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13, we only need to prove that G is an embedded submanifold. Certainly, ϕ is injective as ϕ (X, Z) = ϕ (X , Z ) means X = X , XZ = X Z , and since rank X = k, the latter equation implies Z = Z . Next we prove that ϕ : M × GL(k, R) → G is a homeomorphism. Assume lim ϕ (Xh , Zh ) = lim (Xh , Xh Zh ) h→∞ h→∞ = (X,Y ). Hence lim h→∞ Xh = X. As G is closed in M × M, there exists Z ∈ GL(k, R) such that Y = XZ. We only need to prove that lim h→∞ Zh = Z. Set Xh = (v1,h , . . , vk,h ), X = (v1 , . . , vk ).

18 (a) σ is not an immersion. (b) σ is a non-injective immersion. Fig. 19 (c) σ is an embedding. (d) σ is a non-injective immersion. (c) σ is an immersion as σ (t) = (−2π sin 2π t, 2π cos 2π t, 1) = (0, 0, 0), ∀t ∈ R. 19). (d) σ is an immersion since σ (t) = (−2π sin 2π t, 2π cos 2π t) = (0, 0) for all t, but σ is obviously not injective. Nevertheless, σ (R) is an embedded submanifold. 5 Immersions, Submanifolds, Embeddings and Diffeomorphisms 39 Fig. 20 (e) σ is an embedding. (f) σ is an immersion.

8, there exist local coordinates (x1 , . . , xm ), (y1 , . . , yn ), centered at p0 , q0 in M, N, respectively, such that yi ◦ π = xi , 1 i n. Notice that m n, as π is a submersion. Hence we can define a map σ on the domain of (y1 , . . , yn ) by setting xi ◦ σ = yi 0 if 1 i if n + 1 n i m. Then, for every i = 1, . . , n, we have yi ◦ (π ◦ σ ) = (yi ◦ π ) ◦ σ = xi ◦ σ = yi , thus proving that σ is a local section of π . 1. Prove that if σ is a C∞ curve in the C∞ manifold M, then the tangent vector field σ is a C∞ curve in the tangent bundle T M.

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