We say that two charts$(U,\phi), (V,\theta)$ are smoothly compatible if $U \cap V = \emptyset$ or $ \phi \circ \theta^{-1}$ is smooth. Do we also require that $\theta \circ \phi^{-1}$ to be smooth?
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You can add the requirement that $\theta\circ\phi^{-1}$ be smooth or not. It makes a difference if you want to ask if two particular charts are smoothly compatible. But in the situation you're really interested in (defining a smooth atlas), it doesn't make any difference.
We say that an atlas $\{U_\alpha,\phi_\alpha\}_{\alpha\in A}$ for a topological manifold $M$ is a smooth atlas if for each $\alpha,\beta\in A$, the charts $(U_\alpha,\phi_\alpha)$ and $(U_\beta,\phi_\beta)$ are smoothly compatible. Since this applies equally well to $(U_\beta,\phi_\beta)$ and $(U_\alpha,\phi_\alpha)$, it follows that each transition map $\phi_\alpha\circ \phi_\beta^{-1}$ is in fact a diffeomorphism, even if you adopted the weaker definition of smooth compatibility.
Jack Lee
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Definition: $(U, \phi)$ and $(V,\theta)$ are smoothly compatible if $U \cap V = \emptyset$ or $\phi \circ \theta^{-1}$ is a $C^{\infty}$-diffeomorphism.
Therefore, $\theta \circ \phi^{-1}$ is also smooth by the very definition of a $C^{\infty}$-diffeomorphism.
Hermès
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