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since M is a manifold, we have [tex] {(U_a, x_a)} [/tex] is an atlas for M and i want to show that since f is a diffeomorphism from M to N, [tex] {(U_a, f \circ x_a)} [/tex] is an atlas for N. Since f is a bijection, [tex] f \circ x_a(U) [/tex] cover N. Now what troubles me is that if we take any 2 charts in this atlas, say [tex] f \circ x_a [/tex] and [tex] f \circ x_b [/tex], then [tex] (f \circ x_a)^{-1} \circ (f \circ x_b) = x_a^{-1} \circ f^{-1} \circ f \circ x_b = x_a^{-1} \circ x_b [/tex] and since [tex] x_a [/tex] and [tex] x_b [/tex] belong to an atlas on M, we must have that [tex] x_a^{-1} \circ x_b [/tex] is differentiable. so therefore, it follows that [tex] (f \circ x_a)^{-1} \circ (f \circ x_b) = x_a^{-1} \circ x_b [/tex] is differentiable as well so [tex] {(U_a, f \circ x_a)} [/tex] forms an atlas on N.

It seems like i did not use the fact that f was a diffeomorphism at all! since f and its inverse cancel out, it seems that they don't even enter into the transition function. if i said that f was just a bijection between M and N, wouldn't this same proof work and show that [tex] {(U_a, f \circ x_a)} [/tex] is still an atlas on N?

i am probably missing out on something crucial here and any help clearing this up will be greatly appreciated.