Question

differential TOPOLOGY

7. Zeros of vector fields may be "split" by perturbations just like fixed points of maps. Suppose that x is a zero of u and U is a neighborhood of x in X containing no other zero .of v. Prove that there exists a vector field y, that agrees with v outside some compact subset of U and that has only nondegenerate fixed points inside U. (HINT: Reduce to the case where U is an open subset of Rk. Pick a function p that is 1 near x and O outside a compact subset of U, and define 0.(z)-0(z) + A(z)ā, a E R'. Show that if á is small enough, 7, can be zero only where p = 1. Now demand that -a be a regular value of v.]

Answer 1

. The proof is by induction on n which is the dimension of M. The starting point is n = 0 which is trivial. Therefore suppose that the theorem has been proved for all manifolds of dimensions ≤ n − 1. Next note using the Second Axiom of Countability that it suffices to consider only the special case when f : U −→ R m, U being an open set of R n +, and C is the critical set of f in U. In view of Lemma , we may suppose that U is an open set in the interior of R n +, or in R n. Let D be the set of points in C where the Jacobian matrix J(f) vanishes. We shall show in the next two Lemmas 4.8 and 4.9 that both f(D) and f(C − D) have measure zero in R m. This will complete the proof of the theorem.