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I came to the following proposition:

Let $M$ and $N$ be two smooth manifolds with respective maximal atlases $\mathcal{A}_M,\mathcal{A}_N$. Then a bijection $f:M \rightarrow N$ is a diffeomorphism if and only if the following codition holds:

$$ (U,y)\in\mathcal{A}_N \Leftrightarrow (f^{-1}(U), y\circ f)\in\mathcal{A}_M $$

I can prove the necessary part, but don't know how to prove the sufficiency.

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up vote 1 down vote accepted

We assume that $ \mathcal{A}_{M} $ and $ \mathcal{A}_{N} $ are maximal $ C^{\infty} $-atlases on $ M $ and $ N $ respectively. The proof of sufficiency then proceeds as follows.

The chart maps coming from $ \mathcal{A}_{M} $ and $ \mathcal{A}_{N} $ are automatically diffeomorphisms (to prove this, use the fact that transition functions are smooth by definition). Hence, $ y: U \rightarrow y[U] $ and $ y \circ f: {f^{-1}}[U] \rightarrow y[U] $ are both diffeomorphisms. It thus follows that both $$ f|_{{f^{-1}}[U]} = y^{-1} \circ (y \circ f): {f^{-1}}[U] \rightarrow U $$ and $$ f^{-1}|_{U} = (y \circ f)^{-1} \circ y: U \rightarrow {f^{-1}}[U] $$ are diffeomorphisms, hence smooth functions.

Observe that $ \{ {f^{-1}}[U] \,|\, (U,y) \in \mathcal{A}_{N} \} $ is an open cover of $ M $ and that $ f $ is smooth on each piece of the cover. Hence, by the smooth version of the Pasting Lemma, $ f $ is smooth globally.

Likewise, $ \{ U \,|\, (U,y) \in \mathcal{A}_{N} \} $ is an open cover of $ N $ and $ f^{-1}|_{U} $ is smooth on each piece of the cover. Hence, $ f^{-1} $ is smooth globally.

We therefore conclude that $ f: M \rightarrow N $ is a diffeomorphism.

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Why does chart map need to be diffeomorphism? By definition, they are just bijections between $\mathcal{M}$ and $\mathcal{R}^n$. You hints 'transition functions', but I don't understand. What if there is only one global chart? Then there will be no transition function. – hxhxhx88 Oct 22 '12 at 2:25
Still a little confused. Diffeomorphism is a concept in the smooth manifold category, I think it has nothing to do with a chart map. Actually, your proof give me an insight. By definition, when $f:M\rightarrow N$ satisfies both $\phi\circ f\circ\varphi^{-1}$ and $\varphi\circ f^{-1}\circ\phi^{-1}$ are smooth where $\varphi$ and $\phi$ are charts of $M$ and $N$, it is a diffeomorphism. So we only need to check $y\circ f\circ(y\circ f)^{-1}:\mathbb{R}^n\rightarrow \mathbb{R}^n$ and $(y\circ f)\circ f^{-1}\circ y^{-1}:\mathbb{R}^n\rightarrow \mathbb{R}^n$ to be smooth. – hxhxhx88 Oct 23 '12 at 5:05
Obviously they are, because they are identity.. So I think we do not need the chart map to be smooth..I think maybe we are not agree on the definition of diffeomorphism:) – hxhxhx88 Oct 23 '12 at 5:05
I think we don't need to check all pairs $(\varphi, \phi)$, because by definition for every point $P$ in $M$, there should exists a neighborhood on which $f$ is smooth. Because $f$ in a bijection as assumed, I only need to pick $f^{-1}(U)$ and $U$ with corresponding chart maps. Still don't understand your words: yes, when defining a $C^{\infty}$-atlas, we need $\phi \circ \psi^{-1}$ to be smooth, but it does not imply and require $\phi$ and $\psi$ themselves to be smooth. I'm new to this subject, hope I'm not making your mad :) – hxhxhx88 Oct 23 '12 at 5:34
Yeah, but it doesn't mean $\varphi$ itself should be smooth, we just require the composition $\varphi_1\circ\varphi_2^{-1}$ to be smooth, right?...Actually, what does the smoothness of a chart map mean?.. – hxhxhx88 Oct 23 '12 at 5:43

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