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My question is the same of this Two atlases on a manifold $M$ are equivalent if and only if they determine the same set of smooth functions $f:M\rightarrow\mathbb{R}$ , but i need a more elementary answer. In particular, i have only the definitions of atlases, charts, and smooth functions $f:M\to\mathbb{R}$. And, finally, for me $M$ is simply a topological space.

Let $(U,\phi)$ be a chart of the first atlas $A_1$, and $(V,\psi)$ be a chart of the second atlas $A_2$, such that $U\cap V\ne\emptyset$. I have to show that $$\phi\circ\psi^{-1}\colon \psi(U\cap V)\to\phi(U\cap V)$$ is $C^\infty$ with inverse $C^\infty$.

My hypothesis is that $C^\infty(M,A_1)=C^\infty(M,A_2)$, and i say that $f\colon M\to \mathbb{R}\in C^\infty(M,A_1)$ if and only if for every chart $(U,\phi)\in A_1$ i have $f\circ\phi^{-1}\colon \phi(U)\to\mathbb{R}$ is $C^\infty$ in the usual sense of Analysis.

So my excercise was:

Prove that $C^\infty(M,A_1)=C^\infty(M,A_2)$ if and only if $A_1$ and $A_2$ are compatible, i.e. for every chart $(U,\phi)\in A_1$ and for every chart $(V,\psi)\in A_2$ i have that $U\cap V=\emptyset$, or $U\cap V\ne\emptyset$ and $$\phi\circ\psi^{-1}\colon \psi(U\cap V)\to\phi(U\cap V)$$ is $C^\infty$ with inverse $C^\infty$.

I have succeded in showing the implication: $A_1$ and $A_2$ compatible $\implies$ $C^\infty(M,A_1)=C^\infty(M,A_2)$.

I need the converse.

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  • $\begingroup$ If $(U, \phi)$ is a chart of the first atlas, is the maps $\phi \colon U \to \phi(U)$ smooth ? Yes. Then it should be smooth for the second atlas. $\endgroup$ – Orest Bucicovschi Apr 6 '18 at 18:51
  • $\begingroup$ @orangeskid the map $\phi\colon U\to \phi (U) $ is only homeomorphism, NOT smooth. $\endgroup$ – Minato Apr 8 '18 at 17:31

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