Let $M=\{(x,y,|x|) \in \Bbb R^3 \mid (x,y) \in \Bbb R^2 \}$. Construct a $C^\infty$ atlas $\mathcal{A}$ so that $(M, \overline{\mathcal{A}})$ (where $\overline{\mathcal{A}}$ is the maximal atlas) becomes a differentiable $2$-manifold. Is it diffeomorphic to $\Bbb R^2$? If so, what is the diffeomorphism $f: M \to \Bbb R^2$?
Define $\varphi : M \to \Bbb R^2$ as $\varphi(x,y, |x|) = (x,y)$. Then $\varphi$ is a homeomorphism. And if we define $\mathcal{A}=\{(M, \varphi)\}$ this is an atlas, but is it the maximal atlas?
Also I don't think this is diffeomorphic to $\Bbb R^2$ since the inverse $\varphi^{-1}(x,y)=(x,y,|x|)$ has the last coordinate function as the absolute value function which isn't differentiable at the origin? We need $\varphi^{-1}$ to be differentiable also?
In order to test whether a particular function $f : M \to \mathbb R^2$ is a diffeomorphism, using your atlas $\mathcal A$ which only has a single coordinate function, the requirement is that the composition $$\mathbb R^2 \xrightarrow{\phi^{-1}} M \xrightarrow{f} \mathbb R^2 $$ must be a diffeomorphism.
So, now do this with $f=\phi$. Is the composition $$\mathbb R^2 \xrightarrow{\phi^{-1}} M \xrightarrow{\phi} \mathbb R^2 $$ a diffeomorphism?