=(x^{2}+a_{7}x+a_{8}y+a_{9}z+a_{10},x+b_{10}).$

By composing $F$ with translations we can assume that $a_{7}=a_{10}=b_{10}=0$
and obtain $F=(x^{2}+a_{8}y+a_{9}z,x)$. Assume that $a_{8}\neq 0$ or
$a_{9}\neq 0$, by symmetry we can take $a_{8}\neq 0$. Composing with
$(x,a_{8}^{-1}(y-a_{9}z),z)$ we obtain

(6.48) $F_{58}=(x^{2}+y,x).$

On the other hand, for $a_{8}=a_{9}=0$ we have

(6.49) $F_{59}=(x^{2},x).$

Similarly, for $\tilde{F}=\tilde{F}_{18}$ we obtain

(6.50) $F_{60}=(x^{2}+y,0),\quad F_{61}=(x^{2},0).$

Finally for $F$ equal to $\tilde{F}_{19}$, $\tilde{F}_{20}$ or
$\tilde{F}_{21}$ we must have $F=(\tilde{f}+a_{10},\tilde{g}+b_{10})$, so
after composing with a translation we obtain

(6.51) $F_{62}=(x,y),\quad F_{63}=(x,0),\quad F_{64}=(0,0).$

Acknowledgments. I would like to thank professor Zbigniew Jelonek and
professor Maria Aparecida Soares Ruas for many helpful discussions.

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