Let be a embedded submanifold. Let be the normal bundle of , defined by Thm: is an embedded -dimensional submanifold of . Let be the zero section.
There is a shift map: One can check that for any , is an isomorphism. Hence is a diffeomorphism in a neighborhood of . Since is arbitrary, we have is a diffeomorphism in a neighborhood . More precisely, we want to constraint the shape of . We define
Tubular neighborhood of in is a neighborhood of , that is the diffeomorphic image under of an open subset of the form for some positive continuous function .
See Thm 6.24 for a proof of the existence of a tubular neighborhood. Mainly, one try to construct the 'radius' function and show that it is continuous.
We define a map , that is identity map when restrict to , such a map is called a retraction. , where is the projection map.
Let be a finite dimensional vector space, be two subspaces, we say and are transversal, if .
Consider two embedded submanifolds in . Suppose . If , then intuition tells us that and 'generically' does not intersect(no intersection is a special case of transverse intersection). If , then 'generically' and should intersects transversally, that is, for any , we should have . But what does 'generically' mean? Let's make it more precise.
Let's consider a more general setting. Let be an embedded submanifold, and let be any smooth map. We say that is transverse to , if for every , we have
To understand the following theorem, we need to have
Theorem (6.30): (a) If is a smooth map between smooth manifolds, embedded manifold, and is transverse to , then is an embedded submanifold of with the same codimension as in
(b) If are two embedded submanifolds, that intersects transversely, then is an embedded submanifold of , with .
Sketch of proof: (b) follows from (a) by taking , and , composition of embedding is embedding.
To prove (a), we take a point , and take slice coordinate chart of adapted to , that is, can be written as where . Denote . Then, consider , thus . One want to show that has a regular value, i.e for any , is surjective. That is guaranteed by transversality condition.
Next, one consider the parametrix transversality theorem.
Let be smooth manifolds. A smooth family of maps parametrized by is a smooth map , whereas .
Theorem (6.35) Let be an embedded submanifold, and let be a smooth map. If is transversal to , then for almost every , is transverse to .
Proof: Let . Let , and its restriction to . By Sard theorem, the critical value of is negligible. Suffice to prove that for any point away from the critical value , is transversal to . (remember that if , it is still transverse intersection).
Let's do some simple dimension counting. Suppose , then . Let , . If , then , then for most , and we are done.
Now assume . We can see that is a critical value, if and only if intersects non-transversely (hence intersection is non-empty). .
Theorem 6.36 (Transversality Homotopy Theorem). Suppose and are smooth manifolds and is an embedded submanifold. Every smooth map is homotopic to a smooth map that is transverse to .
Sketch of proof: We are going to let , -dimensional unit open ball, and build a smooth map that is a submersion, hence transverse to . First, we use Whitney embedding theorem to embed to , . Create tubular neighborhood of in . Then, we will wiggle the composition in all the direction, within , then project back to .