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Unformatted text preview: 5.4. DEGREE OF A MAP 135 2. Suppose ϕ 1 ( y ) is infinite. 3. Then by compactness argument ϕ 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 ∈ ϕ 1 ( y ), converging to some x , x k → x , so that ϕ ( x ) = y . Thus x is a regular point of M . 4. Since ϕ * : T x M → T y V is bijective at x , the map ϕ : M → V is a di ff eomorphism in a neighborhood of x . That is, det ϕ *  x , 0. 5. This contradicts the fact that there is sequence of points x k → x such that ϕ ( x k ) = y . Since, otherwise there exist infinitely many points x k ∈ M such that ϕ ( x k ) = ϕ ( x ) = y contradicting the fact that ϕ 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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 Spring '10
 Wong
 Geometry, Topology, vy

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