Example for Atlases that are Not Equivalent? In a lecture on differential geometry, we had the following definition of equivalent atlases:

Two atlases $\mathcal A$ and $\mathcal B$ on $M$ are called equivalent if $\mathcal A \cup \mathcal B$ is an atlas on $\mathcal M$.

The definition of atlas we had is the following:

Let $M$ be a second countable Hausdorff topological space. An $n$-dimensional smooth atlas on $M$ is a collection of maps $$\mathcal A = \left\{ \left(\varphi_i, U_i\right) \mid i\in A\right\}, \quad \varphi_i: U_i\rightarrow \varphi_i(U_i)\subset \mathbb R^n,$$ such that all $U_i \subset M$ are open, all $\varphi_i$ are homeomorphisms, and

*

*$\{U_i, i\in I\}$  is an open covering of $\mathcal M$

*$\varphi_i\circ \varphi_j^{-1}: \varphi_j\left(U_i\cap U_j\right)\rightarrow \varphi_i\left( U_i\cap U_j\right)$ are smooth for all $i, j\in I$.



Question: Let $\mathcal A$ and $\mathcal B$ be two atlases on $M$. If $\mathcal A$ consists of charts $(\varphi_i, U_i)$, and $\mathcal B$ constists of charts $(\psi_i, V_i)$, then $\{U_i, V_{i'}, i\in I, i'\in I'\}$ is obviously still an open covering of $\mathcal M$. What might break is that $\varphi_i\circ \psi_j^{-1}: \psi_j(U_i\cap V_j)\rightarrow \varphi_i(U_i \cap V_j)$ is not smooth anymore.
Does anybody happen to have a concrete counterexample of two atlases $\mathcal A$ and $\mathcal B$ that are not equivalent on a manifold $\mathcal M$?
 A: The standard example is given already in the comment: let $M = \mathbb R$ with the standard topology and $\mathscr A = \{ (\mathbb R, x)\}$ and $\mathscr B = \{ (\mathbb R, x^3)\}$. Then $\mathscr A, \mathscr B$ are both smooth atlas for $\mathbb R$, but they are not compatible since
$$ x\circ  (x^3)^{-1} = x^{1/3}$$
is not smooth.
For a general manifold $M$, let $\mathscr A$ be any atlas on $M$. Let $H: M\to M$ be any homeomorphism. Then
$$\mathscr A^H = \{ (H(U), \varphi \circ H^{-1}) : (U, \varphi ) \in \mathscr A\}$$
is another charts on $M$ which is not compatible with $\mathscr A$ in general (here $H(U) = \{ h(p) : p\in U\}$).
For some explicit example of such $H$ so that $\mathscr A$ and $\mathscr A^H$ are not compatible: Pick any fixed $(U_0, \varphi_0 ) \in \mathscr A$ and let $h : \varphi_0(U_0) \to \varphi_0(U_0)$ be any homeomorphism (but not diffeomorphism) which is identity outside a compact set $K$ in $\varphi_0 (U_0)$. Then define $H: M \to M$ by $H(p)  =\varphi_0^{-1} \circ h\circ \varphi_0 (p)$ when $p \in U_0$ and $H(p) = p$ otherwise).
Then $\mathscr A$ and $\mathscr A^H$ are not compatible: note that $(U_0, \varphi_0)\in \mathscr A$ and $(U_0, \varphi_0 \circ H^{-1}) \in \mathscr A^H$ and
$$ \varphi_0 \circ (\varphi_0 \circ H^{-1})^{-1} = \varphi_0 \circ H \circ \varphi_0^{-1} = h$$
is not a diffeomorphism.
