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Unformatted text preview: Lecture 27 We proved the following Poincare Lemma: Poincare Lemma. Let U be a connected open subset of R n , and let ω ∈ Ω n c ( U ) . The following conditions are equivalent: 1. ω = 0 , U 2. ω = dµ , for some µ ∈ Ω n − 1 ( U ) . c We first proved this for the case U = Int Q , where Q was a rectangle. Then we used this result to generalize to arbitrary open connected sets. We discussed a nice application: proper maps and degree. Let U, V be open subsets of R n , and let f : U → V be a C ∞ map. The map f is proper if for every compact set C ⊆ V , the preimage f − 1 ( C ) is also compact. Hence, if f is proper, then f ∗ Ω c k ( V ) ⊆ Ω c k ( U ) . (5.88) That is, if ω ∈ Ω c k ( V ), then f ∗ ω ∈ Ω c k ( U ), for all k . When k = n , ω ∈ Ω n c ( V ) . (5.89) In which case, we compare ω and f ∗ ω. (5.90) v U Using the Poincare Lemma, we obtain the following theorem. Theorem 5.14. There exists a constant γ f with the property that for all ω ∈ Ω n c ( V ) , f ∗ ω = γ f ω. (5.91) U V We call this constant the degree of f , Definition 5.15. deg( f ) = γ f . (5.92) Let U, V, W be open connected subsets of R n , and let f : U → V and g : V W → be proper C ∞ maps. Then the map g ◦ f : U → W is proper, and deg( g ◦ f ) = deg( f ) deg( g ) . (5.93) Proof Hint: For all ω ∈ Ω c n ( W ), ( g ◦ f ) ∗ ω = f ∗ ( g ∗ ω )....
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 Fall '04
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 Topology, Neighbourhood, Open set, deg

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