What is wrong with my proof? (A problem of tangent bundle) I am proving $TS^1$ is diffeomorphic to $S^1\times\mathbb{R}$. The following is my proof and I think it is wrong, because I only use the fact that $S^1$ is 1-dimentional. However, I do not know how to correct my proof. ($S^1$ is the unit circle).
For any $p\in S^1$, we can choose a chart $(U,\varphi)$ around it. Therefore, every $p\in S^1$ is associated with a vector $v_p^0=\frac{\partial}{\partial x}|_p$ if a chart is given.
Now, a function $F$ from $S^1\times\mathbb{R}$ to $TS^1$ is defined by
$$F(p,\lambda)=(p,\lambda v_p^0)$$
I want to show that $F$ is a diffeomorphism.
Clearly, $F$ is injective. For any $(p,v)\in TS^1$, we have
$$v=v^1\frac{\partial}{\partial x}|_p$$
$$v_p^0=v_p^{0,1}\frac{\partial}{\partial x}|_p$$
under some chart around $p$.
Therefore, we choose $\lambda=v^1/v_p^{0,1}$. $\lambda$ should be independet of choice of charts. Therefore, $F$ is also surjective.
Now choose two charts $(U\times\mathbb{R},\varphi\times i)$ and $(\pi^{-1}(U),\tilde{\varphi})$ for $S^1\times\mathbb{R}$ and $TS^1$, respectively. The expression of $F$ is
\begin{align*}
\hat{F}(q,x)&=\tilde{\varphi}\circ F\circ(\varphi\times i)^{-1}(q,x)\\
   &=\tilde{\varphi}\circ F(p,x)\\
   &=\tilde{\varphi}(p,xv_p^0)\\
   &=(q,xv_p^{0,1})
\end{align*}
$\hat{F}$ is smooth, since $v_p^{0,1}$ is smooth with respect to $p$.
For $F^{-1}$, my proof to show that it is smooth is similar. Therefore, $F$ is diffeomorphism.
However, I do not use any specific property of $S^1$ except that $S^1$ is 1-dimentional.
Is my proof correct? If not, how to correct it?
Thanks!
 A: Your function $F$ isn't well-defined. Since $S^1$ isn't diffeomorphic to $\mathbb{R}$, your coordinate patch $U$ can't contain every point of $S^1$. So, if $p\in S^1\setminus U$ and $\lambda$ is any real number, what is $F(p,\lambda)$?
edit: To answer your question below, consider the coordinate chart $\phi(x,y):=x$ defined on the open set $U:=\{(x,y)\in S^1: y>0\}$, where I'm viewing $S^1$ as a subset of $\mathbb{R}^2$. Let $x$ be the coordinate corresponding to this chart. Then for $p\in U$, your $v_p^0$ is $\frac{\partial}{\partial x}\Bigr|_p$. Now, consider the chart $\psi(x,y):=y$ defined on the open set $V:=\{(x,y)\in S^1: x>0\}$. Letting $y$ be the corresponding coordinate, your $v_p^0$ (which to emphasize the different coordinate chart I'll write as $w_p^0$) is $\frac{\partial}{\partial y}\Bigr|_p$. Are these the same? Well, the transition function from the $x$ coordinate chart to the $y$ coordinate chart is 
$$\psi\circ \phi^{-1}(x)=\psi\left(x,\sqrt{1-x^2}\right) = \sqrt{1-x^2},$$
which has derivative 
$$\frac{-x}{\sqrt{1-x^2}}=\frac{-\sqrt{1-y^2}}{y}.$$
So, for $p\in U\cap V$,
$$v^0_p=\frac{\partial}{\partial x}\Biggr|_p = \frac{-\sqrt{1-y^2}}{y} \frac{\partial}{\partial y}\Biggr|_p\neq \frac{\partial}{\partial y}\Biggr|_p=w_p^0.$$
Thus, $v_p^0$ isn't well-defined.
A: What you need to do is find the right atlas on $S^1$, say, the atlas with two charts given by polar coordinates on $S^1 \setminus \{-1\}$ and polar coordinates on $S^1 \setminus \{1\}$, so that $\tfrac{d}{dx}$ patches together to define a nowhere-vanishing vector field $\xi$ on $S^1$. Given such a vector field $\xi$, you can then define your $F : S^1 \times \mathbb{R} \to TS^1$ in a completely coordinate-independent way by $F(p,\lambda) := [(p,\lambda \xi_p)]$, and now use your concrete atlas on $S^1$ and your concrete nowhere-vanishing vector field $\xi$ to show that $F$ is indeed a diffeomorphism.
