Randell A1: Defn: Suppose f : M → N where M and N are smooth manifolds. f is smooth if for all p ∈ M, ∃ charts (ϕ, U) and (φ, V) and such that p ∈ U, f(p) ∈ V, f(U) ⊂ V and φ ∘ f ∘ ϕ^{−1} is smooth. Note this definition is equivalent to: Suppose f : M → N where M and N are smooth manifolds. f is smooth if for all p ∈ M and for all charts (ϕ, U) and (φ, V) such that p ∈ U, f(p) ∈ V, f(U) ⊂ V, then φ ∘ f ∘ ϕ^{−1} is smooth. Suppose g : M → W and f : W → N are smooth. Let M be an m-manifold, W a k-manifold, and N an n-manifold. Claim f ∘ g : M → N is smooth. Let p ∈ M. g smooth implies ∃ charts (ϕ_{1}, U_{1}) and (φ_{1}, V_{1}) such that p ∈ U_{1}, ϕ_{1}(p) ∈ V_{1}, g(U_{1}) ⊂ V_{1} and φ_{1} ∘ g ∘ ϕ_{1}^{−1} : ϕ(U_{1}) ⊂ R^{m} → φ_{1}(V_{1}) ⊂ R^{k} is smooth. Let (φ_{2}, V_{2}) be a chart such that f(g(p)) ∈ V_{2}. Let V_{3} = f|_{W}^{−1}(V_{2}) ∩ V_{1}. Then g(p) ∈ V_{3} and f(V_{3}) ⊂ V_{2} f smooth implies f is continuous. Thus V_{3} is open in W and (φ_{1}|_{V_{3}}, V_{3}) is a chart. f smooth implies φ_{2} ∘ f ∘ φ_{1}^{−1} : φ_{1}(V_{3}) ⊂ R^{k} → φ_{2}(V_{2}) ⊂ R^{n} is smooth. Thus (φ_{2} ∘ f ∘ φ_{1}^{−1}) ∘ (φ_{1} ∘ g ∘ ϕ_{1}^{−1}) = φ_{2} ∘ f ∘ g ∘ ϕ_{1}^{−1} : ϕ(U_{1}) ⊂ R^{m} → φ_{2}(V_{2}) ⊂ R^{n} is smooth. Thus f ∘ g is smooth

A2 T_{p}(M) = {v : G(p) → R | v is linear and satisfies the Leibniz rule } Given a chart (U, ϕ) at p where ϕ(p) = 0, the standard basis for T_{p}(M) = {v_{1}, ..., v_{m}}, where v_{i} = D_{α_{i}} and for some ϵ > 0, α_{i} : (−ϵ, ϵ) → M, α_{i}(t) = ϕ^{−1}(0, ..., t, ..., 0)

B1) Let U = Γ(f), ϕ : Γ(f) → R^{2}, ϕ(x, y, z) = (x, y). Since the domain of f is R^{2}, Γ(f) is onto. Γ(f) ⊂ Γ(f), Γ(f) is open in Γ(f), and R^{2} is open in R^{2}, thus if ϕ is a homeomorphism, {(ϕ, Γ(f))} is a pre-atlas. Suppose ϕ(x_{1}, y_{1}, z_{1}) = ϕ(x_{2}, y_{2}, z_{2}). Then (x_{1}, y_{1}) = ϕ(x_{1}, y_{1}, z_{1}) = ϕ(x_{2}, y_{2}, z_{2}) = (x_{2}, x_{2}). Also z_{1} = f(x_{1}, y_{1}) = f(x_{2}, x_{2}) = z_{2}. Thus x_{1} = x_{2}, y_{1} = y_{2}, z_{1} = z_{2}. Hence ϕ is 1:1. Let π_{xy} : R^{3} → R^{2}, π_{xy}(x, y, z) = (x, y). If U is open in R^{2}, π_{xy}^{−1}(U) = U × R which is open in R^{3}. Thus π_{xy} is continuous and ϕ = π_{xy}|_{Γ(f)} is continuous. Let V be open in R^{2}. Then π_{xy}|_{Γ(f)}(V) = (V × R) ∩ Γ(f) is open in Γ(f). Thus ϕ : Γ(f) → R^{2} is a homeomorphism. Since Γ(f) has a pre-atlas, Γ(f) is a smooth manifold.