Lemma: Let $M,N$ be smooth manifolds and let $F:M\to N$ be any map. If $(U_{\alpha},\phi_{\alpha})$ are smooth atlases for M and N, respectively, and if for each $\alpha$ and $\beta$, $\psi_{\beta}\circ F \circ\psi^{-1}_\alpha$ is smooth on its domain of definition, then $F$ is smooth.
I am confused about what is meant about smoothness of $f$ in differential topology. I am interpreting as synonym of differentiation.
So if $\psi_{\beta}\circ F \circ\psi^{-1}_\alpha$ is differentiable that means F is differentiable hence smooth.
The domain where $F$ must be differentiable is given by $\psi^{-1}_{\alpha}(\mathbb{R}^m)$ considering M m-dimensional.
And considering another atlas $(U'_{\alpha},\psi'_{\alpha})$ on M.
$F\circ\psi'^{-1}_{\alpha}=F\circ\psi^{-1}_{\alpha}\circ\psi_{\alpha}\circ\psi'^{-1}_{\alpha}$
Since $F\circ\psi^{-1}_{\alpha}$ is smooth and $\psi_{\alpha}\circ\psi'^{-1}_{\alpha}$ is also smooth then $F\circ\psi'^{-1}_{\alpha}$ is smooth. Since $(U'_{\alpha},\psi'_{\alpha})$ is arbitrary then F is smooth on all M.
Considering another atlas $(U'_{\beta},\psi'_{\beta})$ on N.
$\psi'_{\beta} F \psi'^{-1}_{\alpha}=\psi'_{\beta}\circ\psi^{-1}_{\beta}\circ\psi_{\beta}\circ F\circ\psi^{-1}_{\alpha}\circ\psi_{\alpha}\circ\psi'^{-1}_{\alpha}$
is also smooth by the same argument, where I think the coordinate transformation in $N$ is superfluous.
Question:
Am I proving what the statement asks? Am I understanding smoothness in the context of differential manifolds mapping? I would appreciate if someone could clarify me on this topic.
Thanks in advance!