Is $\{(\mathbb R,\phi)\}$ a smooth atlas for $\mathbb R$, where $\phi(x)=\left\{ \begin{array}{ll} x & x<0, \\ 2x& x\geq0~? \end{array} \right.$
Let $M$ be a topological $ n $-manifold. A coordinate chart (or just a chart) on $ M $ is a pair $ (U, \varphi) $, where $ U $ is an open subset of $ M $ and $ \varphi: U \rightarrow \widehat{U} $ is a homeomorphism from $ U $ to an open subset $ \widehat{U}=\varphi(U) \subseteq \mathbb{R}^{n} $.
If $ U $ and $ V $ are open subsets of Euclidean spaces $ \mathbb{R}^{n} $ and $ \mathbb{R}^{m} $, respectively, a function $ F: U \rightarrow V $ is said to be smooth if each of its component functions has continuous partial derivatives of all orders. If in addition $ F $ is bijective and has a smooth inverse map, it is called a diffeomorphism.
Two charts $ (U, \varphi) $ and $ (V, \psi) $ are said to be smoothly compatible if either $ U \cap V=\varnothing $ or the transition map $ \psi \circ \varphi^{-1} $ is a diffeomorphism.
An atlas for $ M $ is a collection of charts whose domains cover $ M $. An atlas $ \mathcal{A} $ is called a smooth atlas if any two charts in $ \mathcal{A} $ are smoothly compatible with each other.
I have carefully checked the definition of each concept related to this question, but I have not found anything contrary to a certain definition. Although it is obvious that $\phi$ itself is not smooth, my understanding is that the atlas $\{(\mathbb R,\phi)\}$ has only a single coordinate chart $(\mathbb R,\phi)$, and no other coordinate chart to determine whether it is smoothly compatible with $(\mathbb R,\phi)$. So, is it a smooth atlas? This is a question out of curiosity. Any help would be appreciated.
Best Answer
Just to get this off of the unanswered list.
For this question, I have asked my professor and teaching assistant. Combining the two comments above (thank you very much, Deane and Kubrick), we can draw the surprising conclusion that $\{(\mathbb R,\phi)\}$ is indeed a smooth atlas for $\mathbb R$ since $\phi\circ \phi^{-1}={\rm id}_{\mathbb{R}}$ although $\phi$ itself is not smooth!