Under the assumptions of Lemma 1.3,
has a finite number of connected components, each of which satisfies the same assumptions as
itself. Thus, with no loss of generality, we will assume that
is connected.

If
and
are
-manifolds of class
with
and
and
are
diffeomorphic, they are also
diffeomorphic ([10, Theorem
, page 57]). Thus, since
is of class
with
it suffices to find a bounded open subset
of
such that
is
and
diffeomorphic to

In a first step, we find a

function

such that

and

on

while

in

,

in

and

This can be done in various ways and even when

However, since

the most convenient argument is to rely on the fact that the signed distance function

is an open neighborhood of
in
This is shown in Gilbarg and Trudinger [11, page 355] and also in Krantz and Parks [12]. Both proofs reveal that
when
that is, when
(Without further assumptions, the
regularity of
breaks down when
)

Let
be nondecreasing and such that
if
and
if
where
is given. Then,
is
in
vanishes only on
, and
on
Furthermore, since
on a neighborhood of
in
and
on a neighborhood of
in
,
remains
after being extended to
by setting
if
and
if

This
satisfies all the required conditions except
Since
for
large enough, this can be achieved by replacing
by
Since
off
it follows from a classical theorem of Whitney [13, Theorem III] (with
in that theorem) that there is a
function
on
of class
in
such that, if
then
if
and
if

Evidently,
does not vanish on
and
has the same sign as
off
, that is,
in
and
in
Furthermore,
for every
so that
for
for some
Upon shrinking
we may assume that
Also,
For convenience, we summarize the relevant properties of
below:

(i)
is
on
and
off
,

(ii)
for
,

(iii)
,

(iv)
,

(v)

Choose
It follows from (v) that
is compact and, from (iii) and (iv), that
if
is small enough (argue by contradiction). Since
by (iii) and (iv) and since
this implies
Thus, by (i) and (ii),
is a
submanifold of
and the boundary of the open set
In fact,
is a
-manifold of class
since, once again by (ii),
lies on one side of its boundary.

We now proceed to show that
is
diffeomorphic to
This will be done by a variant of the procedure used to prove that nearby noncritical level sets on compact manifolds are diffeomorphic. However, since we are dealing with sublevel sets and since critical points will abound, the details are significantly different.

Let

be such that

and

on

Since

on

by (ii), the function

extended by

outside

is a bounded

vector field on

Since

the function

defined by

is well defined and of class
and
is an orientation-preserving
diffeomorphism of
for every
We claim that
produces the desired diffeomorphism from
to

It follows at once from (3.3) that
so that
is decreasing along the flow lines and hence that
maps
into itself for every
Also, if
then
for every
so that
by (iii). If now
then
and
is strictly decreasing for
small enough. It follows that
that is,
for
Altogether, this yields

Suppose now that
Then,
and hence
For
small enough,
and so
for
small enough. In fact, it is obvious that
until
is large enough that
But since
and
is decreasing along the flow lines,
implies
Since
this means that
for some
Call
the first (and, in fact, only, but this is unimportant) time when
From the above,
for
and hence for
since
Then,
for
so that
for
In particular, since
and hence
it follows that
In other words,
Thus,
that is,
If
(so that
and hence
), this yields
On the other hand, if
then
Since
,
is strictly decreasing for
near
and so
whence

The above shows that
maps
into
,
into
, and
into
That it actually maps
*onto*
follows from a Brouwer's degree argument:
is connected and no point of
is in
since, as just noted,
Thus, for
is defined and independent of
Now, choose
so that
Then,
for every
and so
Since
is one to one and orientation-preserving, it follows that
and so
for every
Thus, there is
such that
which proves the claimed surjectivity.

At this stage, we have shown that
is a
diffeomorphism of
mapping
into
,
into
, and
into and onto
It is straightforward to check that such a diffeomorphism also maps
onto
(approximate
by a sequence from
) and hence it is a boundary-preserving diffeomorphism of
onto
This completes the proof of Lemma 1.3.

Remark 3.1.

The
diffeomorphism
above is induced by a diffeomorphism of
but this does not mean that the same thing is true of the
diffeomorphism of Lemma 1.3.