Partial derivatives $\frac{\partial}{\partial x^1},\dots,\frac{\partial}{\partial x^n}$ form a smooth frame on $\mathbb{R}^n$ Tu defines (in An Introduction to Manifolds, pp. 137-138) smooth frames as a collection of smooth sections $s_1,\dots,s_n$, s.t. $s_1(p),\dots,s_n(p)$ form a basis for the fiber $\pi^{-1}(p)$ in the vector bundle $\pi:E \to M$ over an open set $U$, $p \in U$. Then Tu gives an example that the collection of vector fields $\frac{\partial}{\partial x}, \frac{\partial}{\partial y}, \frac{\partial}{\partial z}$ is a smooth frame on $\mathbb{R}^3$. What I'm wondering is that what is the technical argument on the frameness of the collection $\frac{\partial}{\partial x}, \frac{\partial}{\partial y}, \frac{\partial}{\partial z}$, that is, why does each of the partial derivative classify as a section? Are the partial derivatives is sections, because each of the partial derivative is evaluated at a point $p$, so that the partial derivative at a point classifies as a mapping $f(p) = \frac{\partial}{\partial x}|_p =  p$, so that $(\pi \circ f)(p) = p$, with the natural projection mapping of the vector bundle?
There already exists a similar post: Section 12.5 - Tu's Introduction to manifold, smooth frame in $\mathbb{R}^3$, but it didn't answer my question.
 A: The tangent bundle $TM$ was originally defined as the set $\bigcup_{p \in M } T_pM$, the union of all tangent spaces (which are pairwise disjoint due to Tu's definition of $T_pM$). Tu defined a topology and a smooth structure on $TM$ which makes $TM$ a smooth manifold. You should read again p. 131 / 131. He shows that each chart $(U,\phi)$ on $M$ induces a bijection $\tilde \phi : TU \to \phi(U) \times \mathbb R^n$ which allows to topologize $TU$. The inverse $\psi : \phi(U) \times \mathbb R^n \to TU $ of $\tilde \phi$ is given by
$$\psi(\phi(p),c^1,\ldots,c^n) = \sum c^i \frac{\partial}{\partial x^i} \mid_p$$
where the $\frac{\partial}{\partial x^i} \mid_p$ are the derivations introduced on p. 87. Be aware that these depend on the chart $\phi$.
Let us see what happens for $M = \mathbb R^n$. Here we can use the single chart $id : \mathbb R^n \to \mathbb R^n$ and get $\tilde{id} : T\mathbb R^n \to  \mathbb R^n \times \mathbb R^n$. Its inverse $\psi$ is given by
$$\psi(p,c^1,\ldots,c^n) = \sum c^i \frac{\partial}{\partial x^i} \mid_p .$$
But here the $\frac{\partial}{\partial x^i} \mid_p$ are the standard partial derivatives defined for multivariable functions $f : W \to \mathbb R$, where $W$ is an open neigborhood of $p$ in $\mathbb R^n$.
The maps $s_i : \mathbb R^n \to T\mathbb R^n, s_i(p) = \frac{\partial}{\partial x^i} \mid_p$, are in fact smooth sections. By definition $s_i(p) \in T_p\mathbb R^n$. Moreover, the maps $\epsilon_i : \mathbb R^n \to \mathbb R^n \times \mathbb R^n, \epsilon_i(p) = (p,\delta_{i1},\ldots,\delta_{n1})$, with $\delta_{ij} = 0$ for $i \ne j$ and $\delta_{ij} = 1$ for $i =j$, are smooth. Hence $s_i = \psi \circ \epsilon_i$ is smooth. Clearly the $s_i(p) =  \frac{\partial}{\partial x^i} \mid_p$ form a basis of $T_p\mathbb R^n$ because they are the images of the standard basis vectors of $\{p\} \times \mathbb R^n$. Hence the $s_i$ are a (global) frame for $T\mathbb R^n$. Tu simply writes $s_i =  \frac{\partial}{\partial x^i}$.
