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I'm studying the definition of smooth manifolds and constantly thrown off by two atlases, $A=(\mathbb{R}, f:x\mapsto x^{1/3})$ and $B=(\mathbb{R}, f:x\mapsto x)$, each of which constutes of a single chart (I guess this is valid, correct?).

In Tu's An Introduction to Manifolds (second edition), he writes on pages 62 and page 63 the following two results which I cannot seem to digest with regard to $A$ and $B$.

Proposition 6.8. Let $N$ and $M$ be smooth manifolds, and $F: N\to M$ a continuous map. The following are equivalent.

(i) The map $F:N\to M$ is $C^{\infty}$.

(ii) There are atlases $u$ for $N$ and $B$ for $N$ such that for every chart $(U,\phi)$ in $u$ and $(V,\psi)$ in $B$, the map

$$\psi\circ F\circ \phi^{-1}:\phi(U\cap F^{-1}(V))\to \mathbb{R}^m$$ is $C^{\infty}$.

(iii) For every chart $(U,\phi)$ on $N$ and $(V,\psi)$ on $M$, the map $$\psi\circ F\circ \phi^{-1}:\phi(U\cap F^{-1}(V))\to\mathbb{R}^m$$ is $C^{\infty}$.

Does this mean that the smoothness between two manifolds is independent of the choice of maximal atlases? Like, even if we impose some stupid atlas like $A$, it doesn't matter?

Proposition 6.10. If $(U,\phi)$ is a chart on a manifold $M$ of dimension $n$, then the coordinate map $\phi:U\to \phi(U)\subset\mathbb{R}^n$ is a diffeomorphism.

I'm really having trouble with this proposition. For example, $A$ is a chart on the manifold $\mathbb{R}$, but it is not a diffeomorphism at $x=0$. Could someone please explain? Thank you.

Michael Albanese
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able20
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  • I kind of get the idea of where I did think wrong for both questions.... for the first one, I think the smoothness IS dependent on the choice of the maximal atlas. i.e. if we choose a different choice of atlas for say, $N$, the smoothness might change. For the second question, in order to check the smoothness between $\mathbb{R}\to \phi(\mathbb{R})$, I needed note that $\mathbb{R}$ itself is a manifold so I need to look at its image on some chart. But I'm yet to figure out why Lu was able to specifically choose identity chart, because according to the first one it is dependent on the choice – able20 Feb 07 '21 at 07:00
  • of maximal atlas. – able20 Feb 07 '21 at 07:01
  • Note that $M$ and $N$ are smooth manifolds, so we have already chosen maximal atlases for them. In particular, a chart on $M$ means a chart which belongs to the maximal atlas, it is not an arbitrary chart. – Michael Albanese Feb 07 '21 at 11:41
  • Oh, ok. Thanks that addresses my first question. For the second one, do you think that we are assumed to take maximal atlas for $\phi(U)$ as the one including the identity chart $(\phi(U),1_{\phi(U)})$? I am wondering about this because in the proof Tu uses the fact that $\phi(U)$ has the atlas with that single chart(with identity function). But the statement does not specify what atlas we have on $\phi(U)$, so I wanted to know if it's something like a convention to choose atlas that includes identity map – able20 Feb 07 '21 at 11:59
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    Yes, I assume Tu means $\mathbb{R}$ with its usual maximal atlas (the one containing the chart $B$). – Michael Albanese Feb 07 '21 at 12:05
  • Very much appreciate it – able20 Feb 07 '21 at 12:08

1 Answers1

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As $M$ and $N$ are smooth manifolds, they are already equipped with maximal atlases. Therefore the charts in the statement refer to charts from those atlases; Tu makes this explicit in the last sentence of section $5.3$.

As for Proposition $6.10$, Tu is endowing $\phi(U)$ with its usual smooth structure. In the proof, $\phi(U)$ is equipped with the atlas $(\phi(U), \mathbb{1}_{\phi(U)})$.

Michael Albanese
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