The following is Proposition 6.11 in Tu's book on manifolds:
Let $U$ be an open subset of a manifold $M$ of dimension $n$. If $F: U \rightarrow F(U) \subset \mathbb{R}^n$ is a diffeomorphism onto an open subset of $\mathbb{R}^n$, then $(U, F)$ is a chart in the differentiable structure of $M$.
His proof:
For any chart $(U_\alpha, \phi_\alpha)$ in the maximal atlas of $M$, both $\phi_\alpha$ and $\phi_\alpha^{-1}$ are $C^\infty$. As composites of $C^\infty$ maps, both $F \circ \phi_\alpha^{-1}$ and $\phi_\alpha \circ F^{-1}$ are $C^\infty$. Hence $(U, F)$ is compatible with the maximal atlas. By the maximality of the atlas, the chart $(U, F)$ is in the atlas.
The proof considers a coordinate domain of $U_\alpha$, so that the range of $\phi_\alpha^{-1}$ is $U_\alpha$. However $F$ is defined on $U$ not $U_\alpha$, so how are we able to deduce that $F \circ \phi_\alpha^{-1}$ (and similarly $\phi_\alpha \circ F^{-1}$) is well defined? Are we implicitly taking the domain to always be $U \cap U_\alpha$?
Best Answer
Hint
$ \phi_\alpha \circ F^{-1}:F(U)\cap \phi_\alpha (U_\alpha)\to\Bbb R^n$. That's on overlaps. Similarly $ F\circ\phi_\alpha^{-1}:\phi_\alpha(U_\alpha) \cap F(U)\to \Bbb R^n$.
Since this is true for any chart $(\phi_\alpha, U_\alpha) $, by maximality $(F,U)$ is in the atlas. That's it's a chart.