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Wikipedia said

The description of most manifolds requires more than one chart (a single chart is adequate for only the simplest manifolds). A specific collection of charts which covers a manifold is called an atlas. An atlas is not unique as all manifolds can be covered multiple ways using different combinations of charts. Two atlases are said to be equivalent if their union is also an atlas.

I think it means: Let $X$ be a topological space and $\Phi_1=\{\phi_\alpha:U_{\alpha}\to\Bbb R^{n_\alpha}\mid\alpha\in A\}$, $\Phi_2=\{\phi_\beta:U_{\beta}\to\Bbb R^{n_\beta}\mid\beta\in B\}$ be two of its(i.e., $X$'s) topological atlas. Then $\Phi_1$ and $\Phi_2$ are equivalent iff $\Phi_1\cup\Phi_2$ is also a topological atlas of $X$.

My question is: if $\Phi_1$ and $\Phi_2$ are atlases of $X$, which means they both can "cover" $X$, then it is definitely true that $\Phi_1\cup\Phi_2$ can "cover" $X$, isn't it? I can't see the meaning of such definition of equivalence. Where did I make the mistake?

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It's not the covering that's the (potential) problem, it's the compatibility, but actually this won't be an issue for topological atlases.

Let $U_{\alpha_1\alpha_2} = U_{\alpha_1} \cap U_{\alpha_2}$. For the two charts $\phi_{\alpha_1}, \phi_{\alpha_2}$ to be compatible, you need $$\tau_{1,2}=\phi_{\alpha_2} \circ \phi_{\alpha_1}^{-1}: \phi_{\alpha_1}(U_{\alpha_1\alpha_2}) \to \phi_{\alpha_2}(U_{\alpha_1\alpha_2})$$ to be continuous (in fact a homeomorphism, but swapping the indices gives the continuous inverse). But this is automatic, since both $\phi_{\alpha_1}$ and $\phi_{\alpha_2}$ are generally required to be homeomorphisms.

The trouble comes in when you want to look at atlases with more structure. For example, a "smooth atlas" is one where you require the transition maps $\tau_{i,j}$ to be smooth, not just continuous. Then there is the possiblity that two charts will be incompatible.

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  • $\begingroup$ Thanks! Very clear. Instantly clarify my confusions. Good teacher! $\endgroup$ – Eric Jan 30 at 13:43
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Wikipedia's brief overview is not actually the definition of an atlas (or, well, I suppose it might be the definition of a "topological atlas", but no one ever uses those). The full definition of a (smooth) atlas has a crucial extra requirement: for each $\phi:U\to\mathbb{R}^n$ and $\psi:V\to\mathbb{R}^n$ in the atlas, the "transition map" $\psi\circ\phi^{-1}:\phi(U\cap V)\to\psi(U\cap V)$ is required to be smooth. Here $\phi(U\cap V)$ and $\psi(U\cap V)$ are open subsets of $\mathbb{R}^n$, so it makes sense to talk about a smooth (i.e., $C^\infty$) function between them.

So, if $\Phi_1$ and $\Phi_2$ are atlases, the union $\Phi_1\cup\Phi_2$ may not be an atlas, since the transition map between a chart in $\Phi_1$ and a chart in $\Phi_2$ may not be smooth.

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  • $\begingroup$ Thanks a lot! Btw did you mean that the word "topological atlas" was not common in literature? $\endgroup$ – Eric Jan 30 at 13:46
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I'd like to add that "topological equivalence" usually means a certain property of smooth (or Lipschitz etc.) atlases, and the results concerning it can be quite deep. In my opinion, what you had in mind and Wikipedia called "equivalence of atlases" should be systematically called "compatibility of atlases". (As mentioned above, it is trivially satisfied and hence redundant in the category of topological manifolds, but not in smooth, Lipschitz and other categories determined by additional differentiability structures.)

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