I have shown that the graph of the function $f(x)=|x|$ is a manifold covered by one chart. I now need to show that on this smooth structure $\Gamma_f=\{(x,f(x))|x\in{\mathbb{R}}\}$ is diffeomorphic to ${\mathbb{R}}$.
There must be a gap in my understanding because I'm thinking that since |x| is not differentiable at 0 they should not diffeomorphic?
Can someone give me a hint as to what I am missing?
**To clarify what I'm thinking: for $\Gamma_f$ and ${\mathbb{R}}$ the function $\phi:\Gamma_f \to{\mathbb{R}}$ given by $\phi(x,|x|)=x$ with inverse $\phi^{-1}(y)=(y,|y|)$ would make $\Gamma_f$ and ${\mathbb{R}}$ diffeomorphic, but the inverse is not smooth at $x=0$?
Should I create use multiple charts to show they are diffeomorphic? is it because the inverse is in fact diffeomorphic on $\Gamma_f$ somehow? Where am I going wrong?
The graph $\Gamma$ of a continuous function $f : X \to Y$ on a topological space $X$ is always homeomorphic to $X$, so if $X$ happens to be $\Bbb{R}$ then you can use the homeomorphism to make $\Gamma$ into a differentiable 1-manifold that will necessarily be diffeomrphic with $\Bbb{R}$ (and only one chart is needed for this). However, if the range $Y$ of $f$ is also $\Bbb{R}$, the continuity of $f$ is not enough to imply that $\Gamma$ is a smooth sub-manifold of $X \times Y$. That is exactly what is happening in your example: you have equipped the graph of $|\cdot|$ with the structure of a differentiable 1-manifold. Given that structure, the graph is not a smoooth sub-manifold of $\Bbb{R} \times \Bbb{R}$, but that is a property of the embedding of the graph in $\Bbb{R} \times \Bbb{R}$ and not an intrinsic property of the differentiable 1-manifold structure that you have imposed on the graph.