# if M, N are manifolds and f from M to N is a diffeomorphism, then f induces an atlas on N

#### oblixps

##### Member
this is probably trivial and i may be overthinking it but, it's been bugging me for a while now.

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

#### Chris L T521

##### Well-known member
Staff member
this is probably trivial and i may be overthinking it but, it's been bugging me for a while now.

since M is a manifold, we have $${(U_a, x_a)}$$ is an atlas for M and i want to show that since f is a diffeomorphism from M to N, $${(U_a, f \circ x_a)}$$ is an atlas for N. Since f is a bijection, $$f \circ x_a(U)$$ cover N. Now what troubles me is that if we take any 2 charts in this atlas, say $$f \circ x_a$$ and $$f \circ x_b$$, then $$(f \circ x_a)^{-1} \circ (f \circ x_b) = x_a^{-1} \circ f^{-1} \circ f \circ x_b = x_a^{-1} \circ x_b$$ and since $$x_a$$ and $$x_b$$ belong to an atlas on M, we must have that $$x_a^{-1} \circ x_b$$ is differentiable. so therefore, it follows that $$(f \circ x_a)^{-1} \circ (f \circ x_b) = x_a^{-1} \circ x_b$$ is differentiable as well so $${(U_a, f \circ x_a)}$$ 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 $${(U_a, f \circ x_a)}$$ 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.
Is it safe to assume that $M$ and $N$ are smooth manifolds? If so, then I would like to say that $f:M\rightarrow N$ is a $C^{\infty}$-diffeomorphism. You used the fact that $f$ was a diffeo in your transition function computations; $(f\circ x_a)^{-1}$ for example is defined since $f^{-1}$ exists and is smooth (due to $f$ being a diffeo), and since $f$ and $x_a$ are smooth maps, $(f\circ x_a)^{-1}$ is smooth as well; it just so happened that the $f$'s cancelled out in your computation, but you still used the fact it was a diffeo to take the inverses.

I think that's all there is to it really. If you have any additional follow-up questions, feel free to post back!

#### oblixps

##### Member
yes we can assume M and N are smooth manifolds. I'm currently going through doCarmo's Riemannian Geometry and he defines differentiable manifolds saying (this is one part of the definition) that the transition function $$x_a^{-1} \circ x_b$$ for any 2 charts in the atlas must be differentiable.

what i'm not sure about is how come we can say that $$x_a$$ by itself is differentiable? since $$x_a: U_a \rightarrow M$$ and we can't really work with derivatives in some arbitrary set M.

i've been looking around and most definitions i see only have that the transition functions must be differentiable. do we somehow deduce from the definition that the charts themselves are differentiable as well?