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Unformatted text preview: 5.4. DEGREE OF A MAP 135 2. Suppose - 1 ( y ) is infinite. 3. Then by compactness argument - 1 ( y ) has a limit point x M , which is a regular point. Indeed, every sequence has a convergent subsequence. Thus, there is a sequence ( x k ), such that x k - 1 ( y ), converging to some x , x k x , so that ( x ) = y . Thus x is a regular point of M . 4. Since * : T x M T y V is bijective at x , the map : M V is a di ff eomorphism in a neighborhood of x . That is, det * | x , 0. 5. This contradicts the fact that there is sequence of points x k x such that ( x k ) = y . Since, otherwise there exist infinitely many points x k M such that ( x k ) = ( x ) = y contradicting the fact that is bijective. 6. (II). Claim: The point y V has a neighborhood whose inverse image is a disjoint union of neighborhoods of the preimages of y , each of which is di ff eomorphic to V y ....
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This note was uploaded on 11/26/2011 for the course MAT 4821 taught by Professor Wong during the Spring '10 term at FSU.
- Spring '10