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I need to show that :

On any smooth manifold $(M,A)$ all chart maps are $C^{\infty}$-diffeomorphisms.

Definitions : Let $M$ be a Hausdorff second countable topological space. Then a pair $(U, \phi)$ is called a chart for $M$ if $U \in M$ is open and $\phi : U \subset M \to \phi(U) \subset \mathbb{R}^n$ be a homeomorphism. The set $\{\ (U_{\alpha}, \phi_{\alpha}) \}$ is called a topological atlas for $M$ if $\cup U_{\alpha} = M$ and then $(M,A)$ will be called a topological manifold. For $(M,A)$, two charts $(U, \phi), (V, \psi) \in A$ are called $C^{\infty}$-related if $U \cap V = \emptyset$ or if $U \cap V \ne \emptyset$ then $\psi \circ \phi^{-1} : \phi(U \cap V) \subset \mathbb{R}^n \to \psi(U \cap V) \subset \mathbb{R}^n$ be $C^{\infty}$-diffeomorphism. If all charts in $(M,A)$ are pairwise $C^{\infty}$-related then $(M,A)$ is called a smooth manifold.

My attempt based on the definitions above to answer the question in the first paragraph : , The question asks that if for any $(U, \phi) \in A$, $\phi : U \subset M \to \phi(U) \subset \mathbb{R}^n$ is a $C^{\infty}$-diffeomorphism. So because $\phi \circ \phi^{-1} = \operatorname{id}$ is $C^{\infty}$-diffeomorphism so is $\phi$.

Am I in a right track and if so how to make this rigorous?

Added : Can I go with this approach that I add $\operatorname{id}$ to atlas and say that $\phi = \phi \circ id$ so $\phi$ is diffeomorphism?

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  • $\begingroup$ @MarkSaving, A bit confusion I have, first to clear : is the question in OP means to show that $\phi : U \to \phi(U)$ is a diffeomorphism? Is that what the question asks? $\endgroup$
    – user200918
    Commented Dec 8, 2021 at 0:40
  • $\begingroup$ Sorry, I got a bit mixed up. Yes, the task is showing that $\phi : U \to \phi(U)$ is a diffeomorphism where $U$ and $\phi(U)$ are considered smooth manifolds in their own right. $\endgroup$ Commented Dec 8, 2021 at 0:42
  • $\begingroup$ @MarkSaving, OK thanks :) $\endgroup$
    – user200918
    Commented Dec 8, 2021 at 0:44
  • $\begingroup$ See this question and this answer. $\endgroup$
    – Lee Mosher
    Commented Dec 8, 2021 at 0:54
  • $\begingroup$ @LeeMosher I am sorry your answer is ambiguous, the task I need to hand in asks for a rigorous proof. Could you help me more? $\endgroup$
    – user200918
    Commented Dec 8, 2021 at 2:05

3 Answers 3

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Consider $\mathcal{A}_U = \{(U,\phi)\}$. It is a smooth atlas on $U$ (show it), and therefore, $(U,\mathcal{A}_U)$ is a smooth manifold.

Similarly, $\mathcal{A}_{\phi(U)} = \{\phi(U),\mathrm{id}_{\phi(U)}\}$, is a smooth atlas on $\phi(U)$ and it follows that $(\phi(U),\mathcal{A}_{\phi(U)})$ is a smooth manifold.

Let us look at the homeomorphism $\phi \colon U \to \phi(U)$ through the only charts of our atlases. It is given by $$ \mathrm{id}_{\phi(U)}\circ \phi \circ (\phi)^{-1} = \mathrm{id}_{\phi(U)} $$ and is therefore smooth. Similarly, the inverse map $\phi^{-1} \colon \phi(U) \to U$ reads, in the charts, as follows $$ \phi \circ \phi^{-1} \circ (\mathrm{id}_{\phi(U)})^{-1} = \mathrm{id}_{\phi(U)} $$ which is again smooth. It follows that $\phi\colon U \to \phi(U)$ is a smooth homeomorphism with smooth inverse between two smooth manifolds, and is hence a diffeomorphism.

Edit

Smoothness is a notion defined for functions from some open subset of $\Bbb R^p$ with range into some other open subset of $\Bbb R^q$. In order to extend this definition for maps between manifolds, one has to find some trick to go back to the Euclidean setting. The trick is the following.

Let $f\colon M \to N$ be a map between smooth manifolds. If $(U,\phi)$ is a chart of $M$, $(V,\psi)$ a chart of $N$, and if $f(U) \subset V$, then the map $$ \psi \circ f \circ \phi^{-1} \colon \phi(U) \to \psi(V) $$ goes from an open subset of an Euclidean space onto another one. Here, we have a notion of smoothness.

Definition. A map $f\colon M\to N$ is said to be smooth if for any charts $(U,\phi)$ and $(V,\psi)$ as above, the map $\psi\circ f \circ \phi^{-1} \colon \phi(U) \to \psi(V)$ is smooth.

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  • $\begingroup$ Thanks a lot. I know the first two paragraphs, after that you use some definitions that I don't know, because the lecturer gave different definitions so why I have included them in OP. Also can a solution be build up for an idea that I said in "added" part of OP? $\endgroup$
    – user200918
    Commented Dec 8, 2021 at 11:30
  • $\begingroup$ @TheMagicMountain What definition of smoothness did the lecturer give you for a map between manifolds? $\endgroup$
    – Didier
    Commented Dec 8, 2021 at 12:24
  • $\begingroup$ I translated all the related definitions of the lecture in OP, unfortunately not more than that is given. Since $\mathrm{id}_{\phi(U)}\circ \phi \circ (\phi)^{-1}$ is not in the notes so I am a bit confused where does that come from? Thanks $\endgroup$
    – user200918
    Commented Dec 8, 2021 at 12:28
  • $\begingroup$ Well, there clearly is something missing: the definition of a diffeomorphism between smooth manifolds, relying itself on the definition of a smooth map between manifolds. One cannot say $\phi \colon U \to \varphi(U)$ is a diffeomorphism of smooth manifolds if one cannot define what this mean! $\endgroup$
    – Didier
    Commented Dec 8, 2021 at 12:40
  • $\begingroup$ @TheMagicMountain I added the definition of smoothness for maps between manifold in an edit. $\endgroup$
    – Didier
    Commented Dec 8, 2021 at 12:46
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Building on the answer by Lee Mosher, you have the following definition:

A map $f$ from an open subset $U\subset M$ to $\mathbb{R}^k$ is smooth if and only if its composition with each chart inverse $\phi^{-1}$ associated to $W$ with $U\cap W \neq \emptyset$ is smooth as map between Euclidean spaces. If our manifold is $n$-dimensional, this is saying to look at $$ f\circ\phi^{-1}: \phi(U\cap W)\subset\mathbb{R}^n\to \mathbb{R}^k. $$

Since your map $f$ is a chart map, what can you say about the expression above? Which definition out of the ones you wrote down does it look like?

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Let us go step by step.

We need to establish (recall) some preliminary results.

Step 1. It is easy to show that if $M$ is a smooth manifold with atlas $\{\ (U_{\alpha}, \phi_{\alpha}) \}$, and $W$ is an open set in $M$, then $W$ is a smooth manifold with atlas $\{\ (W\cap U_{\alpha}, \phi_{\alpha}|_{W\cap U_{\alpha}}) \}$

Step 2. Let $M$ be smooth manifold with atlas $\{\ (U_{\alpha}, \phi_{\alpha}) \}$, $N$ be smooth manifold with atlas $\{\ (V_{\beta}, \psi_{\beta}) \}$. Then

A function $f: M \to N$ is $C^{\infty}$ if $f$ is continuous and, for all $\alpha$ and $\beta$, the function $\psi_\beta \circ f \circ \phi_\alpha^{-1}: \phi_\alpha(U_\alpha \cap f^{-1}(V_\beta)) \to \psi_\beta({V_\beta})$ is $C^{\infty}$.

Note that $\phi_\alpha(U_\alpha \cap f^{-1}(V_\beta))$ is an open set in $\Bbb R^n$ and $\psi_\beta({V_\beta})$ is an open set in $\Bbb R^k$ (for some $k$, dimension of $N$), so it is clear what means $\psi_\beta \circ f \circ \phi_\alpha^{-1}: \phi_\alpha(U_\alpha \cap f^{-1}(V_\beta)) \to \psi_\beta({V_\beta})$ to be $C^{\infty}$.

Note also that, if $U_\alpha \cap f^{-1}(V_\beta) = \emptyset$, then $\phi_\alpha(U_\alpha \cap f^{-1}(V_\beta))= \emptyset$ and $\psi_\beta \circ f \circ \phi_\alpha^{-1}: \phi_\alpha(U_\alpha \cap f^{-1}(V_\beta)) \to \psi_\beta({V_\beta})$ is trivially $C^{\infty}$.

Step 3. The space $\Bbb R^n$ is smooth manifold with the standard (trivial) atlas $\{(\Bbb R^n, id_{\Bbb R^n} )\}$. So, by Step 1, if $W$ is an open set in $\Bbb R^n$, $W$ is a smooth manifold with atlas $\{(W, id_{W} )\}$.

So, in Step 2, if $V$ is an open set of $\Bbb R^k$ and $N= V$, we can simplify the definition of $f$ being $C^{\infty}$.

In fact, let $M$ be smooth manifold with atlas $\{\ (U_{\alpha}, \phi_{\alpha}) \}$, then

A function $f: M \to V$ is $C^{\infty}$ if $f$ is continuous and, for all $\alpha$, $ f \circ \phi_\alpha^{-1}: \phi_\alpha({U_\alpha} \cap f^{-1}(V)) \to V $ is $C^{\infty}$.

and

A function $g: V \to M$ is $C^{\infty}$ if $g$ is continuous and, for all $\alpha$, $\phi_\alpha \circ g : V \cap g^{-1}(U_\alpha) \to \phi_\alpha(U_\alpha)$ is $C^{\infty}$.

Step 4. We want to prove that,

for each $(U, \phi) \in A$, $\phi : U \subset M \to \phi(U) \subset \mathbb{R}^n$ is a $C^{\infty}$-diffeomorphism.

It means $\phi$ is bijective and $\phi$ and $\phi^{-1}$ are $C^{\infty}$-functions. The proof that $phi$ is trivial from the fact that $(U, \phi)$ is a chart.

Now, to prove that $\phi : U \to \phi(U) \subset \mathbb{R}^n$ is a $C^{\infty}$, we must consider U as smooth manifold (by Step1) and, since $\phi(U)$ is an open set in $\Bbb R^n$ and $\phi^{-1}(\phi(U))=U$, then by Step 3, it is enough to prove that , for all $\alpha$, $\phi \circ \phi_\alpha^{-1} : \phi_\alpha(U_\alpha \cap U) \to \phi(U) \subset \mathbb{R}^n$ is $C^{\infty}$. If $U_\alpha \cap U =\emptyset$, it is trivially true that $\phi \circ \phi_\alpha^{-1}$ is $C^{\infty}$. If $U_\alpha \cap U \neq \emptyset$, it is a trivial consequence of the charts $(U,\phi)$ and $(U_\alpha,\phi_\alpha)$ being $C^{\infty}$-related.

Now, to prove that $\phi^{-1} : \phi(U) \subset \mathbb{R}^n \to U$ is a $C^{\infty}$, is similar. We must consider U as smooth manifold (by Step1) and, since $\phi(U)$ is an open set in $\Bbb R^n$ and, for any $\alpha$, $(\phi^{-1})^{-1}(U_\alpha)=(\phi^{-1})^{-1}(U \cap U_\alpha)=\phi(U \cap U_\alpha)$, by Step 3, it is enough to prove that , for all $\alpha$, $\phi_\alpha\circ \phi^{-1} : \phi(U \cap U_\alpha) \to \phi_\alpha(U_\alpha)$ is $C^{\infty}$. If $U\cap U_\alpha =\emptyset$, it is trivially true that $\phi_\alpha \circ \phi^{-1}$ is $C^{\infty}$. If $U\cap U_\alpha \neq \emptyset$, it is a trivial consequence of the charts $(U_\alpha,\phi_\alpha)$ and $(U,\phi)$ are $C^{\infty}$-related.

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