Preliminary exam prep
Let X,Y be manifolds embedded in Rn, and f:X→Y a map, with dfx:TxX→Tf(x)Y the induced map on tangent spaces.
Definition: The map f is a
Definition: The manifolds X and Y are transverse if TpX⊕TpY≅Rn for every p∈X∩Y. The map f and Y are transverse if im(f) and Y are transverse.
Note that being transverse (or transversal) is a symmetric, but not a reflexive, nor a transitive relation. Recall that a regular value of f is y∈Y such that dfx:TxX→Tf(x)Y is surjective for all x∈f−1(y). If y is not in the image of f, then f−1(y) is empty, so y is trivially a regular value. Every value that is not a regular value is a critical value.
Theorem: (Preimage theorem) For every regular value y of f, the subset f−1(y)⊂X is a submanifold of X of dimension dim(X)−dim(Y).
Now let M be a submanifold of Y.
Corollary: If f is transverse to M, then f−1(M) is a manifold, with codimY(M)=codimX(f−1(M)).
Theorem: (Transversality theorem) Let {gs:X→Y | s∈S} be a smooth family of maps. If g:X×S→Y is transverse to M, then for almost every s∈S the map gs is transverse to M.
If we replace f with df, and ask that it be transverse to M, then df|s is also transverse to M.
Example: Consider the map gs:X→Rn given by gs(X)=i(X)+s=X+s, where i is the embedding of X into Rn. Since g(X×Rn)=Rn and g varies smoothly in both variables, we have that g is transverse to X. Hence by the transversality theorem, X is transverse to its translates X+s for almost all s∈Rn.
Theorem: (Sard) For f smooth and ∂Y=∅, almost every y∈Y is a regular value of f and f|∂X. Equivalently, the set of critical values of f has measure zero.
Resources: Guillemin and Pollack (Differential topology, Chapters 1, 2), Lee (Introduction to smooth manifolds, Chapter 6)
Definition: The map f is a
- homeomorphism if it is continuous and has a continuous inverse,
- diffeomorphism if it is smooth and has a smooth inverse,
- injection if f(a)=f(b) implies a=b,
- immersion if dfx is injective for all x∈X,
- embedding if it is an immersion and dfx is a homeomorphism onto its image,
- submersion if dfx is surjective for all x∈X.
Definition: The manifolds X and Y are transverse if TpX⊕TpY≅Rn for every p∈X∩Y. The map f and Y are transverse if im(f) and Y are transverse.
Note that being transverse (or transversal) is a symmetric, but not a reflexive, nor a transitive relation. Recall that a regular value of f is y∈Y such that dfx:TxX→Tf(x)Y is surjective for all x∈f−1(y). If y is not in the image of f, then f−1(y) is empty, so y is trivially a regular value. Every value that is not a regular value is a critical value.
Theorem: (Preimage theorem) For every regular value y of f, the subset f−1(y)⊂X is a submanifold of X of dimension dim(X)−dim(Y).
Now let M be a submanifold of Y.
Corollary: If f is transverse to M, then f−1(M) is a manifold, with codimY(M)=codimX(f−1(M)).
Theorem: (Transversality theorem) Let {gs:X→Y | s∈S} be a smooth family of maps. If g:X×S→Y is transverse to M, then for almost every s∈S the map gs is transverse to M.
If we replace f with df, and ask that it be transverse to M, then df|s is also transverse to M.
Example: Consider the map gs:X→Rn given by gs(X)=i(X)+s=X+s, where i is the embedding of X into Rn. Since g(X×Rn)=Rn and g varies smoothly in both variables, we have that g is transverse to X. Hence by the transversality theorem, X is transverse to its translates X+s for almost all s∈Rn.
Theorem: (Sard) For f smooth and ∂Y=∅, almost every y∈Y is a regular value of f and f|∂X. Equivalently, the set of critical values of f has measure zero.
Resources: Guillemin and Pollack (Differential topology, Chapters 1, 2), Lee (Introduction to smooth manifolds, Chapter 6)
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