# The necessary and sufficient condition for diffeomorphism

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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## 1 Answer

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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