# Two atlases on a manifold $M$ are equivalent if and only if they determine the same set of smooth functions $f:M\rightarrow\mathbb{R}$

Suppose $\{\phi_\alpha\}_{\alpha\in\mathcal{A}}$ and $\{\phi_\beta\}_{\beta\in\mathcal{B}}$ are two smooth atlases on a topological manifold $M$. My definition of two such atlases being equivalent is that their union $\{\phi_\alpha\}_{\alpha\in\mathcal{A}\cup\mathcal{B}}$ is also a smooth atlas, that is, the transition maps between the charts of different atlases are smooth.

I have shown that if two atlases are equivalent, then they determine the same set of smooth functions $f:M\rightarrow\mathbb{R}$ - i.e. $f$ is smooth with respect to one chart if and only if it is smooth with respect to the other. But I do not know how to prove the converse statement. I would like to do something along the lines of $(f\circ\phi_\beta^{-1})^{-1}\circ(f\circ\phi_\alpha^{-1})=\phi_\beta\circ\phi_\alpha^{-1}$ and conclude from that that the RHS is smooth since both bracketed functions on the LHS are, but I know I can't take the inverse on the LHS like this,since there is no guarantee $f$ is invertible. Any help would be appreciated.

Suppose $\varphi_\beta\circ\varphi_\alpha^{-1}$ isn't smooth. Then one of $x_i\circ\varphi_\beta$ isn't smooth in $(\varphi_\alpha)_{\alpha\in A}$, while it's clearly smooth in $(\varphi_\beta)_{\beta\in B}$. Choose a sufficiently small compact neighborhood $K$ of a point in which it isn't smooth. Then $x_i\circ\varphi_\beta|_K$ can be smoothly extended to the whole $M$ and is the function you want.
• I guess you mean "Choose a sufficiently small compact neighborhood $K$ of a point in which it wasn't smooth ... " – KonKan Sep 25 '16 at 1:54