# Equivalence of Milnor and Milnor-Lê fibrations for real analytic maps

José Luis Cisneros-Molina  Instituto de Matemáticas, Unidad Cuernavaca
Universidad Nacional Autónoma de México
Avenida Universidad s/n, Colonia Lomas de Chamilpa
Cuernavaca, Morelos, Mexico<EMAIL_ADDRESS>and  Aurélio Menegon
Universidade Federal da Paraíba
Departamento de Matemática
CEP 58051-900, João Pessoa - PB, Brazil<EMAIL_ADDRESS>Man approaches
the unattainable truth through a succession of errors.
Aldous Huxley

###### Abstract.

In [22] Milnor proved that a real analytic map
$f\colon(\mathbbm{R}^{n},0)\to(\mathbbm{R}^{p},0)$, where $n\geq p$, with an
isolated critical point at the origin has a fibration on the tube
$f|\colon\mathbbm{B}_{\varepsilon}^{n}\cap
f^{-1}(\mathbbm{S}_{\delta}^{p-1})\to\mathbbm{S}_{\delta}^{p-1}$. Constructing
a vector field such that, (1) it is transverse to the spheres, and (2) it is
transverse to the tubes, he “inflates” the tube to the sphere, to get a
fibration $\varphi\colon\mathbbm{S}_{\varepsilon}^{n-1}\setminus
f^{-1}(0)\to\mathbbm{S}^{p-1}$, but the projection is not necessarily given by
$f/\|f\|$ as in the complex case.

In the case $f$ has isolated critical value, in [9] it was proved that if the
fibres inside a small tube are transverse to the sphere
$\mathbbm{S}_{\varepsilon}$, then it has a fibration on the tube. Also in [9],
the concept of $d$-regularity was defined, it turns out that $f$ is
$d$-regular if and only if the map
$f/\|f\|\colon\mathbbm{S}_{\varepsilon}^{n-1}\setminus
f^{-1}(0)\to\mathbbm{S}^{p-1}$ is a fibre bundle equivalent to the one on the
tube.

In this article, we prove the corresponding facts in a more general setting:
if a locally surjective map $f$ has a linear discriminant $\Delta$ and a
fibration on the tube $f|\colon\mathbbm{B}_{\varepsilon}^{n}\cap
f^{-1}(\mathbbm{S}_{\delta}^{p-1}\setminus\Delta)\to\mathbbm{S}_{\delta}^{p-1}\setminus\Delta$,
then $f$ is $d$-regular if and only if the map
$f/\|f\|\colon\mathbbm{S}_{\varepsilon}^{n-1}\setminus
f^{-1}(\Delta)\to\mathbbm{S}^{p-1}\setminus\mathcal{A}$ (with $\mathcal{A}$
the radial projection of $\Delta$ on $\mathbbm{S}^{p-1}$) is a fibre bundle
equivalent to the one on the tube. We do this by constructing a vector field
$\tilde{w}$ which inflates the tube to the sphere in a controlled way, it
satisfies properties analogous to the vector field constructed by Milnor in
the complex setting: besides satisfying (1) and (2) above, it also satisfies
that $f/\|f\|$ is constant on the integral curves of $\tilde{w}$.

This is a corrected version of the article published in Internat. J. Math.,
30(14):1950078, 1-25, 2019, where in the proof of Theorem 3.7 (Theorem 3.8
here) two inequalities were used that do not hold in general. Such
inequalities are not essential for the proof and Theorem 3.7 can be proved
without them.

###### Key words and phrases:

Milnor Fibration, Milnor-Lê Fibration, $d$-regularity, equivalence of
fibrations, spherefication

###### Key words and phrases:

Milnor Fibration, Milnor-Lê Fibration, $d$-regularity, equivalence of
fibrations, spherefication

###### 2010 Mathematics Subject Classification:

Primary 32S55, 58K05, 58K15

The first author is Regular Associate of the Abdus Salam International Centre
for Theoretical Physics, Trieste, Italy. Supported by CONACYT 253506.

## Preface

This is a corrected version of the article [7] published in Internat. J.
Math., 30(14):1950078, 1-25, 2019, where in the proof of Theorem 3.7 (Theorem
3.8 here) two inequalities were used that do not hold in general. Such
inequalities are not essential for the proof and Theorem 3.7 can be proved
without them. Here we give such proof, which just needs the complementary
Proposition 3.6. This proposition also simplifies the proof of Lemma 3.7 since
it implies that Case 2.B with $\mu(x)<0$ cannot occur, so the the lengthy
construction for that case given in the original article is not necessary.
Hence, all the results in the original article [7] are valid.

We also take the opportunity to add Proposition 2.3, Corollary 2.4 and
Corollary 2.5, which give a formula for the differential of the spherification
map and some results about it. They are not neccesary for the results of the
article but can be useful and help to describe in a more precise way the
vector fields constructed (see Remark 3.10). An errata has also being sent to
the journal.

##  1\. Introduction

Milnor Fibration Theorem is an important result in singularity theory. It is
about the topology of the fibres of holomorphic functions near its critical
points. To each singular point of a complex hypersurface it associates a fibre
bundle, known as the Milnor Fibration. For an overview of its origin,
generalizations and connections with other branches of mathematics we
recommend the recent survey article by Seade [30].

Let $f\colon(\mathbbm{C}^{n},\underline{0})\to(\mathbbm{C},0)$ be a
holomorphic map with a critical point at the origin
$\underline{0}\in\mathbbm{C}^{n}$. The Milnor Fibration [22, Theorem 4.8] is
given by

$\phi:=\frac{f}{|f|}:\mathbbm{S}_{\varepsilon}\setminus
K\longrightarrow\mathbbm{S}^{1}\,,$ (1.1)

where $K$ is the link of $f$ at $\underline{0}$, that is
$K=f^{-1}(0)\cap\mathbbm{S}_{\varepsilon}$ with $\mathbbm{S}_{\varepsilon}$
being a sufficiently small sphere around $\underline{0}$.

There is a second fibre bundle called the Milnor-Lê Fibration; this is a fibre
bundle on a Milnor tube $N(\varepsilon,\delta)=\mathbbm{B}_{\varepsilon}\cap
f^{-1}(\partial\mathbbm{D}_{\delta})$ where $\mathbbm{B}_{\varepsilon}$ is the
ball bounded by $\mathbbm{S}_{\varepsilon}$, $\partial\mathbbm{D}_{\delta}$ is
the circle bounding the disc $\mathbbm{D}_{\delta}$ of radius $\delta$ centred
at the origin of $\mathbbm{C}$. When $0<\delta\ll\varepsilon$ the map $f$
induces a fibration

$f\colon N(\varepsilon,\delta)\rightarrow\partial\mathbbm{D}_{\delta}.$ (1.2)

Its existence was proved by Milnor [22, Theorem 11.2] for the case of isolated
singularity and by Lê in [16, Theorem (1.1)] for the general case, both using
Ehresmann Fibration Theorem for manifolds with boundary.

Fibrations (1.1) and (1.2) are equivalent, this is proved using a smooth
vector field on $\mathbbm{B}_{\varepsilon}\setminus f^{-1}(0)$ constructed by
Milnor [22, Lemma 5.9] for an arbitrary holomorphic map
$f\colon(\mathbbm{C}^{n},\underline{0})\to(\mathbbm{C},0)$ with the following
properties:

  1. (1)

It is transverse to the spheres centred at the origin contained in
$\mathbbm{B}_{\varepsilon}$,

  2. (2)

It is transverse to the Milnor tubes,

  3. (3)

Given an integral curve $p(t)$ of such vector field $\frac{f(p(t))}{\lVert
f(p(t))\rVert}$ is constant.

With this vector field the Milnor tube can be “inflated” to the sphere, i. e.,
following the flow of the vector field one gets a diffeomorphism between the
Milnor tube and the complement on the sphere of a neighbourhood of the link.

Milnor also proved a Fibration Theorem for real singularities [21, Theorem 2]
or [22, Theorem 11.2]. Let $n\geq p$ and consider a real analytic map
$f\colon(\mathbbm{R}^{n},\underline{0})\to(\mathbbm{R}^{p},0)$ with an
isolated critical point at $\underline{0}$. Let
$\mathbbm{B}_{\varepsilon}^{n}$ be a ball of radius $\varepsilon>0$
sufficiently small and, as before, let $0<\delta\ll\varepsilon$. First, Milnor
proved that there is a fibration on the Milnor tube
$N(\varepsilon,\delta)=\mathbbm{B}_{\varepsilon}^{n}\cap
f^{-1}(\mathbbm{S}_{\delta}^{p-1})$

$f\colon N(\varepsilon,\delta)\rightarrow\mathbbm{S}_{\delta}^{p-1},$ (1.3)

where $\mathbbm{S}_{\delta}^{p-1}$ is the sphere of radius $\delta$ centred at
the origin of $\mathbbm{R}^{p}$. Then he proved that on the sphere
$\mathbbm{S}_{\varepsilon}^{n-1}=\partial\mathbbm{B}_{\varepsilon}^{n}$ one
has a fibre bundle

$\varphi\colon\mathbbm{S}_{\varepsilon}^{n-1}\setminus K\to\mathbbm{S}^{p-1}$
(1.4)

where $K=f^{-1}(0)\cap\mathbbm{S}_{\varepsilon}^{n-1}$ is the link. This was
done constructing a vector field on
$\mathbbm{B}_{\varepsilon}^{n}\setminus\\{0\\}$ such that:

  1. (1)

It is transverse to the spheres centred at the origin contained in
$\mathbbm{B}_{\varepsilon}^{n}$,

  2. (2)

It is transverse to the Milnor tubes,

and “inflating” the Milnor tube to the sphere. Milnor pointed out that his
real Fibration Theorem has some weaknesses: the condition that $f$ has an
isolated critical point is very restrictive, and since the vector field to
“inflate” the tube to the sphere does not satisfy property (3) as in the
complex case, the projection map $\varphi$ is not necessarily given by
$\frac{f}{\lVert f\rVert}$. In fact, in [22, page 99] Milnor gives an example
of a map whose fibration on the sphere does not have projection
$\frac{f}{\lVert f\rVert}$.

From these remarks two natural questions arose:

  1. (i)

When does the fibration on the sphere have projection given by
$\frac{f}{\lVert f\rVert}$?

  2. (ii)

Is it possible to relax the condition that $f$ has isolated critical point and
still have fibrations on the tube and on the sphere?

Question (i) for $f$ with isolated critical point was studied by several
authors [13, 14, 29, 27, 26, 1]. In [2] dos Santos considered the following
question

  1. (iii)

If the fibration on the tube and the fibration on the sphere given by
$\frac{f}{\lVert f\rVert}$ both exist, are they equivalent?

where he answered it positively for $p=2$ and $f$ quasi-homogeneous.

For question (ii), the natural generalization was to consider real analytic
maps $f\colon(\mathbbm{R}^{n},\underline{0})\to(\mathbbm{R}^{p},0)$ with an
isolated critical value, as in the case of holomorphic functions. There are
several works in this direction [24, 9, 10, 19, 3]. To have the fibration
(1.3) on the tube in a ball $\mathbbm{B}_{\varepsilon}^{n}$ of radius
$\varepsilon>0$ sufficiently small, it is necessary that $f$ has the
transversality property: there exists a solid Milnor tube
$\widehat{N}(\varepsilon,\delta)=\mathbbm{B}_{\varepsilon}^{n}\cap
f^{-1}(\mathbbm{B}_{\delta}^{p})$, with $0<\delta\ll\varepsilon$, such that
all the fibres in the tube are transverse to $\mathbbm{S}_{\varepsilon}$, to
be able to apply Ehresmann Fibration Theorem for manifolds with boundary.
Examples of maps with the transversality property are maps with the Thom
$a_{f}$ property [9, Proposition 5.1, Remark 5.7]. Having the fibration on the
tube, one can use Milnor’s vector field to inflate the tube to the sphere to
get a fibration on the sphere, but again, not necessarily with projection
$\frac{f}{\lVert f\rVert}$. The reason why the fibration on the sphere of
Milnor’s example [22, page 99] does not have projection $\frac{f}{\lVert
f\rVert}$ is because the map

$\phi:=\frac{f}{\lVert f\rVert}\colon\mathbbm{S}_{\varepsilon}^{n-1}\setminus
K\to\mathbbm{S}^{p-1}$ (1.5)

is not a submersion, which is a necessary condition to be a smooth fibre
bundle. In [9, Definition 2.4] the concept of $d$-regularity was introduced
(see Section 2 below). In [9, Proposition 3.2-(4)] it was proved that the map
$f$ is $d$-regular if and only if the map (1.5) is a submersion, so
$d$-regularity is a necessary condition for (1.5) to be a fibre bundle. On the
other hand, [9, Lemma 5.2] states that $f$ is $d$-regular if and only if there
exists a vector field on $\mathbbm{B}_{\varepsilon}^{n}\setminus f^{-1}(0)$
which satisfies properties (1), (2) and (3) above, which implies that
“inflating” the tube to the sphere we have that (1.5) is a fibre bundle which
is equivalent to the fibre bundle (1.3), so $d$-regularity is a necessary and
sufficient condition for (1.5) to be a fibre bundle. Thus, the existence of
such vector field answers questions (i) and (iii) for
$f\colon(\mathbbm{R}^{n},\underline{0})\to(\mathbbm{R}^{p},0)$ with isolated
critical value, satisfying the transversality property.

However, Brodersen [11] and Hansen [23] pointed out that the argument of the
proof of [9, Lemma 5.2] is not complete, and Hansen in [23] gives sufficient
conditions for the existence of such vector field. Since then, other authors
[25, 5, 28] have also given sufficient conditions.

In this article we prove of the existence of a vector field satisfying
properties (1), (2) and (3) if and only if $f$ is $d$-regular in the more
general setting of real analytic maps with linear discriminant considered in
[8]. Let $f\colon(\mathbbm{B}_{\varepsilon}^{n},0)\to(\mathbbm{R}^{p},0)$ be a
real analytic map with linear discriminant $\Delta$, i. e., $\Delta$ is a
union of line segments (see Section 5 for the precise definition). In order to
simplify notation, suppose that $f$ is locally surjective (see Remark 1.1
below).

Suppose that $f$ satisfies the transversality property, then there exists
$\delta>0$ sufficiently small such that the restriction of $f$ to the tube

$f|\colon\mathbbm{B}_{\varepsilon}^{n}\cap
f^{-1}(\mathbbm{S}_{\delta}^{p-1}\setminus\mathcal{A}_{\delta})\to(\mathbbm{S}_{\delta}^{p-1}\setminus\mathcal{A}_{\delta})$
(1.6)

is a fibre bundle where
$\mathcal{A}_{\delta}:=\Delta\cap\mathbbm{S}^{p-1}_{\delta}$ (see [8, Theorem
2.7]).

In [8, Theorem 3.9] it is also proved that if $f$ is $d$-regular then one has
a fibre bundle

$f/\|f\|\colon(\mathbbm{S}_{\varepsilon}^{n-1}\setminus
f^{-1}(\Delta))\to(\mathbbm{S}^{p-1}\setminus\mathcal{A})$ (1.7)

where $\mathcal{A}$ is the radial projection of $\mathcal{A}_{\delta}$ on
$\mathbbm{S}^{p-1}$.

The aim of the present article is to give a construction of a vector field
with properties (1), (2) and (3) which works in any open set of
$\mathbbm{R}^{n}$ where the maps $f$ and $f/\|f\|$ are submersions. Applying
this construction to maps $f$ with isolated critical value we get a proof of
[9, Lemma 5.2] which follows the idea of the original (incomplete) proof. Also
this construction of the vector field allows us to prove the equivalence of
fibrations (1.6) and (1.7) for real analytic maps with arbitrary linear
discriminant.

######  Remark 1.1.

Throughout this paper, we will assume that $f$ is locally surjective, that is,
the image of $f$ contains an open neighbourhood of the origin in
$\mathbbm{R}^{p}$, and we shall not mention it all the time. Nevertheless, it
is easy to see that in the general case the same results hold if one
intersects the bases of the locally trivial fibrations with their image. This
choice is to avoid a heavy notation.

##  2\. The canonical pencil, $d$-regularity and the spherefication

Let $U$ be an open neighbourhood of $\underline{0}\in\mathbbm{R}^{n}$, $n>p$,
and $f\colon(U,\underline{0})\to(\mathbbm{R}^{p},0)$ a non-constant analytic
map with a critical point at $\underline{0}\in\mathbbm{R}^{n}$ and
$0\in\mathbbm{R}^{p}$ is an isolated critical value. Equip $V=f^{-1}(0)$ with
a Whitney stratification and let $\mathbbm{B}_{\varepsilon}^{n}$ be an open
ball in $\mathbbm{R}^{n}$, centred at $\underline{0}$, of sufficiently small
radius $\varepsilon$, so that every sphere in this ball, centred at
$\underline{0}$, meets transversely every stratum of $V$, if $V$ is not an
isolated point at the origin. Such a ball exists by [22, Corollary 2.9] if $f$
has an isolated critical point at $\underline{0}$, and by the Bertini-Sard
theorem in [31] in general.

We define a family of real analytic spaces as follows. For each
$\ell\in\mathbbm{RP}^{p-1}$, consider the line
$\mathcal{L}_{\ell}\subset\mathbbm{R}^{p}$ passing through the origin
corresponding to $\ell$ and set

$X_{\ell}=\\{x\in U\,|\,f(x)\in\mathcal{L}_{\ell}\\}\,.$

If we let $\mathcal{L}_{\ell}^{\perp}$ be the hyperplane orthogonal to
$\mathcal{L}_{\ell}$, let
$\varpi_{\ell}\colon\mathbbm{R}^{p}\to\mathcal{L}_{\ell}^{\perp}$ be the
orthogonal projection, and we set $h_{\ell}=\varpi_{\ell}\circ f\,$, then
$X_{\ell}$ is the vanishing set of $h_{\ell}$, which is real analytic. Hence
$\\{X_{\ell}\\}$ is a family of real analytic varieties parameterized by
$\mathbbm{RP}^{p-1}$.

Notice that the singular points of $X_{\ell}$ are contained in $V$. Hence each
$X_{\ell}\setminus V$ is either a smooth submanifold of
$U\subset\mathbbm{R}^{n}$ of dimension $n-p+1$ or empty. The union of all the
$X_{\ell}$’s is all of $U$ and one has $V=\cap X_{\ell}=X_{\ell_{1}}\cap
X_{\ell_{2}}$ for each pair $\ell_{1}\neq\ell_{2}$.

######  Remark 2.1.

We notice that each $X_{\ell}$ is naturally the union of three sets: the
points $x\in U$ such that $f(x)=0$, i.e., $x\in V$, and the points $x\in U$
such that $f(x)$ is in one of the two half lines of
$\mathcal{L}_{\ell}\setminus\\{0\\}$. Write this as:

$X_{\ell}=E_{\ell}^{+}\cup V\cup E_{\ell}^{-}\,.$ (2.1)

The family $\\{X_{\ell}\,|\,\ell\in\mathbbm{RP}^{p-1}\\}$ is the canonical
pencil of $f$. Let $\rho$ be a metric in $\mathbbm{R}^{n}$ induced by some
positive definite quadratic form. The map $f$ is said to be $d$-regular at
$\underline{0}$ (with respect to the metric $\rho$) if there exists
$\varepsilon>0$ such that every sphere (for the metric $\rho$) of radius
$\leq\varepsilon$ centred at $\underline{0}$ meets every $X_{\ell}\setminus V$
transversely whenever the intersection is not empty. If the metric is fixed we
just say that $f$ is $d$-regular.

In this article we consider $\rho$ as the Euclidean metric in
$\mathbbm{R}^{n}$ and we only consider $d$-regular maps with respect to this
metric.

###  2.1. The spherefication

Consider the maps $\Phi\colon U\setminus V\to\mathbbm{S}^{p-1}$ and
$\;\mathfrak{F}\colon U\setminus V\to\mathbbm{R}^{p}\setminus\\{0\\}$ defined
by

$\Phi(x)=\frac{f(x)}{\lVert
f(x)\rVert}\quad\hbox{and}\quad\mathfrak{F}(x)=\lVert x\rVert\Phi(x).$ (2.2)

For each $\ell\in\mathbbm{RP}^{p-1}$ one has that the line
$\mathcal{L}_{\ell}$ in $\mathbbm{R}^{p}$ meets the sphere $\mathbbm{S}^{p-1}$
in two antipodal points, say $\pm y_{\ell}$, and one has:

$\Phi^{-1}(y_{\ell})=E_{\ell}^{+}\quad\text{and}\quad\Phi^{-1}(-y_{\ell})=E_{\ell}^{-},$

where $E_{\ell}^{+}$ (resp. $E_{\ell}^{-}$) is the inverse image under $f$ of
one of the two half lines in $\mathcal{L}_{\ell}\setminus\\{0\\}$.

Notice that given
$y\in\mathcal{L}_{\ell}\setminus\\{0\\}\subset\mathbbm{R}^{p}\setminus\\{0\\}$,
the fibre $\mathfrak{F}^{-1}(y)$ is the intersection of $E_{\ell}^{\pm}$ with
the sphere of radius $\lVert y\rVert$ centred at $\underline{0}$. Hence,
$\mathfrak{F}$ carries spheres in $\mathbbm{R}^{n}$ of radius $r>0$ into
spheres in $\mathbbm{R}^{p}$ of the same radius, and each $X_{\ell}$ is also a
union of fibres of $\mathfrak{F}$, just as it is a union of fibres of $f$. The
analytic map $\mathfrak{F}$ is called the spherefication of $f$.

As we mentioned in the Introduction, a map $f$ is $d$-regular if and only if
the map $\frac{f}{\lVert f\rVert}$ is a submersion. This is given by the
following proposition together with other characterizations of $d$-regularity
in terms of the spherefication.

######  Proposition 2.2 ([9, Proposition 3.2]).

Let
$f\colon(\mathbbm{B}_{\varepsilon}^{n},\underline{0})\to(\mathbbm{R}^{p},0)$
be a real analytic map with isolated critical value at the origin. Set
$V=f^{-1}(0)$ and
$K_{\varepsilon^{\prime}}=V\cap\mathbbm{S}_{\varepsilon^{\prime}}^{n-1}$. The
following are equivalent:

  1. (A)

The map $f$ is $d$-regular at $0$.

  2. (B)

For each sphere $\mathbbm{S}_{\varepsilon^{\prime}}^{n-1}$ in
$\mathbbm{R}^{n}$ centred at $\underline{0}$ of radius
${\varepsilon^{\prime}}<\varepsilon$, the restriction map
$\mathfrak{F}_{\varepsilon^{\prime}}\colon\mathbbm{S}_{\varepsilon^{\prime}}^{n-1}\setminus
V\to\mathbbm{S}^{p-1}_{{\varepsilon^{\prime}}}$ of $\mathfrak{F}$ is a
submersion.

  3. (C)

The spherefication map $\mathfrak{F}$ is a submersion at each
$x\in\mathbbm{B}_{\varepsilon}^{n}\setminus V$.

  4. (D)

The map $\phi=\frac{f}{\lVert
f\rVert}\colon\mathbbm{S}_{{\varepsilon^{\prime}}}^{n-1}\setminus
K_{{\varepsilon^{\prime}}}\longrightarrow\mathbbm{S}^{p-1}$ is a submersion
for every sphere $\mathbbm{S}_{{\varepsilon^{\prime}}}^{n-1}$ with
${\varepsilon^{\prime}}<\varepsilon$.

The following proposition relates the differential of the spherefication map
$\mathfrak{F}$ and the differential of $f$.

######  Proposition 2.3.

Let $x\in U\setminus V$ and $v\in T_{x}(U\setminus V)$. Then

$D\mathfrak{F}_{x}(v)=\frac{\langle x,v\rangle}{\lVert
x\rVert^{2}}\mathfrak{F}(x)-\frac{f(x)}{\lVert
f(x)\rVert^{2}}\langle\mathfrak{F}(x),Df_{x}(v)\rangle+\frac{\lVert
x\rVert}{\lVert f(x)\rVert}Df_{x}(v).$

###### Proof.

Set

$\displaystyle x$
$\displaystyle=(x_{1},\dots,x_{n})\in\mathbbm{B}_{\varepsilon}^{n}\setminus
V,$ $\displaystyle v$ $\displaystyle=(v_{1},\dots,v_{n})\in
T_{x}(\mathbbm{B}_{\varepsilon}^{n}\setminus V),$ $\displaystyle w$
$\displaystyle=(w_{1},\dots,w_{p})=D\mathfrak{F}_{x}(v),$ $\displaystyle u$
$\displaystyle=(u_{1},\dots,u_{p})=Df_{x}(v).$

Using the quotient rule for the derivative we get

$\frac{\partial}{\partial x_{j}}\biggl{(}\frac{\lVert x\rVert}{\lVert
f(x)\rVert}\biggr{)}=\frac{x_{j}}{\lVert x\rVert\lVert
f(x)\rVert}-\frac{\lVert x\rVert}{\lVert
f(x)\rVert^{3}}\biggl{(}\sum_{l=1}^{p}f_{l}(x)\frac{\partial
f_{l}(x)}{\partial x_{j}}\biggr{)}.$ (2.3)

Write $f(x)=(f_{1}(x),\dots,f_{p}(x))$; by the definition of the
spherefication map (2.2) we have that

$\mathfrak{F}(x)=\bigl{(}\mathfrak{F}_{1}(x),\dots,\mathfrak{F}_{p}(x)\bigr{)}=\biggl{(}\frac{\lVert
x\rVert}{\lVert f(x)\rVert}f_{1}(x),\dots,\frac{\lVert x\rVert}{\lVert
f(x)\rVert}f_{p}(x)\biggr{)}.$

Using (2.3) and the product rule we obtain

$\frac{\partial\mathfrak{F}_{i}(x)}{\partial
x_{j}}=\frac{x_{j}f_{i}(x)}{\lVert x\rVert\lVert f(x)\rVert}-\frac{\lVert
x\rVert}{\lVert f(x)\rVert^{3}}\biggl{(}\sum_{l=1}^{p}f_{l}(x)\frac{\partial
f_{l}(x)}{\partial x_{j}}\biggr{)}f_{i}(x)+\frac{\lVert x\rVert}{\lVert
f(x)\rVert}\frac{\partial f_{i}(x)}{\partial x_{j}}.$

Recall that

$u_{i}=\sum_{j=1}^{n}\frac{\partial f_{i}(x)}{\partial x_{j}}v_{j}.$ (2.4)

Then using (2.4) we have:

$\displaystyle w_{i}$
$\displaystyle=\sum_{j=1}^{n}\frac{\partial\mathfrak{F}_{i}(x)}{\partial
x_{j}}v_{j}$
$\displaystyle=\sum_{j=1}^{n}\left[\frac{f_{i}(x)x_{j}v_{j}}{\lVert
x\rVert\lVert f(x)\rVert}-\frac{\lVert x\rVert}{\lVert
f(x)\rVert^{3}}\biggl{(}\sum_{l=1}^{p}f_{l}(x)\frac{\partial
f_{l}(x)}{\partial x_{j}}\biggr{)}f_{i}(x)v_{j}+\frac{\lVert x\rVert}{\lVert
f(x)\rVert}\frac{\partial f_{i}(x)}{\partial x_{j}}v_{j}\right]$
$\displaystyle=\frac{f_{i}(x)}{\lVert x\rVert\lVert
f(x)\rVert}\sum_{j=1}^{n}x_{j}v_{j}-\frac{\lVert x\rVert}{\lVert
f(x)\rVert^{3}}f_{i}(x)\sum_{l=1}^{p}f_{l}(x)\sum_{j=1}^{n}\frac{\partial
f_{l}(x)}{\partial x_{j}}v_{j}+\frac{\lVert x\rVert}{\lVert f(x)\rVert}u_{i}$
$\displaystyle=\frac{\langle x,v\rangle}{\lVert x\rVert\lVert
f(x)\rVert}f_{i}(x)-\frac{\lVert x\rVert}{\lVert
f(x)\rVert^{3}}f_{i}(x)\sum_{l=1}^{p}f_{l}(x)u_{l}+\frac{\lVert
x\rVert}{\lVert f(x)\rVert}u_{i}$ $\displaystyle=\frac{\langle
x,v\rangle}{\lVert x\rVert\lVert f(x)\rVert}f_{i}(x)-\frac{\lVert
x\rVert}{\lVert f(x)\rVert^{3}}f_{i}(x)\langle
f(x),Df_{x}(v)\rangle+\frac{\lVert x\rVert}{\lVert f(x)\rVert}u_{i}$
$\displaystyle=\frac{\langle x,v\rangle}{\lVert x\rVert\lVert
f(x)\rVert}f_{i}(x)-\frac{f_{i}(x)}{\lVert
f(x)\rVert^{2}}\langle\mathfrak{F}(x),Df_{x}(v)\rangle+\frac{\lVert
x\rVert}{\lVert f(x)\rVert}u_{i}\,.$ (2.5)

Hence the formula of the proposition follows. ∎

######  Corollary 2.4.

Let $x\in U\setminus V$ and $v\in T_{x}(U\setminus V)$. Then

$\lVert D\mathfrak{F}_{x}(v)\rVert^{2}=\frac{\langle x,v\rangle^{2}}{\lVert
x\rVert^{2}}+\frac{\lVert x\rVert^{2}\lVert
Df_{x}(v)\rVert^{2}-\langle\mathfrak{F}(x),Df_{x}(v)\rangle^{2}}{\lVert
f(x)\rVert^{2}}\,.$

###### Proof.

From (2.5) we have that

$\displaystyle w_{i}^{2}$ $\displaystyle=\frac{\langle x,v\rangle^{2}}{\lVert
x\rVert^{2}\lVert f(x)\rVert^{2}}f_{i}(x)^{2}+\frac{f_{i}(x)^{2}}{\lVert
f(x)\rVert^{4}}\langle\mathfrak{F}(x),Df_{x}(v)\rangle^{2}+\frac{\lVert
x\rVert^{2}}{\lVert f(x)\rVert^{2}}u_{i}^{2}$
$\displaystyle\qquad-2\frac{\langle x,v\rangle}{\lVert x\rVert\lVert
f(x)\rVert^{3}}\langle\mathfrak{F}(x),Df_{x}(v)\rangle
f_{i}(x)^{2}+2\frac{\langle x,v\rangle}{\lVert f(x)\rVert^{2}}f_{i}(x)u_{i}$
$\displaystyle\qquad-2\frac{\lVert x\rVert}{\lVert
f(x)\rVert^{3}}\langle\mathfrak{F}(x),Df_{x}(v)\rangle f_{i}(x)u_{i}.$

Therefore the corollary follows. ∎

Let $y\in\mathbbm{R}^{p}$ and let $\mathbbm{S}^{p-1}_{\lVert y\rVert}$ be the
sphere in $\mathbbm{R}^{p}$ with centre at the origin which contains $y$. Let
$\ell_{y}$ be the line through the origin in $\mathbbm{R}^{p}$ generated by
$y$. Hence, the tangent space $T_{y}\mathbbm{R}^{p}$ decomposes as the direct
sum of the orthogonal subspaces

$T_{y}\mathbbm{R}^{p}=T_{y}\mathbbm{S}^{p-1}_{\lVert y\rVert}\oplus\ell_{y}.$

Thus, any vector $u\in T_{y}\mathbbm{R}^{p}$ can be written in a unique way as
$u=s+r$, with $s\in T_{y}\mathbbm{R}^{p}$ and $r\in\ell_{y}$. We call $s$ and
$r$ respectively, the spherical and radial parts of $u$.

######  Proposition 2.5.

Let $x\in U\setminus V$, $v\in T_{x}U$ and consider $D\mathfrak{F}_{x}(v)$.
Then

  1. I)

$\frac{\langle x,v\rangle}{\lVert x\rVert^{2}}\mathfrak{F}(x)$ is the radial
part of $D\mathfrak{F}_{x}(v)$.

  2. II)

$-\frac{f(x)}{\lVert
f(x)\rVert^{2}}\langle\mathfrak{F}(x),Df_{x}(v)\rangle+\frac{\lVert
x\rVert}{\lVert f(x)\rVert}Df_{x}(v)$ is the spherical part of
$D\mathfrak{F}_{x}(v)$.

###### Proof.

Part I) is obvious since it is a multiple of $\mathfrak{F}(x)$. For part II)
it is enough to see that its inner product with $f(x)$ vanishes.

$-\langle f(x),\frac{f(x)}{\lVert
f(x)\rVert^{2}}\langle\mathfrak{F}(x),Df_{x}(v)\rangle\rangle+\langle
f(x),\frac{\lVert x\rVert}{\lVert f(x)\rVert}Df_{x}(v)\rangle=\\\
-\langle\mathfrak{F}(x),Df_{x}(v)\rangle+\langle\frac{\lVert x\rVert}{\lVert
f(x)\rVert}f(x),Df_{x}(v)\rangle=0.$

∎

Proposition 2.5 gives a criterium for a vector $v\in
T_{x}\mathbbm{B}_{\varepsilon}^{n}$ to be tangent to the corresponding
$E_{\ell}^{\pm}$

######  Corollary 2.6.

Let $x\in U\setminus V$ and $v\in T_{x}U$. Then $v\in
T_{x}E_{\ell_{\mathfrak{F}(x)}}^{\pm}$ if and only if

$Df_{x}(v)=\frac{f(x)}{\lVert f(x)\rVert^{2}}\langle f(x),Df_{x}(v)\rangle.$

###### Proof.

We have that $v\in T_{x}E_{\ell_{\mathfrak{F}(x)}}^{\pm}$ if and only if the
spherical part of $D_{x}\mathfrak{F}(v)$ is zero. Then the formula follows
from Proposition 2.5-II. ∎

##  3\. Construction of a vector field satisfying (1), (2) and (3)

In this section we construct a vector field satisfying (1), (2) and (3). First
in Subsection 3.1 we prove some auxiliary lemmas, in particular, a lemma about
liftings to $\mathbbm{R}^{n}$ of the gradient vector field of a function
$g\colon\mathbbm{R}^{p}\to\mathbbm{R}$ via a submersion
$f\colon\mathbbm{R}^{n}\to\mathbbm{R}^{p}$. Then, in Subsection 3.2 we apply
the lemmas to a real analytic $d$-regular map
$f\colon\mathbbm{R}^{n}\to\mathbbm{R}^{p}$ with isolated critical point, to
construct the aforementioned vector field.

###  3.1. Lifting of gradients

Consider open subsets $U\subset\mathbbm{R}^{n}$ and
$U^{\prime}\subset\mathbbm{R}^{p}$. Let $f\colon U\to U^{\prime}$ and $g\colon
U^{\prime}\to\mathbbm{R}$ be submersions. Consider the function $h=g\circ
f\colon U\to\mathbbm{R}$ which is a submersion, being the composition of two
submersions. Let $\nabla g$ and $\nabla h$ be the gradients of $g$ and $h$
respectively, which are non zero everywhere. Let $x\in U$ and let
$t=h(x)=g(f(x))$, we have that $t$ is a regular value of $g$ and $h$, so
$N=g^{-1}(t)$ and $M=f^{-1}(N)=h^{-1}(t)$ are submanifolds of codimension $1$
of $U^{\prime}$ and $U$ respectively. Since $f$ is a submersion, it is
transverse to $N$ and therefore we also have $M=f^{-1}(N)=h^{-1}(t)$ and

$T_{x}M=Df_{x}^{-1}(T_{f(x)}N).$ (3.1)

######  Lemma 3.1.

Let $x\in U$. Then there exists a unique vector $v(x)\in T_{x}M$ orthogonal to
$\ker Df_{x}$ such that

$Df_{x}(v(x)+\alpha(x)\nabla h(x))=\nabla g(f(x)).$ (3.2)

for some positive function $\alpha\colon U\to\mathbbm{R}_{+}$.

###### Proof.

The gradient $\nabla g(f(x))$ is normal to $T_{f(x)}N$. Denote by $G$ the line
in $T_{f(x)}U^{\prime}$ generated by $\nabla g(f(x))$. Analogously the
gradient $\nabla h(x)$ is normal to $T_{x}M$. Denote by $H$ the line in
$T_{x}U$ generated by $\nabla h(x)$. Hence, we have the following isomorphisms

$T_{x}U\simeq T_{x}M\oplus H$

and

$T_{f(x)}U^{\prime}\simeq T_{f(x)}N\oplus G.$

On the other hand, we have that $\ker Df_{x}=T_{x}f^{-1}(f(x))\subset T_{x}M$,
so there is an isomorphism:

$T_{x}U\cong T_{x}f^{-1}(f(x))\oplus\tilde{M}\oplus H$

where $\tilde{M}$ denotes the orthogonal complement of $T_{x}f^{-1}(f(x))$ in
$T_{x}M$. Hence $Df_{x}$ induces an isomorphism:

$Df_{x}|_{\tilde{M}\oplus H}\colon\tilde{M}\oplus H\to T_{f(x)}N\oplus G.$
(3.3)

By (3.1) $Df_{x}|_{\tilde{M}}\colon\tilde{M}\to T_{f(x)}N$ is an isomorphism,
therefore $Df_{x}(\nabla h(x))$ has the form

$Df_{x}(\nabla h(x))=u(x)+\lambda(x)\nabla g(f(x)),$ (3.4)

with $u(x)\in T_{f(x)}N$ and for some $\lambda(x)\neq 0$ (otherwise (3.3)
would not be an isomorphism). Set $\alpha(x)=\frac{1}{\lambda(x)}$, by the
isomorphism (3.3) there exist a unique vector $v(x)$ in $\tilde{M}$ such that
$Df_{x}(v(x))=-\alpha(x)u(x)$, then

$Df_{x}(\alpha(x)\nabla h(x)+v(x))=\frac{1}{\lambda(x)}u(x)+\nabla
g(f(x))-\frac{1}{\lambda(x)}u(x)=\nabla g(f(x)).$

It only remains to check that $\alpha(x)>0$ for every $x\in U$. By the chain
rule we have that

$\nabla h(x)=\nabla g(f(x))\cdot Df_{x}.$ (3.5)

Taking the inner product with $\nabla g(f(x))$ in (3.2) we have

$\displaystyle\nabla g(f(x))\cdot Df_{x}(v(x)+\alpha(x)\nabla h(x))$
$\displaystyle=\nabla g(f(x))\cdot\nabla g(f(x))$ $\displaystyle\nabla
g(f(x))\cdot\bigl{(}-\alpha(x)u(x)\bigr{)}+\alpha(x)\nabla g(f(x))\cdot
Df_{x}(\nabla h(x))$ $\displaystyle=\nabla g(f(x))\cdot\nabla g(f(x))$
$\displaystyle\alpha(x)\nabla h(x)\cdot\nabla h(x)$
$\displaystyle=\lVert\nabla g(f(x))\rVert^{2},\qquad\text{by \eqref{eq:c.r},}$
$\displaystyle\alpha(x)\lVert\nabla h(x)\rVert^{2}$
$\displaystyle=\lVert\nabla g(f(x))\rVert^{2},$

therefore

$\alpha(x)=\frac{\lVert\nabla g(f(x))\rVert^{2}}{\lVert\nabla
h(x)\rVert^{2}}>0.$ (3.6)

∎

######  Corollary 3.2.

Let $f\colon U\to U^{\prime}$, $g\colon U^{\prime}\to\mathbbm{R}$ and
$h=g\circ f$ be as in Lemma 3.1. Let $x\in U$ and let $L_{x}$ be a subspace of
$T_{x}M$ of codimension $1$ in $T_{x}M$, which is transverse to
$T_{x}f^{-1}(f(x))$ in $T_{x}M$. Then there exists a unique vector
$\bar{v}(x)\in L_{x}$ orthogonal to $L_{x}\cap T_{x}f^{-1}(f(x))$ such that

$Df_{x}(\bar{v}(x)+\alpha(x)\nabla h(x))=\nabla g(f(x)).$ (3.7)

for some positive function $\alpha\colon U\to\mathbbm{R}_{+}$.

###### Proof.

We have the isomorphism

$T_{x}M\cong T_{x}f^{-1}(f(x))\oplus\tilde{M}$ (3.8)

where $\tilde{M}$ denotes the orthogonal complement of $T_{x}f^{-1}(f(x))$ in
$T_{x}M$. Consider the projection

$\pi\colon T_{x}f^{-1}(f(x))\oplus\tilde{M}\to\tilde{M},$ (3.9)

and its restriction to $L_{x}$

$\pi|_{L_{x}}\colon L_{x}\to\tilde{M}.$ (3.10)

We claim that (3.10) is onto. Recall that $\dim T_{x}M=n-1$, and $\dim
L_{x}=n-2$, since $L_{x}$ is a codimension $1$ subspace of $T_{x}M$. The
kernel of (3.10) is $L_{x}\cap T_{x}f^{-1}(f(x))$, which has dimension
$n-p-1$, since $\dim T_{x}f^{-1}(f(x))=n-p$ and by hypothesis $L_{x}$ and
$T_{x}f^{-1}(f(x))$ intersect transversely in $T_{x}M$. Thus, by the rank-
nullity theorem, the rank of (3.10) is $n-2-(n-p-1)=p-1$, and since
$\dim\tilde{M}=p-1$ (3.10) is onto as claimed. Since
$\ker\pi|_{L_{x}}=L_{x}\cap T_{x}f^{-1}(f(x))$ the orthogonal complement of
$L_{x}\cap T_{x}f^{-1}(f(x))$ in $L_{x}$ is sent isomorphically onto
$\tilde{M}$ by $\pi|_{L_{x}}$.

Let $v(x)\in\tilde{M}$ be the vector given by Lemma 3.1, let $\bar{v}(x)\in
L_{x}$ the unique vector orthogonal to $L_{x}\cap T_{x}f^{-1}(f(x))$ such that
$\pi(\bar{v}(x))=v(x)$. By isomorphism (3.8) the vector $\bar{v}(x)$ has the
form

$\bar{v}(x)=k(x)+v(x),\quad\text{with $k(x)\in T_{x}f^{-1}(f(x))$.}$ (3.11)

Therefore

$Df_{x}(\bar{v}(x)+\alpha(x)\nabla h(x))=Df_{x}(v(x)+\alpha(x)\nabla
h(x))=\nabla g(f(x)).$

∎

######  Lemma 3.3.

Let $r,s\in\mathbbm{R}^{n}$ such that they do not point in opposite
directions. Let $v_{1},v_{2}\in\mathbbm{R}^{n}$ be orthogonal vectors to both
$r$ and $s$. Then the vectors $v_{1}+\alpha r$ and $v_{2}+\beta s$, with
$\alpha,\beta\in\mathbbm{R}^{+}$, cannot point in opposite directions.

###### Proof.

Suppose $v_{1}+\alpha r$ and $v_{2}+\beta s$, with
$\alpha,\beta\in\mathbbm{R}^{+}$ point in opposite directions, then there
exists $0<\lambda\in\mathbbm{R}$ such that

$v_{1}+\alpha r=-\lambda(v_{2}+\beta s).$

By hypothesis, the subspace generated by $r$ and $s$ is orthogonal to the
subspace generated by $v_{1}$ and $v_{2}$, so there is no contribution by $r$
and $s$ on the subspace generated by $v_{1}$ and $v_{2}$ and we need to have
$v_{1}=-\lambda v_{2}$, which implies that $\alpha r=-\lambda\beta s$, which
contradicts the fact that $r$ and $s$ do not point in opposite directions. ∎

######  Lemma 3.4.

Suppose $r,s\in\mathbbm{R}^{n}$ point in opposite directions. Let $0\neq
k\in\mathbbm{R}^{n}$ be orthogonal to $r$ (and therefore also to $s$). Then
the vectors $r+k$ and $s$ cannot point in opposite directions.

###### Proof.

Since $r$ and $s$ point in opposite directions there exists
$0<\lambda\in\mathbbm{R}$ such that $r=-\lambda s$. Suppose $r+k$ and $s$
point in opposite directions, then there exists $0<\mu\in\mathbbm{R}$ such
that $r+k=-\mu s$, then $-\lambda s+k=-\mu s$, hence $k=(\lambda-\mu)s$. But
$k$ and $s$ are orthogonal, so we need to have $\lambda=\mu$ and $k=0$ which
is a contradiction. ∎

###  3.2. Construction of the vector field

Let $U$ be an open neighbourhood of $\underline{0}\in\mathbbm{R}^{n}$. Let
$f\colon(U,0)\to(\mathbbm{R}^{p},0)$ be a real analytic map, with isolated
critical value at $0\in\mathbbm{R}^{p}$. Also suppose $f$ is $d$-regular and
set $V=f^{-1}(0)$. Let $g\colon\mathbbm{R}^{p}\to\mathbbm{R}$ be given by
$g(y)=\lVert y\rVert^{2}$. We have that the restriction $f\colon U\setminus
V\to\mathbbm{R}^{p}\setminus\\{0\\}$ is a submersion. On the other hand, since
$f$ is $d$-regular, by Proposition 2.2(C) the spherefication
$\mathfrak{F}\colon U\setminus V\to\mathbbm{R}^{p}\setminus\\{0\\}$ of $f$ is
also a submersion. Define the maps

$\displaystyle h$ $\displaystyle=g\circ f\colon U\setminus V\to\mathbbm{R},$
$\displaystyle\mathfrak{H}$ $\displaystyle=g\circ\mathfrak{F}\colon U\setminus
V\to\mathbbm{R}.$

Notice that by the definitions

$\displaystyle h(x)$ $\displaystyle=\lVert f(x)\rVert^{2},$ (3.11)
$\displaystyle\mathfrak{H}(x)$
$\displaystyle=\lVert\mathfrak{F}(x)\rVert^{2}=\left\|\lVert
x\rVert\frac{f(x)}{\lVert f(x)\rVert}\right\|^{2}=\lVert
x\rVert^{2}\frac{\lVert f(x)\rVert^{2}}{\lVert f(x)\rVert^{2}}=\lVert
x\rVert^{2}.$

Let $\mathbbm{B}_{\varepsilon}^{n}$ be an open ball in $U$, centred at
$\underline{0}$, of sufficiently small radius $\varepsilon$ as in Section 2.
Let $x\in\mathbbm{B}_{\varepsilon}^{n}\setminus V$ and let $\delta^{2}=h(x)$,
then $h^{-1}(\delta^{2})=\mathbbm{B}_{\varepsilon}^{n}\cap
f^{-1}(\mathbbm{S}_{\delta}^{k-1}\setminus\\{0\\})$, that is, the fibres of
$h$ are the tubes. On the other hand, let $\nu^{2}=\mathfrak{H}(x)$, then
$\mathfrak{H}^{-1}(\nu^{2})=\mathbbm{S}_{\nu}^{n-1}\setminus V$, that is, the
fibres of $\mathfrak{H}$ are the spheres.

Consider the corresponding gradient vector fields $\nabla h(x)$ and
$\nabla\mathfrak{H}(x)=2x$ which are non-zero for any $x\in U\setminus V$. The
vector field $\nabla h$ is normal to the Milnor tubes, while the vector field
$\nabla\mathfrak{H}$ is normal to the spheres. The gradient $\nabla g$ is the
radial vector field $\nabla g(y)=2y$ on $\mathbbm{R}^{p}$.

######  Proposition 3.5.

There exists $0<\varepsilon^{\prime}\leq\varepsilon$ such that for
$x\in\mathbbm{B}_{\varepsilon^{\prime}}^{n}\setminus V$ the vector fields
$\nabla h(x)$ and $\nabla\mathfrak{H}(x)$ cannot point in exactly opposite
directions at any point.

###### Proof.

Apply [22, Corollary 3.4] to the non-negative analytic functions $h$ and
$\mathfrak{H}$ but using the Analytic Curve Selection Lemma (see Proposition
2.2 of [6] or (2.1) of [17]). ∎

Let $f\colon(\mathbbm{B}_{\varepsilon}^{n},0)\to(\mathbbm{R}^{p},0)$ be a real
analytic map, with isolated critical value at $0\in\mathbbm{R}^{p}$. Also
assume that $\varepsilon>0$ is small enough such that Proposition 3.5 holds.
Let $\\{X_{\ell}\\}_{\ell\in\mathbbm{RP}^{p-1}}$ be the canonical pencil of
$f$ and consider the decomposition $X_{\ell}=E_{\ell}^{+}\cup V\cup
E_{\ell}^{-}$ of each of its elements given in (2.1). From now on, to simplify
notation we will write $E_{\ell}$ instead of $E_{\ell}^{\pm}.$

Let $x\in\mathring{\mathbbm{B}}_{\varepsilon}^{n}\setminus V$, let

$\displaystyle w_{f}(x)$ $\displaystyle=v_{f}(x)+\alpha(x)\nabla
h(x),\quad\text{and}$ (3.12) $\displaystyle w_{\mathfrak{F}}(x)$
$\displaystyle=v_{\mathfrak{F}}(x)+\beta(x)\nabla\mathfrak{H}(x)=v_{\mathfrak{F}}(x)+\nabla\mathfrak{H}(x),$
(3.13)

be, respectively, the normal liftings of the radial vector field $\nabla g$
obtained from Lemma 3.1 applied to $h$ and $\mathfrak{H}$, with $\langle
v_{f}(x),\nabla h(x)\rangle=0$, $\langle
v_{\mathfrak{F}}(x),\nabla\mathfrak{H}(x)\rangle=0$, $\alpha(x)>0$ and
$\beta(x)>0$. We have that $\beta(x)=1$, by equation (3.6) and the definitions
of $\mathfrak{F}$ and $\mathfrak{H}$ (or see equation (3.32) below). Since
both are liftings of $\nabla g$, both are tangent to the corresponding
$E_{\ell}$; $w_{f}(x)$ is normal to the fibres of $f$ and to the Milnor tubes,
while $w_{\mathfrak{F}}(x)$ is normal to the fibres of $\mathfrak{F}$ and to
the spheres.

Let us analyse how are $w_{f}(x)$ and $w_{\mathfrak{F}}(x)$ on the set of
points where the tangent space of the fibre of $f$ coincides with the tangent
space of the fibre of $\mathfrak{F}$.

Let $x\in\mathring{\mathbbm{B}}^{n}_{\varepsilon}\setminus V$, denote by
$\ell_{x}$ the ray from $0\in\mathbbm{R}^{p}$ passing by $f(x)$. Let $P_{x}$
be the normal space of $E_{\ell_{x}}$ at $x$, the $p$-dimensional subspace
$\langle P_{x},\nabla h(x)\rangle$ generated by $P_{x}$ and $\nabla h(x)$ is
the normal space to the fibre $f^{-1}(f(x))$ at $x$. If the vector
$\nabla\mathfrak{H}(x^{\prime})$ is in $\langle P_{x},\nabla h(x)\rangle$,
then the tangent spaces to the fibres of $f$ and $\mathfrak{F}$ at $x$
coincide. Let $M(f)$ be the set defined by

$\displaystyle M(f)$
$\displaystyle=\\{\,{x\in\mathring{\mathbbm{B}}^{n}_{\varepsilon}\setminus
V}\mid{\nabla\mathfrak{H}(x)\in\langle P_{x},\nabla h(x)\rangle}\,\\},$
$\displaystyle=\\{\,{x\in\mathring{\mathbbm{B}}^{n}_{\varepsilon}\setminus
V}\mid{T_{x}f^{-1}(f(x))=T_{x}\mathfrak{F}^{-1}(\mathfrak{F}(x))}\,\\}.$

Let $F\colon U\to\mathbbm{R}^{p+1}$ be the map defined by $F(x)=(f(x),\lVert
x\rVert^{2})$. We have that $M(f)$ is the set of critical points of $F$. It is
a semi-analytic set with semi-analytic components each of which having $0$ in
its closure if $\varepsilon$ is sufficiently small.111We learned this
description from [11].

Let $x\in M(f)$, the tangent space
$\mathbf{T}_{x}=T_{x}f^{-1}(f(x))=T_{x}\mathfrak{F}^{-1}(\mathfrak{F}(x))$ is
a subspace of $T_{x}E_{\ell_{x}}$ with codimension $1$. Since
$w_{f}(x),w_{\mathfrak{F}}(x)\in T_{x}E_{\ell_{x}}$ and both of them are
orthogonal to $\mathbf{T}_{x}$, they are collinear, that is,

$\displaystyle w_{\mathfrak{F}}(x)$
$\displaystyle=\mu(x)w_{f}(x),\qquad\text{with $\mu(x)\neq 0$.}$ (3.14) Hence
we have $\displaystyle w_{f}(x)$
$\displaystyle=\frac{1}{\mu(x)}w_{\mathfrak{F}}(x).$ (3.15)

Restricting $x$ to a component of $M(f)$ we have that either $\mu(x)>0$ or
$\mu(x)<0$ for every $x$ in such component.

By (3.12), (3.14), (3.13) and (3.15) we have that

$\displaystyle\langle w_{\mathfrak{F}}(x),\nabla h(x)\rangle$
$\displaystyle=\alpha(x)\mu(x)\lVert\nabla h(x)\rVert^{2}$ (3.16)
$\displaystyle\langle w_{f}(x),\nabla\mathfrak{H}(x)\rangle$
$\displaystyle=\frac{1}{\mu(x)}\lVert\nabla\mathfrak{H}(x)\rVert^{2}.$ (3.17)

######  Proposition 3.6.

Let $x\in M(f)$ and let $w_{f}(x)$ and $w_{\mathfrak{F}}(x)$ be the normal
liftings of the radial vector field $\nabla g$ obtained by applying Lemma 3.1
to $h$ and $\mathfrak{H}$ respectively. Then $w_{f}(x)$ and
$w_{\mathfrak{F}}(x)$ cannot point in opposite directions, that is,
$\mu(x)>0$, for $\varepsilon$ sufficiently small.

###### Proof.

Let $x\in M(f)$ and let $\delta^{2}=\lVert f(x)\rVert^{2}=h(x)$. Consider the
open solid Milnor tube

$\mathring{N}(\varepsilon,\delta):=\mathbbm{B}_{\varepsilon}^{n}\cap
f^{-1}(\mathring{\mathbbm{B}}_{\delta}^{p}).$

Consider the inverse image of the open interval $(-\delta^{2},\delta^{2})$
under the map $h$. We have that

$h^{-1}((-\delta^{2},\delta^{2}))=h^{-1}([0,\delta^{2}))=\mathring{N}(\varepsilon,\delta)$

and $x$ is in the closure of $\mathring{N}(\varepsilon,\delta)$. Let
$\mathring{\mathbbm{B}}^{n}_{\lVert x\rVert}$ be the open ball centered at $0$
of radius $\lVert x\rVert$ and let $W$ be the open set given by the
intersection

$W=\mathring{N}(\varepsilon,\delta)\cap\mathring{\mathbbm{B}}^{n}_{\lVert
x\rVert},$

which is nonempty since $\underline{0}\in W$. By definition we have that

$h(W)\subset[0,\delta^{2}).$ (3.18)

If $\lVert x\rVert$ is sufficiently small we have that $x$ is in the closure
of $W$.

Suppose $w_{f}(x)$ and $w_{\mathfrak{F}}(x)$ point in opposite directions,
that is, $\mu(x)<0$ in (3.14), then by (3.17) we have that $\langle
w_{f}(x),\nabla\mathfrak{H}(x)\rangle<0$.

Since $w_{f}(x)\in T_{x}E_{\ell_{x}}$ there exist a curve
$\beta\colon(-\eta,\eta)\to E_{\ell_{x}}$ such that $\beta(0)=x$ and
$\beta^{\prime}(0)=w_{f}(x)$. By (3.12) and (3.17) we have that

$\displaystyle\frac{d}{dt}h(\beta(t))|_{t=0}$ $\displaystyle=\langle
w_{f}(x),\nabla h(x)\rangle=\alpha(x)\lVert\nabla h(x)\rVert^{2}>0,$ (3.19)
$\displaystyle\frac{d}{dt}\mathfrak{H}(\beta(t))|_{t=0}$
$\displaystyle=\langle
w_{f}(x),\nabla\mathfrak{H}(x)\rangle=\frac{1}{\mu(x)}\lVert\nabla\mathfrak{H}(x)\rVert^{2}<0.$
(3.20)

From (3.19) we have that $h(\beta(t))=\lVert f(\beta(t))\rVert^{2}$ is an
increasing function, thus we have that

$\lVert f(\beta(t))\rVert^{2}>\lVert f(\beta(0))\rVert^{2}=\lVert
f(x)\rVert^{2}=\delta^{2},\quad\text{for $t>0$.}$ (3.21)

From (3.20) we have that
$\mathfrak{H}(\beta(t))=\lVert\mathfrak{F}(\beta(t))\rVert^{2}=\lVert\beta(t)\rVert^{2}$
is a decreasing function, thus we have

$\lVert\beta(t)\rVert^{2}<\lVert\beta(0)\rVert^{2}=\lVert
x\rVert^{2},\quad\text{for $t>0$.}$ (3.22)

The vector $w_{f}(x)$ is transverse to the Milnor tube
$N(\varepsilon,\delta)=h^{-1}(\delta^{2})$ which is the boundary of the open
Milnor tube $\mathring{N}(\varepsilon,\delta)$ and by (3.13)

$\langle w_{\mathfrak{F}}(x),2x\rangle=\langle
w_{\mathfrak{F}}(x),\nabla\mathfrak{H}(x)\rangle=\lVert\nabla\mathfrak{H}(x)\rVert^{2}>0,$

so $w_{\mathfrak{F}}(x)$ is transverse to the sphere $\mathbbm{S}_{\lVert
x\rVert}^{n-1}$ pointing outwards. By hypothesis, $w_{f}(x)$ points in the
opposite direction of $w_{\mathfrak{F}}(x)$, hence, $w_{f}(x)$ is transverse
to the sphere $\mathbbm{S}_{\lVert x\rVert}^{n-1}$ pointing inwards. Since $x$
is in the closure of $W$, we have that

$\beta(t)\in W,\quad\text{for $t>0$,}$ (3.23)

but then (3.21) and (3.18) give a contradiction with the continuity of the map
$h$. Therefore, $w_{f}(x)$ and $w_{\mathfrak{F}}(x)$ cannot point in opposite
directions. Since $x$ is an arbitrary point in $M(f)$, $w_{f}(x)$ and
$w_{\mathfrak{F}}(x)$ must point in the same direction in all the components
of $M(f)$ in a sufficiently small sphere $\mathbbm{B}_{\varepsilon}^{n}$. ∎

######  Lemma 3.7.

Let $g\colon\mathbbm{R}^{p}\to\mathbbm{R}$ given by $g(y)=\lVert y\rVert^{2}$.
If the map $f$ is $d$-regular, then for every
$x\in\mathring{\mathbbm{B}}_{\varepsilon}^{n}\setminus V$ there is a
neighbourhood $U_{x}$ of $x$ and vector fields $w_{f}$ and $w_{\mathfrak{F}}$
in $U_{x}$, which are differentiable liftings of the radial vector field
$\nabla g$ by $f$ and its spherefication $\mathfrak{F}$, respectively, such
that:

  1. (a)

$w_{f}$ is tangent to each $E_{\ell}$, it is transverse to the fibres of $f$
and therefore, it is transverse to the tubes.

  2. (b)

$w_{\mathfrak{F}}$ is tangent to each $E_{\ell}$, it is transverse to the
fibres of $\mathfrak{F}$, and therefore to the spheres.

  3. (c)

At every point $x^{\prime}\in U_{x}$ the vectors $w_{f}(x^{\prime})$ and
$w_{\mathfrak{F}}(x^{\prime})$ do not point in opposite directions.

###### Proof.

Recall that by Proposition 2.2-(C) the spherefication $\mathfrak{F}$ is also a
submersion in $\mathring{\mathbbm{B}}_{\varepsilon}^{n}\setminus V$. Set
$h=g\circ f$ and $\mathfrak{H}=g\circ\mathfrak{F}$ as before.

#### Case 1

Let $x\in\mathring{\mathbbm{B}}_{\varepsilon}^{n}\setminus V$ be such that the
vectors $\nabla h(x)$ and $\nabla\mathfrak{H}(x)$ are colinear. Let $w_{f}(x)$
and $w_{\mathfrak{F}}(x)$ be the normal liftings of the radial vector field
$\nabla g$ obtained by applying Lemma 3.1 to $h$ and $\mathfrak{H}$
respectively. By Proposition 3.5, $\nabla h(x)$ and $\nabla\mathfrak{H}(x)$
point in the same direction and $v_{f}(x)$ and $v_{\mathfrak{F}}(x)$ are
orthogonal to them. Thus, by Lemma 3.3, $w_{f}(x)$ and $w_{\mathfrak{F}}(x)$
cannot point in opposite directions, and the lemma is proved in this case.

#### Case 2

The vectors $\nabla h(x)$ and $\nabla\mathfrak{H}(x)$ are linearly
independent. In this case the tube $N_{x}=\mathbbm{B}_{\varepsilon}^{n}\cap
f^{-1}(\mathbbm{S}_{\lVert f(x)\rVert}^{k-1})$ and the sphere
$S_{x}=\mathbbm{S}_{\lVert x\rVert}^{n-1}$, are transverse at $x$. Hence, its
intersection $N_{x}\cap S_{x}$ is a submanifold of codimension $2$ of
$U\subset\mathbbm{R}^{n}$. Suppose $x\in E_{\ell}$ and, as before, denote by
$P_{x}$ the normal space of $E_{\ell}$ at $x$, so we have that $\dim
P_{x}=p-1$. The tube $N_{x}$ and $E_{\ell}$ are transverse at $x$ since $f$ is
a submersion, hence $\nabla h(x)$ is not in $P_{x}$ and the $p$-dimensional
subspace $\langle P_{x},\nabla h(x)\rangle$ generated by $P_{x}$ and $\nabla
h(x)$ is the normal space to the fibre $f^{-1}(f(x))$ at $x$. Since $f$ is
$d$-regular, $E_{\ell}$ and $S_{x}$ are transverse, so $\nabla\mathfrak{H}(x)$
is not in $P_{x}$ and the $p$-dimensional subspace $\langle
P_{x},\nabla\mathfrak{h}(x)\rangle$ generated by $P_{x}$ and
$\nabla\mathfrak{h}(x)$ is the normal space to the fibre
$\mathfrak{F}^{-1}(\mathfrak{F}(x))$.

We have two subcases.

#### Subcase A

The vector $\nabla\mathfrak{H}(x)$ is not in $\langle P_{x},\nabla
h(x)\rangle$. Then $N_{x}\cap S_{x}$ is transverse to the fibre
$f^{-1}(f(x))$, i.e., $T_{x}(N_{x}\cap S_{x})$ is transverse to
$T_{x}f^{-1}(f(x))$. By Corollary 3.2 there exists a lifting of $\nabla
g(f(x))$ by $f$,

$w_{f}(x)=\bar{v}_{f}(x)+\alpha(x)\nabla h(x)\,,$

with $\bar{v}_{f}(x)\in T_{x}(N_{x}\cap S_{x})$. Notice that the hypothesis
$\nabla\mathfrak{H}\notin\langle P_{x},\nabla h(x)\rangle$ implies $\nabla
h(x)\notin\langle P_{x},\nabla\mathfrak{H}(x)\rangle$, so $N_{x}\cap S_{x}$ is
also transverse to the fibre $\mathfrak{F}^{-1}(\mathfrak{F}(x))$. Hence
$T_{x}(N_{x}\cap S_{x})$ is transverse to
$T_{x}\mathfrak{F}^{-1}(\mathfrak{F}(x))$. By Corollary 3.2 there exists a
lifting of $\nabla g(\mathfrak{F}(x))$ by $\mathfrak{F}$,

$w_{\mathfrak{F}}(x)=\bar{v}_{\mathfrak{F}}(x)+\beta(x)\nabla\mathfrak{H}(x)\,,$

with $\bar{v}_{\mathfrak{F}}(x)\in T_{x}(N_{x}\cap S_{x})$. Since
$\bar{v}_{f}(x)$ and $\bar{v}_{\mathfrak{F}}(x)$ are in $T_{x}(N_{x}\cap
S_{x})$, they are orthogonal to $\nabla h(x)$ and $\nabla\mathfrak{H}(x)$,
which do not point in opposite directions. Then Lemma 3.3 implies that
$w_{f}(x)$ and $w_{\mathfrak{F}}(x)$ do not point in opposite directions, and
the lemma is proved in this case.

#### Subcase B

The vector $\nabla\mathfrak{H}(x^{\prime})$ is in $\langle P_{x},\nabla
h(x)\rangle$. Then $x\in M(f)$. Consider the normal liftings $w_{f}(x)$ and
$w_{\mathfrak{F}}(x)$ of the radial vector field $\nabla g$ obtained by
applying Lemma 3.1 to $h$ and $\mathfrak{H}$ respectively. We have that
$w_{f}(x)$ and $w_{\mathfrak{F}}(x)$ are collinear and by Proposition 3.6, if
$\varepsilon$ is small enough, they point in the same direction. Then there
exists a neighbourhood $U_{x}$ of $x$ such that $w^{\prime}_{f}(x^{\prime})$
and $w_{\mathfrak{F}}(x^{\prime})$ do not point in opposite directions for any
$x^{\prime}\in U_{x}$ and we are done. ∎

######  Theorem 3.8 ([9, Lemma 5.2]).

The map $f$ is $d$-regular, if and only if there exists an analytic vector
field $\tilde{w}$ on $\mathring{\mathbbm{B}}_{\varepsilon}\setminus V$ which
has the following properties:

  1. (1)

It is radial, i.e., it is transverse to all spheres in
$\mathring{\mathbbm{B}}_{\varepsilon}$ centred at $0$.

  2. (2)

It is transverse to all the tubes $f^{-1}(\mathbbm{S}_{\delta}^{k-1})$.

  3. (3)

It is tangent to each $E_{\ell}$, whenever it is not empty.

###### Proof.

It is clear that properties (1) and (3) imply $d$-regularity. For the proof
that $d$-regularity implies the existence of a vector field as stated, as in
[22, Lemma 5.9], it suffices to construct such vector field locally, and then
use a partition of unity to get the desired global vector field.

For every $x\in\mathring{\mathbbm{B}}^{n}_{\varepsilon}\setminus V$ let
$U_{x}$ be the neighbourhood of $x$ and $w_{f}$ and $w_{\mathfrak{F}}$ the
vector fields in $U_{x}$ given by Lemma 3.7. Remember that in all cases the
vectors $w_{f}(x^{\prime})$ and $w_{\mathfrak{F}}(x^{\prime})$ have the form

$\displaystyle w_{f}(x^{\prime})$
$\displaystyle=\hat{v}_{f}(x^{\prime})+\alpha(x^{\prime})\nabla
h(x^{\prime}),$ (3.24) $\displaystyle w_{\mathfrak{F}}(x^{\prime})$
$\displaystyle=\hat{v}_{\mathfrak{F}}(x^{\prime})+\nabla\mathfrak{H}(x^{\prime}),$
(3.25)

with $\langle\hat{v}_{f}(x^{\prime}),\nabla h(x^{\prime})\rangle=0$,
$\langle\hat{v}_{\mathfrak{F}}(x^{\prime}),\nabla\mathfrak{H}(x^{\prime})\rangle=0$,
$\alpha(x^{\prime})>0$ and $\beta(x^{\prime})>0$ for all $x^{\prime}\in
U_{x}$.

Now on each $U_{x}$ let $\tilde{w}$ be the vector field given by

$\tilde{w}(x^{\prime})=\lVert\nabla\mathfrak{H}(x^{\prime})\rVert
w_{f}(x^{\prime})+\alpha(x^{\prime})\lVert\nabla h(x^{\prime})\rVert
w_{\mathfrak{F}}(x^{\prime}).$ (3.26)

Let us check that it has the desired properties. Since $w_{f}(x^{\prime})$ and
$w_{\mathfrak{F}}(x^{\prime})$ are both tangent to $E_{\ell}$, then
$\tilde{w}(x^{\prime})$ is also tangent to $E_{\ell}$, so it satisfies
property (3).

By (3.26) and (3.25) we have that

$\displaystyle\langle\tilde{w}$ $\displaystyle(x^{\prime}),\nabla
h(x^{\prime})\rangle=\alpha(x^{\prime})\lVert\nabla\mathfrak{H}(x^{\prime})\rVert\lVert\nabla
h(x^{\prime})\rVert^{2}+\alpha(x^{\prime})\lVert\nabla
h(x^{\prime})\rVert\langle w_{\mathfrak{F}}(x^{\prime}),\nabla
h(x^{\prime})\rangle,$ (3.27)
$\displaystyle=\alpha(x^{\prime})\lVert\nabla\mathfrak{H}(x^{\prime})\rVert\lVert\nabla
h(x^{\prime})\rVert^{2}+\alpha(x^{\prime})\lVert\nabla
h(x^{\prime})\rVert\langle\hat{v}_{\mathfrak{F}}(x^{\prime})+\nabla\mathfrak{H}(x^{\prime}),\nabla
h(x^{\prime})\rangle,$
$\displaystyle=\alpha(x^{\prime})\lVert\nabla\mathfrak{H}(x^{\prime})\rVert\lVert\nabla
h(x^{\prime})\rVert^{2}+\alpha(x^{\prime})\lVert\nabla
h(x^{\prime})\rVert\langle\nabla\mathfrak{H}(x^{\prime}),\nabla
h(x^{\prime})\rangle,$ $\displaystyle\hskip
122.34692pt+\alpha(x^{\prime})\lVert\nabla
h(x^{\prime})\rVert\langle\hat{v}_{\mathfrak{F}}(x^{\prime}),\nabla
h(x^{\prime})\rangle,$
$\displaystyle=\alpha(x^{\prime})\lVert\nabla\mathfrak{H}(x^{\prime})\rVert\lVert\nabla
h(x^{\prime})\rVert^{2}+\alpha(x^{\prime})\lVert\nabla\mathfrak{H}(x^{\prime})\rVert\lVert\nabla
h(x^{\prime})\rVert^{2}\cos\vartheta,$ $\displaystyle\hskip
122.34692pt+\alpha(x^{\prime})\lVert\nabla
h(x^{\prime})\rVert\langle\hat{v}_{\mathfrak{F}}(x^{\prime}),\nabla
h(x^{\prime})\rangle,$
$\displaystyle=\alpha(x^{\prime})\lVert\nabla\mathfrak{H}(x^{\prime})\rVert\lVert\nabla
h(x^{\prime})\rVert^{2}(1+\cos\vartheta)+\alpha(x^{\prime})\lVert\nabla
h(x^{\prime})\rVert\langle\hat{v}_{\mathfrak{F}}(x^{\prime}),\nabla
h(x^{\prime})\rangle,$ (3.28)

where $\vartheta$ is the angle between $\nabla\mathfrak{H}(x^{\prime})$ and
$\nabla h(x^{\prime})$.

By an analogous computation, by (3.26) and (3.24) we have

$\displaystyle\langle\tilde{w}$
$\displaystyle(x^{\prime}),\nabla\mathfrak{H}(x^{\prime})\rangle=\lVert\nabla\mathfrak{H}(x^{\prime})\rVert\langle
w_{f}(x^{\prime}),\nabla\mathfrak{H}(x^{\prime})\rangle+\alpha(x)\lVert\nabla
h(x^{\prime})\rVert\lVert\nabla\mathfrak{H}(x^{\prime})\rVert^{2},$ (3.29)
$\displaystyle=\alpha(x^{\prime})\lVert\nabla
h(x^{\prime})\rVert\lVert\nabla\mathfrak{H}(x^{\prime})\rVert^{2}(1+\cos\vartheta)+\lVert\nabla\mathfrak{H}(x^{\prime})\rVert\langle\hat{v}_{f}(x^{\prime}),\nabla\mathfrak{H}(x^{\prime})\rangle,$
(3.30)

where, again, $\vartheta$ is the angle between
$\nabla\mathfrak{H}(x^{\prime})$ and $\nabla h(x^{\prime})$.

Let us analyse these inner products for the different cases considered in
Lemma 3.7.

#### Case 1

Let $x\in\mathring{\mathbbm{B}}_{\varepsilon}^{n}\setminus V$ be such that the
vectors $\nabla h(x)$ and $\nabla\mathfrak{H}(x)$ are colinear. In this case,
in (3.24) and (3.25), both $\hat{v}_{f}(x)$ and $\hat{v}_{\mathfrak{F}}(x)$
are orthogonal to $\nabla\mathfrak{H}(x)$ and $\nabla h(x)$. Hence (3.28) and
(3.30) become

$\displaystyle\langle\tilde{w}(x),\nabla h(x)\rangle$
$\displaystyle=\alpha(x)\lVert\nabla\mathfrak{H}(x)\rVert\lVert\nabla
h(x)\rVert^{2}(1+\cos\vartheta)>0,$
$\displaystyle\langle\tilde{w}(x),\nabla\mathfrak{H}(x)\rangle$
$\displaystyle=\alpha(x)\lVert\nabla
h(x)\rVert\lVert\nabla\mathfrak{H}(x)\rVert^{2}(1+\cos\vartheta)>0,$

since $\cos\vartheta>-1$, otherwise, the vectors $\nabla\mathfrak{H}(x)$ and
$\nabla h(x)$ would point in opposite directions which is impossible by
Proposition 3.5. Thus, there exist an open neighbourhood $U_{x}$ of $x$ where
these inner products are positive.

#### Case 2

The vectors $\nabla h(x)$ and $\nabla\mathfrak{H}(x)$ are linearly
independent. In this case the tube $N_{x}=\mathbbm{B}_{\varepsilon}^{n}\cap
f^{-1}(\mathbbm{S}_{\lVert f(x)\rVert}^{k-1})$ and the sphere
$S_{x}=\mathbbm{S}_{\lVert x\rVert}^{n-1}$, are transverse at $x$. Hence, its
intersection $N_{x}\cap S_{x}$ is a submanifold of codimension $2$ of
$U\subset\mathbbm{R}^{n}$.

We have two subcases.

#### Subcase A

The vector $\nabla\mathfrak{H}(x)$ is not in $\langle P_{x},\nabla
h(x)\rangle$. Then $N_{x}\cap S_{x}$ is transverse to the fibre
$f^{-1}(f(x))$, i.e., $T_{x}(N_{x}\cap S_{x})$ is transverse to
$T_{x}f^{-1}(f(x))$. By the construction of $w_{f}(x)$ and
$w_{\mathfrak{F}}(x)$ in Lemma 3.7 in this subcase, we also have that in
(3.24) and (3.25), both $\hat{v}_{f}(x)$ and $\hat{v}_{\mathfrak{F}}(x)$ are
orthogonal to $\nabla\mathfrak{H}(x^{\prime})$ and $\nabla h(x^{\prime})$.
Hence, we get the same conclusion as in Case 1.

#### Subcase B

The vector $\nabla\mathfrak{H}(x)$ is in $\langle P_{x},\nabla h(x)\rangle$.
Then $x\in M(f)$. In this case $w_{f}(x)$ and $w_{\mathfrak{F}}(x)$ are
collinear and by Proposition 3.6, if $\varepsilon$ is small enough, they point
in the same direction, that is, in (3.14) $\mu(x)>0$.

By (3.27), (3.16), (3.29) and (3.17) we have

$\displaystyle\langle\tilde{w}(x),\nabla h(x)\rangle$
$\displaystyle=\alpha(x)\lVert\nabla\mathfrak{H}(x)\rVert\lVert\nabla
h(x)\rVert^{2}+\alpha(x)^{2}\mu(x)\lVert\nabla h(x)\rVert^{3}>0,$
$\displaystyle\langle\tilde{w}(x),\nabla\mathfrak{H}(x)\rangle$
$\displaystyle=\frac{1}{\mu(x)}\lVert\nabla\mathfrak{H}(x)\rVert^{3}+\alpha(x)\lVert\nabla
h(x)\rVert\lVert\nabla\mathfrak{H}(x)\rVert^{2}>0.$

Hence $\langle\tilde{w}(x),\nabla h(x)\rangle>0$ and
$\langle\tilde{w}(x),\nabla\mathfrak{H}(x)\rangle>0$ in all the cases, which
implies that $\tilde{w}$ is transverse to the tubes and to the spheres,
satisfying properties (2) and (1). Once we have the vector field $\tilde{w}$
defined in each neighbourhood $U_{x}$ using a partition of unity we get a
global vector field $\tilde{w}$ on
$\mathring{\mathbbm{B}}_{\varepsilon}\setminus V$ with the desired properties.
∎

######  Remark 3.9.

Remember that the vector fields $w_{f}$ and $w_{\mathfrak{F}}$ have,
respectively, the form given in (3.24) and (3.25). We can compute the precise
values of $\alpha$ and $\beta$ as follows. We have that

$Df_{x}(w_{f}(x))=\nabla g(f(x)),$

taking the inner product with $\nabla g(f(x))$

$\displaystyle\langle\nabla g(f(x)),Df_{x}(w_{f}(x))\rangle$
$\displaystyle=\lVert\nabla g(f(x))\rVert^{2}$ $\displaystyle\langle\nabla
h(x),w_{f}(x)\rangle$ $\displaystyle=\lVert 2f(x)\rVert^{2}$ by the chain
rule, $\displaystyle\alpha(x)\lVert\nabla h(x)\rVert^{2}$
$\displaystyle=4\lVert f(x)\rVert^{2}$ by (3.24) $\displaystyle\alpha(x)$
$\displaystyle=\frac{4\lVert f(x)\rVert^{2}}{\lVert\nabla h(x)\rVert^{2}}$
(3.31)

Analogously, we have that

$D\mathfrak{F}_{x}(w_{\mathfrak{F}}(x))=\nabla g(\mathfrak{F}(x)),$

taking the inner product with $\nabla g(\mathfrak{F}(x))$

$\displaystyle\langle\nabla
g(\mathfrak{F}(x)),D_{\mathfrak{F}}(w_{\mathfrak{F}}(x))\rangle$
$\displaystyle=\langle\nabla g(\mathfrak{F}(x)),\nabla
g(\mathfrak{F}(x))\rangle$
$\displaystyle\langle\nabla\mathfrak{H}(x),w_{\mathfrak{F}}(x)\rangle$
$\displaystyle=\lVert\nabla g(\mathfrak{F}(x))\rVert^{2}$ by the chain rule,
$\displaystyle\beta(x)\lVert\nabla\mathfrak{H}(x)\rVert^{2}$
$\displaystyle=\lVert 2\mathfrak{F}(x)\rVert^{2}$ by (3.25)
$\displaystyle\beta(x)4\lVert x\rVert^{2}$ $\displaystyle=4\lVert x\rVert^{2}$
$\displaystyle\beta(x)$ $\displaystyle=1.$ (3.32)

######  Remark 3.10.

Let $x\in M(f)$, then we have that $w_{f}(x)$ and $w_{\mathfrak{F}}(x)$ are
collinear, that is $w_{\mathfrak{F}}(x)=\mu(x)w_{f}(x)$, with $\mu(x)\neq 0$
(see (3.14)). Using the differential of the spherification map given in
Proposition 2.3 we can compute the value of $\mu(x)$. Since $w_{f}(x)\in
T_{x}E_{\ell_{x}}$ by Proposition 2.5 we have

$D\mathfrak{F}_{x}(w_{f}(x))=\frac{\langle x,w_{f}(x)\rangle}{\lVert
x\rVert^{2}}\mathfrak{F}(x)=\frac{\langle 2x,w_{f}(x)\rangle}{\lVert
2x\rVert^{2}}2\mathfrak{F}(x)=\frac{\langle\nabla\mathfrak{H}(x),w_{f}(x)\rangle}{\lVert\nabla\mathfrak{H}(x)\rVert^{2}}2\mathfrak{F}(x),$

thus,

$D\mathfrak{F}_{x}\left(\frac{\lVert\nabla\mathfrak{H}(x)\rVert^{2}}{\langle\nabla\mathfrak{H}(x),w_{f}(x)\rangle}w_{f}(x)\right)=2\mathfrak{F}(x)=\nabla
g(\mathfrak{F}(x)),$

and we have that

$\mu(x)=\frac{\lVert\nabla\mathfrak{H}(x)\rVert^{2}}{\langle\nabla\mathfrak{H}(x),w_{f}(x)\rangle}.$
(3.33)

On the other hand, by Corollary 2.6 we have that

$\displaystyle Df_{x}(w_{\mathfrak{F}}(x))$ $\displaystyle=\frac{f(x)}{\lVert
f(x)\rVert^{2}}\langle
f(x),Df_{x}(w_{\mathfrak{F}}(x))\rangle=\frac{2f(x)}{4\lVert
f(x)\rVert^{2}}\langle 2f(x),Df_{x}(w_{\mathfrak{F}}(x))\rangle$
$\displaystyle=\frac{\langle\nabla
g(f(x)),Df_{x}(w_{\mathfrak{F}}(x))\rangle}{4\lVert f(x)\rVert^{2}}\nabla
g(f(x))=\frac{\langle\nabla h(x),w_{\mathfrak{F}}(x)\rangle}{4\lVert
f(x)\rVert^{2}}\nabla g(f(x))$

thus,

$Df_{x}\left(\frac{4\lVert f(x)\rVert^{2}}{\langle\nabla
h(x),w_{\mathfrak{F}}(x)\rangle}w_{\mathfrak{F}}(x)\right)=\nabla g(f(x))$

hence by (3.15) we have that

$\frac{1}{\mu(x)}=\frac{4\lVert f(x)\rVert^{2}}{\langle\nabla
h(x),w_{\mathfrak{F}}(x)\rangle}.$ (3.34)

From (3.33) and (3.34) we have that

$\langle\nabla\mathfrak{H}(x),w_{f}(x)\rangle\langle\nabla
h(x),w_{\mathfrak{F}}(x)\rangle=4\lVert
f(x)\rVert^{2}\lVert\nabla\mathfrak{H}(x)\rVert^{2},$ (3.35)

which implies that for any $x\in M(f)$ the inner products
$\langle\nabla\mathfrak{H}(x),w_{f}(x)\rangle$ and $\langle\nabla
h(x),w_{\mathfrak{F}}(x)\rangle$ always have the same sign, as we also can see
from equations (3.16) and (3.17). By Proposition 3.6, if $\varepsilon$ is
small enough, these inner products are always positive.

######  Remark 3.11.

Let us revise what we have done. In Lemma 3.7 we gave an open cover
$\mathcal{U}=\\{U_{\lambda}\\}_{\lambda\in\Lambda}$ of
$\mathbbm{B}_{\varepsilon}^{n}\setminus V$ and in each open set $U_{\lambda}$
we constructed the vector field $w_{f,\lambda}$ satisfying (a), the vector
field $w_{\mathfrak{F},\lambda}$ satisfying (b) and they also satisfy (c). Let
$\\{\rho_{\lambda}\\}_{\lambda\in\Lambda}$ be a partition of unity
subordinated to the open cover $\mathcal{U}$. In Theorem 3.8 we constructed in
each open set $U_{\lambda}$ the vector field (using (3.32))

$\tilde{w}_{\lambda}(x)=\|\nabla\mathfrak{H}(x)\|w_{f,\lambda}(x)+\alpha(x)\|\nabla
h(x)\|w_{\mathfrak{F},\lambda}(x).$ (3.36)

which satisfies properties (1), (2) and (3) and using the partition of unity
we get the following global vector field with the same properties

$\displaystyle\tilde{w}(x)$
$\displaystyle=\sum_{\lambda\in\Lambda}\rho_{\lambda}(x)\tilde{w}_{\lambda}(x)=\sum_{\lambda\in\Lambda}\rho_{\lambda}(x)\bigl{(}\|\nabla\mathfrak{H}(x)\|w_{f,\lambda}(x)+\alpha(x)\|\nabla
h(x)\|w_{\mathfrak{F},\lambda}(x)\bigr{)},$
$\displaystyle=\|\nabla\mathfrak{H}(x)\|\sum_{\lambda\in\Lambda}\rho_{\lambda}(x)w_{f,\lambda}(x)+\alpha(x)\|\nabla
h(x)\|\sum_{\lambda\in\Lambda}\rho_{\lambda}(x)w_{\mathfrak{F},\lambda}(x).$
(3.37)

Is it easy to check that the vector fields
$\sum_{\lambda\in\Lambda}\rho_{\lambda}(x)w_{f,\lambda}(x)$ and
$\sum_{\lambda\in\Lambda}\rho_{\lambda}(x)w_{\mathfrak{F},\lambda}(x)$ are,
respectively, liftings to $\mathbbm{B}_{\varepsilon}^{n}\setminus V$, by $f$
and $\mathfrak{F}$ of the radial vector field $\nabla g$, and their weighted
sum (3.37) gives the vector field $\tilde{w}$, as it is mentioned in the proof
of [9, Lemma 5.2]. In other words, it turns out that with the proof of Theorem
3.8, the (incomplete) proof of [9, Lemma 5.2] can be seen as a sketch of a
proof.

######  Remark 3.12.

Notice that the construction of the vector fields $w_{f}$ and
$w_{\mathfrak{F}}$ with properties (a), (b) and (c) works in any open set
$\mathcal{U}$ of $\mathbbm{R}^{n}$ where $f$ and its spherefication
$\mathfrak{F}$ are submersions, and defining the vector field $\tilde{w}$ as
in (3.26) in the proof of Theorem 3.8 we get a vector field on $\mathcal{U}$
satisfying properties (1), (3) and (2). By Proposition 2.2 the spherefication
$\mathfrak{F}$ is a submersion if and only if the map $f/\|f\|$ is a
submersion, thus, the construction of the vector field $\tilde{w}$ works on
any open set $\mathcal{U}$ of $\mathbbm{R}^{n}$ where $f$ and $f/\|f\|$ are
submersions, as it was mentioned in the Introduction.

##  4\. Equivalence of the fibration on tube and on the sphere

In this section, using Theorem 3.8, we give for completeness the proof of the
equivalence between the fibration on the tube and the fibration on the sphere
for a real analytic map with isolated critical value.

Let $U$ be an open neighbourhood of $\underline{0}\in\mathbbm{R}^{n}$, $n>p$,
and $f\colon(U,\underline{0})\to(\mathbbm{R}^{p},0)$ a non-constant analytic
map defined on $U$ with a critical point at $\underline{0}\in\mathbbm{R}^{n}$
and $0\in\mathbbm{R}^{p}$ is an isolated critical value. Also assume that $f$
is locally surjective (see Remark 1.1). Let $\mathbbm{B}_{\varepsilon}^{n}$ be
a closed ball in $\mathbbm{R}^{n}$, centred at $\underline{0}$, of
sufficiently small radius $\varepsilon$, so that every sphere in this ball,
centred at $\underline{0}$, meets transversely every stratum of a Whitney
stratification of $V$, and such that Proposition 3.5 holds.

We say that $f$ satisfies the transversality property if there exists
$0<\delta\ll\varepsilon$ such that for every $y\in\mathbbm{B}_{\delta}^{p}$
the fibre $f^{-1}(y)$ meets $\mathbbm{S}_{\varepsilon}^{n-1}$ transversely.
Since we are assuming $f$ locally surjective we can take
$\mathbbm{B}_{\delta}^{p}\subset\operatorname{Im}(f|_{\mathbbm{B}_{\varepsilon}})$.

######  Remark 4.1.

If we work with a germ $f\colon(U,\underline{0})\to(\mathbbm{R}^{p},0)$
instead of with just a map, we need the following stronger transversality
property. We say that the germ $f$ satisfies the transversality property if
for any representative $f$ there exists a ball
$\mathbbm{B}_{\varepsilon_{0}}^{n}$ as above such that for every
$0<\varepsilon<\varepsilon_{0}$ there exists a real number
$\delta=\delta(\varepsilon)$ such that $0<\delta\ll\varepsilon$ and for every
$y\in\mathbbm{B}_{\delta}$ the fibre $f^{-1}(y)$ meets
$\mathbbm{S}_{\varepsilon}^{n-1}$ transversely.

######  Remark 4.2.

In the case that $f$ is not locally surjective, for the transversality
property we need to ask that every
$y\in\mathbbm{B}_{\delta}\cap\operatorname{Im}(f|_{\mathbbm{B}_{\varepsilon}})$
the fibre $f^{-1}(y)$ meets $\mathbbm{S}_{\varepsilon}^{n-1}$ transversely.
But if we consider that an empty fibre intersects transversely the sphere
$\mathbbm{S}_{\varepsilon}^{n-1}$ we can state in general the transversality
property as above.

######  Proposition 4.3.

If $f$ satisfies the transversality property, then the restriction of $f$ to
the Milnor tube

$f\colon N(\varepsilon,\delta)\rightarrow\mathbbm{S}_{\delta}^{p-1},$ (4.1)

is a smooth fibre bundle.

###### Proof.

Since $f$ satisfies the transversality property we can apply Ehresmann
Fibration Theorem for Manifolds with Boundary (see for instance [15, p. 23]).
∎

Consider the projection
$\pi\colon\mathbbm{B}_{\delta}^{p}\setminus\\{0\\}\to\mathbbm{S}^{p-1}$ given
by $\pi(y)=\frac{y}{\lVert y\rVert}$. We have that $\pi$ is a trivial fibre
bundle over $\mathbbm{S}^{p-1}$. Let $\tilde{f}$ be the composition of the
fibre bundle (4.1) with $\pi$, since the restriction of $\pi$ to
$\mathbbm{S}_{\delta}^{p-1}$ is a diffeomorphism, we get an equivalent fibre
bundle

$\tilde{f}\colon N(\varepsilon,\delta)\rightarrow\mathbbm{S}^{p-1}.$ (4.2)

Notice that $\tilde{f}$ is the restriction of the map $\Phi$ given in (2.2) to
the Milnor tube $N(\varepsilon,\delta)$.

######  Theorem 4.4.

Suppose $f\colon(U,\underline{0})\to(\mathbbm{R}^{p},0)$ satisfies the
transversality property. Then $f$ is $d$-regular if and only if the map

$\phi:=\frac{f}{\lVert f\rVert}\colon\mathbbm{S}_{\varepsilon}^{n-1}\setminus
K\to\mathbbm{S}^{p-1}$ (4.3)

is a smooth fibre bundle which is equivalent to the fibre bundle (4.2).

###### Proof.

Since $f$ has the transversality property, by Proposition 4.3 we have the
fibration (4.2) on the tube. If $f$ is $d$-regular, by Theorem 3.8 there
exists a vector field $\tilde{w}$ satisfying properties (1), (2) and (3). The
flow associated to $\tilde{w}$ defines a diffeomorphism $\tau$ between the
tube $N(\varepsilon,\delta)$ and $\mathbbm{S}_{\varepsilon}^{n-1}\setminus
f^{-1}(\mathring{\mathbbm{B}}_{\delta})$, where
$\mathring{\mathbbm{B}}_{\delta}$ is the open ball in $\mathbbm{R}^{p}$
centred at $0$ of radius $\delta$, “inflating” the tube to the sphere in the
following way: for a point $x\in N(\varepsilon,\delta)$ follow the integral
curve of $\tilde{w}$ which passes through $x$ until it reaches
$\mathbbm{S}_{\varepsilon}^{n-1}$ at some point $\hat{x}$, which exists and it
is unique because $\tilde{w}$ satisfies properties (1) and (2). Define
$\tau(x)=\hat{x}$. By property (3) the integral curves of $\tilde{w}$ lie on
the $E_{\ell}$, so we have that
$\tilde{f}(x)=\Phi(x)=\Phi(\hat{x})=\phi(\hat{x})$. Hence the diffeomorphism
$\tau$ gives an equivalence of fibre bundles $\tilde{f}=\phi\circ\tau$

${N(\varepsilon,\delta)}$${\mathbbm{S}_{\varepsilon}\setminus
f^{-1}(\mathring{\mathbbm{B}}_{\delta})}$${\mathbbm{S}^{p-1}}$$\scriptstyle{\tau}$$\scriptstyle{\tilde{f}}$$\scriptstyle{\phi}$

Now we extend the fibre bundle

$\phi\colon\mathbbm{S}_{\varepsilon}\setminus
f^{-1}(\mathring{\mathbbm{B}}_{\delta})\to\mathbbm{S}^{p-1}$ (4.4)

to $\mathbbm{S}_{\varepsilon}\setminus K$ as follows. Since
$\mathbbm{S}_{\varepsilon}\cap f^{-1}(\mathbbm{B}_{\delta}^{p})$ is compact,
the restriction of $f$

$f\colon\mathbbm{S}_{\varepsilon}\cap
f^{-1}(\mathbbm{B}_{\delta}^{p}\setminus\\{0\\})\to\mathbbm{B}_{\delta}^{p}\setminus\\{0\\}$

is a proper submersion, and by Ehresmann Fibration Theorem, it is a smooth
fibre bundle. Composing it with $\pi$, we get a fibre bundle given by the
restriction of $\phi$

$\phi|=\pi\circ f\colon\mathbbm{S}_{\varepsilon}\cap
f^{-1}(\mathbbm{B}_{\delta}^{p}\setminus\\{0\\})\to\mathbbm{S}^{p-1},$ (4.5)

since the composition of two smooth fibre bundles is again a smooth fibre
bundle (see [8, Lemma A.1], [20, Corollary 7] or [12]). We can glue the fibre
bundles (4.4) and (4.5) since they coincide in the intersection and both have
projection $\phi$, to get the smooth fibre bundle (4.3).

On the other hand, if (4.3) is a smooth fibre bundle, then it is a submersion,
and by Proposition 2.2 $f$ is $d$-regular. ∎

##  5\. Real analytic maps with linear discriminant

In this section, we see that Theorem 3.8 also holds for analytic maps with
linear discriminant and we use it to prove the equivalence of the fibration on
the tube and the fibration on the sphere for real analytic maps with arbitrary
linear discriminant (Theorem 5.9 below). This is done in Subsection 5.1.

###  5.1. Equivalence of fibrations for maps with arbitrary linear
discriminant

We start recalling some definitions and results from [8]. Let
$f\colon(\mathbbm{R}^{n},0)\to(\mathbbm{R}^{p},0)$ be an analytic map with a
critical point at $0$. Assume that $f$ is locally surjective (see Remark 1.1).
Equip $\mathbbm{R}^{n}$ with a Whitney stratification adapted to
$V=f^{-1}(0)$, and let $\mathbbm{B}_{\varepsilon_{0}}^{n}$ be a closed ball in
$\mathbbm{R}^{n}$, centred at $\underline{0}$, of sufficiently small radius
$\varepsilon_{0}>0$, so that every sphere in this ball, centred at
$\underline{0}$, meets transversely every stratum of $V$. In what follows, for
$0<\varepsilon<\varepsilon_{0}$ we shall consider the restriction
$f_{\varepsilon}$ of $f$ to the closed ball
$\mathbbm{B}_{\varepsilon}^{n}\subset\mathbbm{B}_{\varepsilon_{0}}^{n}$.
Denote by $\Sigma_{\varepsilon}$ the critical set of $f_{\varepsilon}$ and by
$\Delta_{\varepsilon}:=f_{\varepsilon}(\Sigma_{\varepsilon})$ the discriminant
of $f_{\varepsilon}$.

######  Remark 5.1.

  * $(i)$

The discriminant $\Delta_{\varepsilon}$ is a subanalitic set and it may depend
on the choice of the radius $\varepsilon$, as showed in [4].

  * $(ii)$

The results of this article extend to map-germs
$f\colon(\mathbbm{R}^{n},0)\to(\mathbbm{R}^{p},0)$ if we consider $f$ in the
class of nice analytic map-germs defined in [4, Definition 2.2] for which the
discriminant is a well-defined set-germ at $0$, so it does not depend on the
radius $\varepsilon$.

Let $0<\varepsilon<\varepsilon_{0}$, we say that the ball
$\mathbbm{B}_{\varepsilon}^{n}$ has the transversality property if there exist
$0<\delta\ll\varepsilon$ such that for every
$y\in\mathbbm{B}_{\delta}^{p}\setminus\Delta_{\varepsilon}$ the fibre
$f^{-1}(y)$ is transverse to the sphere $\mathbbm{S}_{\varepsilon}$. Since we
are assuming $f$ locally surjective we can take
$\mathbbm{B}_{\delta}^{p}\subset\operatorname{Im}(f_{\varepsilon})$.

######  Remark 5.2.

If we work with a nice analytic map-germ
$f\colon(U,\underline{0})\to(\mathbbm{R}^{p},0)$ (see Remark 5.1), the
discriminant does not depend on the radius $\varepsilon$ and we just denote it
by $\Delta$. In this case we need the following stronger transversality
property. We say that $f$ satisfies the transversality property if for every
$0<\varepsilon<\varepsilon_{0}$ there exists a real number
$\delta=\delta(\varepsilon)$ such that $0<\delta\ll\varepsilon$ and for every
$y\in\mathbbm{B}_{\delta}\setminus\Delta$ the fibre $f^{-1}(y)$ meets
$\mathbbm{S}_{\varepsilon}^{n-1}$ transversely.

######  Remark 5.3.

In the case that $f$ is not locally surjective for the transversality property
we need to ask that every
$y\in(\mathbbm{B}_{\delta}^{p}\setminus\Delta_{\varepsilon})\cap\operatorname{Im}(f_{\varepsilon})$
the fibre $f^{-1}(y)$ meets $\mathbbm{S}_{\varepsilon}^{n-1}$ transversely.
But if we consider that an empty fibre intersects transversely the sphere
$\mathbbm{S}_{\varepsilon}^{n-1}$ we can state in general the transversality
property as above.

Suppose that $f$ has the transversality property, then Proposition 4.3
generalizes so that the restriction

$f_{\varepsilon}|\colon\mathbbm{B}_{\varepsilon}^{n}\cap
f^{-1}(\mathbbm{S}_{\delta}^{p-1}\setminus\Delta_{\varepsilon})\to\mathbbm{S}_{\delta}^{p-1}\setminus\Delta_{\varepsilon}$
(5.1)

is a smooth fibre bundle [8, Theorem 2.7].

Let $0<\varepsilon<\varepsilon_{0}$, we say that $f$ has linear discriminant
in the ball $\mathbb{B}_{\varepsilon}^{n}$ if $\Delta_{\varepsilon}$ is a
union of line-segments with one endpoint at $0\in\mathbb{R}^{p}$. We say that
$\eta>0$ is a linearity radius for $\Delta_{\varepsilon}$ if each of these
line-segments intersect $\mathbbm{S}_{\eta}^{p-1}$, that is, if

$\displaystyle\Delta_{\varepsilon}\cap\mathbbm{B}_{\eta}^{p}=\operatorname{Cone}\big{(}\Delta_{\varepsilon}\cap\mathbbm{S}_{\eta}^{p-1}\big{)}\,.$

The case when $f$ has $0\in\mathbbm{R}^{p}$ as isolated critical value is
considered to have linear discriminant with
$\Delta_{\varepsilon}\cap\mathbbm{S}_{\eta}^{p-1}=\emptyset$.

Let $f\colon(\mathbbm{B}_{\varepsilon_{0}}^{n},0)\to(\mathbbm{R}^{p},0)$ be an
analytic map with linear discriminant in the ball
$\mathbb{B}_{\varepsilon}^{n}$ and consider a linearity radius $\eta>0$ for
$f$. Set:

$\mathcal{A}_{\eta}:=\Delta_{\varepsilon}\cap\mathbbm{S}_{\eta}^{p}.$

Let $\pi_{\eta}\colon\mathbbm{S}_{\eta}^{p}\to\mathbbm{S}^{p-1}$ be the radial
projection onto the unit sphere $\mathbbm{S}^{p-1}$ and set
$\mathcal{A}=\pi_{\eta}(\mathcal{A}_{\eta})$. For each point
$\vartheta\in\mathbbm{S}_{\eta}^{p-1}$, let
$\mathcal{L}_{\vartheta}\subset\mathbbm{R}^{p}$ be the open ray in
$\mathbbm{R}^{p}$ from the origin that contains the point $\vartheta$. Set:

$E_{\vartheta}:=f_{\varepsilon}^{-1}(\mathcal{L}_{\vartheta}).$

Notice that $E_{\vartheta}$ is a smooth manifold in
$\mathbbm{B}_{\varepsilon}^{n}$, for any $\vartheta$ in
$\mathbbm{S}_{\eta}^{p-1}\setminus\mathcal{A}_{\eta}$.

We say that $f$ is $d$-regular (in the ball $\mathbbm{B}_{\varepsilon}^{n}$)
if $E_{\vartheta}$ intersects the sphere
$\mathbbm{S}_{\varepsilon^{\prime}}^{n-1}$ transversely in $\mathbbm{R}^{n}$,
for every $\varepsilon^{\prime}$ with $0<\varepsilon^{\prime}\leq\varepsilon$
and for every
$\vartheta\in\mathbbm{S}_{\eta}^{p-1}\setminus\mathcal{A}_{\eta}$.

######  Remark 5.4.

In the case that $f$ has an isolated critical value at $0\in\mathbbm{R}^{p}$,
$f$ is $d$-regular (in the definition given in Section 2) if and only if there
exists $0<\varepsilon<\varepsilon_{0}$ such that $f$ is $d$-regular in the
ball $\mathbbm{B}_{\varepsilon}^{n}$.

######  Remark 5.5.

Let $f\colon(\mathbbm{B}_{\varepsilon_{0}}^{n},0)\to(\mathbbm{R}^{p},0)$ with
$n\geq p\geq 2$ be a locally surjective analytic map. Let
$0<\varepsilon<\varepsilon_{0}$ and suppose the ball
$\mathbb{B}_{\varepsilon}^{n}$ has the transversality property and $f$ has
linear discriminant in $\mathbb{B}_{\varepsilon}^{n}$. Notice that in this
case we can take the linearity radius to be the $\delta$ in the definition of
the transversality property.

The following examples are also from [8].

######  Example 5.6.

The real analytic map $f\colon(\mathbbm{R}^{4},0)\to(\mathbbm{R}^{3},0)$

$f(x,y,z,w):=(x^{2}-y^{2}z,y,w)\,.$

has the plane $\\{x=y=0\\}$ in $\mathbbm{R}^{4}$ as its critical set, its
discriminant is the axis $\\{u_{1}=u_{2}=0\\}$ in $\mathbbm{R}^{3}$, so it is
linear. One can check that $f$ is not $d$-regular.

######  Example 5.7.

Let ${\mathbbm{K}}$ be either $\mathbbm{R}$ or $\mathbbm{C}$. Let
$(f,g):{\mathbbm{K}}^{n}\to{\mathbbm{K}}^{2}$ be a ${\mathbbm{K}}$-analytic
map of the form:

$(f,g)=\left(\sum_{i=1}^{n}a_{i}x_{i}^{p}\,,\,\sum_{i=1}^{n}b_{i}x_{i}^{p}\right)\,,$

where $p\geq 2$ is an integer and $a_{i},b_{i}\in{\mathbbm{K}}$ are constants
in generic position, i. e., the origin is in the convex hull of the points
$(a_{i},b_{i})$ (which guarantees that the link of $V(f)$ is non-empty) and no
two of the points $(a_{i},b_{i})$ are linearly dependent, that is,
$a_{i}b_{j}\neq a_{j}b_{i}$, for any $i\neq j$ (Weak Hyperbolicity
Hypothesis). Its critical set $\Sigma$ is given by the coordinate axis of
${\mathbbm{K}}^{n}$, the discriminant $\Delta$ is linear and it is $d$-regular
(see [8, Examples 3.6 and 3.7]). For $p=2$ this maps were studied by S. López
de Medrano in [18].

Let $0<\varepsilon<\varepsilon_{0}$ be such that $f$ has linear discriminant
and it is $d$-regular in the ball $\mathbbm{B}_{\varepsilon}^{n}$. Consider
the maps $\Phi\colon\mathbbm{B}_{\varepsilon}^{n}\setminus
f^{-1}(\Delta_{\varepsilon})\to\mathbbm{S}^{p-1}\setminus\mathcal{A}$ and
$\mathfrak{F}\colon\mathbbm{B}_{\varepsilon}^{n}\setminus
f^{-1}(\Delta_{\varepsilon})\to\mathbbm{R}^{p}\setminus\Delta_{\varepsilon}$
given in (2.2) but restricted to $\mathbbm{B}_{\varepsilon}^{n}\setminus
f^{-1}(\Delta_{\varepsilon})$. Hence Proposition 2.2 generalizes in a
straightforward way for maps with linear discriminant, removing
$f^{-1}(\Delta_{\varepsilon})$ from the domain of the maps involved instead of
only removing $V$ [8, Proposition 3.8]. Thus, the maps $\mathfrak{F}$ and
$\phi=\Phi|\colon\mathbbm{S}_{\varepsilon}^{n-1}\setminus
f^{-1}(\Delta_{\varepsilon})\to\mathbbm{S}^{p-1}\setminus\mathcal{A}$ are
submersions if and only if $f$ is $d$-regular. In fact, we have the following
theorem.

######  Theorem 5.8 ([8, Theorem 3.9]).

Let $f\colon(\mathbbm{B}_{\varepsilon_{0}}^{n},0)\to(\mathbbm{R}^{p},0)$ with
$n\geq p\geq 2$ be a locally surjective analytic map. Let
$0<\varepsilon<\varepsilon_{0}$ and suppose the ball
$\mathbb{B}_{\varepsilon}^{n}$ has the transversality property and $f$ has
linear discriminant in $\mathbb{B}_{\varepsilon}^{n}$. If $f$ is $d$-regular
then the restriction of $\Phi$ given by

$\phi=\Phi|\colon\mathbbm{S}_{\varepsilon}^{n-1}\setminus
f^{-1}(\Delta)\to\mathbbm{S}^{p-1}\setminus\mathcal{A}$ (5.2)

is a smooth locally trivial fibration.

The construction of the vector field $\tilde{w}$ of Theorem 3.8 works in any
open set of $\mathbbm{R}^{n}$ where $f$ and the spherefication $\mathfrak{F}$
are submersions (see Remark 3.12), so in Subsection 3.2 we can replace
$\mathring{\mathbbm{B}}_{\varepsilon}^{n}\setminus V$ by
$\mathring{\mathbbm{B}}_{\varepsilon}^{n}\setminus
f^{-1}(\Delta_{\varepsilon})$. Using this generalization of Theorem 3.8 we
prove the equivalence of fibrations (5.1) and (5.2) for real analytic maps
with arbitrary linear discriminant as in Theorem 4.4.

######  Theorem 5.9.

Let $f\colon(\mathbbm{B}_{\varepsilon_{0}}^{n},0)\to(\mathbbm{R}^{p},0)$ with
$n\geq p\geq 2$ be a locally surjective analytic map. Let
$0<\varepsilon<\varepsilon_{0}$ and suppose the ball
$\mathbb{B}_{\varepsilon}^{n}$ has the transversality property, $f$ has linear
discriminant and it is $d$-regular in $\mathbb{B}_{\varepsilon}^{n}$. Then the
fibre bundles:

$\displaystyle\tilde{f}:=\pi_{\delta}\circ
f_{\varepsilon}|\colon\mathbbm{B}_{\varepsilon}^{n}\cap
f^{-1}(\mathbbm{S}_{\delta}^{p-1}\setminus\mathcal{A}_{\delta})\to\mathbbm{S}^{p-1}\setminus\mathcal{A}$
and $\displaystyle\phi\colon\mathbbm{S}_{\varepsilon}^{n-1}\setminus
f^{-1}(\Delta)\to\mathbbm{S}^{p-1}\setminus\mathcal{A}$

are equivalent, where
$\pi_{\delta}\colon\mathbbm{S}_{\delta}^{p-1}\to\mathbbm{S}^{p-1}$ is the
radial projection.

######  Remark 5.10.

In Theorem 5.9 we take the linearity radius to be $\delta$ as in Remark 5.5.

######  Remark 5.11.

In the PhD thesis [25] and in the articles [4, 5] the authors study nice real
analytic maps-germs $f\colon(\mathbbm{R}^{n},0)\to(\mathbbm{R}^{p},0)$ with
discriminant of positive dimension obtaining some results in common with [8].
They prove the existence of the fibration on the tube (5.1) if $f$ satisfies a
condition equivalent to the transversality property [4, Lemma 3.3]. They also
consider real analytic map-germs with linear discriminant (they call it with
radial discriminant) and if they have fibration on the tube and are
$d$-regular (they call it $\rho$-regular) then the fibration on the sphere
(5.2) exists. Then they consider the problem of finding conditions for which
the two fibrations are equivalent [5, Problem 4.1] and prove that the problem
reduces to find a vector field which satisfies properties (1), (3) and (2)
(which they call a good vector field or Milnor vector field). The authors
conjecture that if $f$ is a real analytic map which is $d$-regular and for
which both fibrations (5.1) and (5.2) exists, then they are equivalent [5,
Equivalence Conjecture 4.4]. Theorem 5.9 above proves that the conjecture is
true.

######  Remark 5.12.

In [25, 5] the authors prove that for fixed
$\vartheta\in\mathbbm{S}_{\eta}^{p-1}\setminus\mathcal{A}_{\eta}$ the
corresponding fibres of fibrations (5.1) and (5.2) are diffeomorphic but
without being able to prove the equivalence of fibrations since they could not
find a radius $\varepsilon_{1}$ which works for every
$\vartheta\in\mathbbm{S}_{\eta}^{p-1}\setminus\mathcal{A}_{\eta}$.

## Acknowledgments

The first author was supported by CONACYT 253506 and a UNAM-DGAPA-PASPA
sabbatical scholarship. He wants to thank Prof. Hans Brodersen for the helpful
email correspondence. Both authors want to thank José Seade and Jawad Snoussi
for many enlightening conversations.

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