# Smoothness of a manifold implies all maps of its altas' charts are diffeomorphism?

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?

• @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?
– user200918
Commented Dec 8, 2021 at 0:40
• 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. Commented Dec 8, 2021 at 0:42
• @MarkSaving, OK thanks :)
– user200918
Commented Dec 8, 2021 at 0:44
• See this question and this answer. Commented Dec 8, 2021 at 0:54
• @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?
– user200918
Commented Dec 8, 2021 at 2:05

## 3 Answers

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.

• 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?
– user200918
Commented Dec 8, 2021 at 11:30
• @TheMagicMountain What definition of smoothness did the lecturer give you for a map between manifolds? Commented Dec 8, 2021 at 12:24
• 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
– user200918
Commented Dec 8, 2021 at 12:28
• 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! Commented Dec 8, 2021 at 12:40
• @TheMagicMountain I added the definition of smoothness for maps between manifold in an edit. Commented Dec 8, 2021 at 12:46

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?

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.