In Lee's Introduction to Smooth Manifolds, he says that
…the tangent bundle of a smooth manifold has a natural structure as a smooth manifold in its own right. The natural coordinates we constructed on $TM$ make it look, locally, like the Cartesian product of an open subset of $M$ with $\mathbb{R}^n$.
He defines the natural coordinates on $TM$ to be $(x^i, v^i)$ given by $\tilde{\varphi}: \pi^{-1}(U) \rightarrow \mathbb{R}^{2n}$, where
$$\tilde{\varphi}\Big(v^i \frac{\partial}{\partial x_i}\Big) = (x^1(p), \ldots, x^n(p), v^1, \ldots, v^n), \quad p \in M $$
and $\pi: TM \rightarrow M$ is the natural projection map $\pi(p, v) = p$.
To me, it seems that we use $\tilde{\varphi}$ to make $TM$ locally look like an open subset of the cartesian product of $\mathbb{R}$ with itself, so $\mathbb{R}^{2n}$. How does this imply that $TM$ locally looks like an open subset of $M \times \mathbb{R}$?
Best Answer
$TM$ is dimension $2n$, so locally it looks like $U\times \mathbb{R}^{n}$ (not $U\times \mathbb{R}$), where $U\subset M$ can be identified with an open subset in $\mathbb{R}^{n}$. On this open subset, the first $n$ coordinates are position coordinates $x^1,...,x^n$, while the second $n$ coordinates are velocity coordinates $\partial_1,...,\partial_n$.