**Week 5**

1. Definition (Transversal intersection). Consider a smooth map
*f*: *M* --> *N* and suppose *N*
has a submanifold *P* of co-dimension *k*.
This means that *P* is covered by coordinate charts (for *N*)
*h*: *U* --> R^*n* such that *P* intersect
*U* is (*p* o *h*)^(-1)(0), where
*p*: R^*n* --> R^*k* is projection on the first
*k* coordinates. We will
say that *f* is transverse to *P* (or that *f*(*M*)
intersects *P*
transversely) if each such (*p* o *h* o *f*):
*f*^(-1)(*U*) --> R^*k* has 0
as a regular value.

Proposition. In that case, *f*^(-1)(*P*) is a smooth submanifold (in
*M*) of codimension *k*.

Our next goal is a proposition which states that any smooth
map R^*n* --> R^*p* can be approximated arbitrarily closely by one
which has 0 as a regular value. This is the core of the proof
of a theorem we will skip, which states that any *f*: *M*
--> *N* as
above can be approximated arbitrarily closely by one which
in transverse to *P* (see Milnor,
Thm. 1.35).

Theorem (Milnor, Lemma 1.19)
In *M*(*n*,*p*), the
*n**p*-dimensional vectorspace of *n* x *p*
matrices with real entries, the
subspace *M*(*n*,*p*;*k*) of matrices of rank *k*
is a smooth submanifold of dimension *k*(*n*+*p*-*k*).

Examples: In *M*(2,2) the space of matrices of rank 1 has dimension 3.
The three dimensions can be described as follows. One for the null-space,
a point in R*P*^1; one for the image, a point in R*P*^1;
and one for the
scaling between the 1-dimensional spaces (null-space complement)-->
(image). Similarly *M*(3,3;1) has dimension 5.

Definition: A subset *A* of R^*n* has *measure zero* if it can
be covered by a countable collection of cubes of arbitrarily
small total volume.

Class exercise: The *x*-axis in R^2 has measure zero.

Proposition: (Milnor, Lemma 1.14)
If *f*:R^*n* --> R^*n* is smooth, and *A* in R^*n*
has measure
zero, then *f*(*A*) has measure zero.

2. Theorem (Milnor, Thm. 1.21)
Consider a smooth *f*: *U* --> R^*p*, where
*U* is open in R^*n*. Then for any epsilon > 0
there exist an *n* x *p* matrix *A* and a
*p*-vector *B*, both with
maximum entry < epsilon, such that the map
*g*: *U* --> R^*p*
defined by *g*(*x*) = *f*(*x*) + *A*x* * +
*B* has 0 as a regular value.

(Milnor shows
that for each *k* < *p* the map *F*_*k*: *M*
(*n*,*p*;*k*) x *U* --> *M*(*n*,*p*)
x R^*p*
given by *F*_*k*(*K*,*x*) =
(*K*-D*f*(*x*),-*f*(*x*)-
(*K*-D*f*(*x*))*x*) has image of
measure zero; it follows that there exist *A*,*B* within epsilon
of (0,0) and not in the image of any *F*_*k*. Use these to define
*g*(*x*) as above, and suppose 0 was *not* a regular value
for *g*. Then there would exist an *x* in *U* with
*g*(*x*)=0 and
D*g*(*x*) of rank *k* < *p*.
Since D*g*(*x*) = D*f*(*x*) + *A*, this means
that *A* is of the form *K* - D*f*(*x*),
where *K* is in *M*(*n*,*p*;*k*); since
*g*(*x*) =
*f*(*x*) + *A**x* + *B* = *f*(*x*) +
(*K*-D*f*(*x*))*x* +*B* = 0, this means
that *B* = -*f*(*x*)-(*K*-D*f*(*x*))*x*,
so (*A*,*B*) is in the
image of *F*_*k*, contrary to our choice.)

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