I'm starting to learning the subject and I would appreciate some feedback to see if my understanding is correct. So I want to prove that the extended real line $\overline{\mathbb{R}}=\mathbb{R}\cup\{\infty,-\infty\}$ is a smooth manifold. That $\overline{\mathbb{R}}$ is a topological manifold and that it is Hausdorff I think it's easy to prove. If I understand the concepts correctly, for each point $x\in\overline{\mathbb{R}}$ I need to find a neighborhood $U_x$ of $x$ and a homeomorphism $\phi_x:U_x\rightarrow\phi(U_x)\subseteq\mathbb{R}$, and then I need that the transition maps $\phi_x\cdot\phi_y^{-1}:\phi_y(U_x\cap U_y)\rightarrow\phi_x(U_x\cap U_y) $ between this homeomorphisms to be $C^\infty$.
So I did the following:
For $x\in\mathbb{R}$ I can take $U=\mathbb{R}$ as the open set and
\begin{align} \phi_x=id: & \mathbb{R}\rightarrow\mathbb{R}\\ & x\mapsto id(x)=x \end{align}
as the homeomorphism between them. If $x=\infty$ (conversely for $-\infty$) then I can take $U_\infty=(1,\infty]$ as the open set and map it to $[0,1)\subseteq\mathbb{R}$ using the bijection $\phi_\infty(x)=\frac{1}{x}$. Now $\phi_\infty$ is continuous because any open set in $[0,1)$ can be written as the union of open intervals, furthermore the preimage of an union is the union of the preimages, and finally $\phi_\infty^{-1}((a,b))=(1/b,1/a)$ which is an open set in $\overline{R}$. Also, $\phi_\infty^{-1}=id$ which is also continuous, therefore $\phi_\infty$ is an homeomorphism between $(1,\infty]$ and $\phi((1,\infty])=[0,1)\subseteq\mathbb{R}$.
To conclude, the transition maps between points in $\mathbb{R}$ are again the identity, which is $C^\infty$. Furthermore for all $x\in\mathbb{R}$ we have $\phi_x\cdot\phi_\infty^{-1}:\phi_\infty(U_x\cap U_\infty)\rightarrow\phi_x(U_\infty\cap U_x)=id$, therefore $\phi_x\cdot\phi_\infty^{-1}\in C^\infty$.
I appreciate any feedback! Thanks! :)