# Regarding the smooth structure on the tangent bundle of a manifold

I have a question regarding the proof of the fact that the tangent bundle of a smooth manifold is itself a smooth manifold.

We usually start from considering the canonical projection $$\pi:TM\rightarrow M.$$ For a chart $(U,\phi)$ on $M$, we have $\pi^{-1}(U)\subset TM$. Then a new map $$\widetilde{\phi}:\pi^{-1}(U)\rightarrow \mathbb{R}^{2n}$$ is defined by $\widetilde{\phi}\big(v^i\frac{\partial}{\partial x^i}\rvert_p\big)=(x^1(p),\dots,x^n(p),v^1,\dots,v^n)$, where $\phi=(x^1,\dots,x^n)$.

But isn't $v^i\frac{\partial}{\partial x^i}\rvert_p$ a tangent vector? At the same time, $\widetilde{\phi}$ is supposed to be applied to elements of $\pi^{-1}(U)\subset TM$, i.e., to tangent spaces. I don't get it. Moreover, even if I accept that there is nothing strange about it, why does this tangent vector has such form (i.e., $v^i\frac{\partial}{\partial x^i}$)?

• Do you understand that the tangent space $T_xM$ at a point $x$ is $n$-dimensional, where $n$ is the dimension of $M$? Do you know what is $\frac{\partial}{\partial x^i}$? – user99914 Aug 30 '16 at 19:25
• @ArcticChar: I know that the $n$ elements $\frac{\partial}{\partial x^i} \rvert_x$ (for $i=1,\dots, n$) of $T_xM$ form its basis. Okay, the second question is eliminated (provided the argument of $\widetilde{\phi}$ is a sum)... – Cary Aug 30 '16 at 19:44