By Renteln P.
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Extra info for Solution manual for Manifolds, tensors, and forms
Ir j1 ... ir j1 ... ir j1 ... ir j1 ... ir j1 ... ir j1 ... js (+X js ,k )∂i1 ⊗ · · · ⊗ ∂ir ⊗ d x i1 ⊗ · · · ⊗ d x js−1 ⊗ d x k .
Lastly, if U is open in Z then by the continuity of g, g −1 (U ) = π −1 ◦ f −1 (U ) is open in X . So by the definition of the quotient topology, f −1 (U ) is open in Y , so f is continuous. 14 It suffices to show that the Jacobian matrix equals the matrix ( f i j ) representing f itself relative to the standard bases. By linearity, f (x) = f ( x jej) = ei f i j x j , j ij so the i th coordinate function is f i (x) = gives j f i j x j . Taking partial derivatives ∂fi = f i j. 15 Following the hint, we observe that f ◦ f −1 = id implies that (D f )(x) · (D f −1 )(y) = I , which shows that (D f −1 )(y) = (D f (x))−1 .
From the definition of the determinant, n B(e1 ∧ e2 ∧ · · · ∧ en ) = (det B)e1 ∧ e2 ∧ · · · ∧ en . Thus, tr n B = det B, because the trace just sums all the components of the linear operator that map the basis elements to themselves. ) But, from the definition, n B(e1 ∧ e2 ∧ · · · ∧ en ) = Be1 ∧ Be2 ∧ · · · ∧ Ben = (I +z A)e1 ∧ (I +z A)e2 ∧ · · · ∧ (I +z A)en . Now we are finished, because the coefficient of z r consists of a sum of all possible (ordered) wedge products of r Aei ’s and n − r ei ’s.