I have been trying to produce an example of two incompatible atlases on $\mathbb R$. But no success. Could someone help me please? All my example seem compatible. For example, $A_1 = \{((-\infty,1), \mathrm{id}_{(-\infty,1)}) , ((-1,\infty), \mathrm{id}_{(-1,\infty)}\}$ and $A_2=\{(\mathbb R, x \mapsto 2x)\}$. This is only my simplest example I also tried with a similar atlas but involving $x \mapsto \sin x$ and a similar one involving $x \mapsto x^2$.
Please: Could someone show me an example of two incompatible topological atlases on $\mathbb R$?
All the topological atlases will be topologically compatible ( that is, the coordinate changes will be homeomorphisms) because the charts are local homeomorphisms with open subsets of $\mathbb{R}$ ( so the topology is already taken care of).
However, if you just consider a chart as a bijection, without regard for the topology of $\mathbb{R}$ then it is easy to find non-equivalent atlases. Take each consisting of one chart. Consider two bijections $f$, $g$ from $\mathbb{R}$ to $\mathbb{R}$ so that the composition $f\circ g^{-1}$ is not continuous. OK, make $f= 1_{\mathbb{R}}$ and $g(x)= x$ if $x \in \mathbb{Q}$ and $-x$ if $x \in \mathbb{R} \backslash \mathbb{Q}$.
Perpahs more interesting is two topological atlases with for the usual topology that are not $C^{1}$ equivalent, like @Najib Idrissi: suggested. Take $f(x) = x$, $g(x) = \sqrt[3]{x}$.
However, one should note that any two $C^{\alpha}$ structures on $\mathbb{R}$ are still diffeomorphic, perhaps not under the map $1_{\mathbb{R}}$.