Diffeomorphism between $TS^1$ and $S^1\times\mathbb{R}$

Question

I want to show that there exists a diffeomorphic $\phi$ such that the following diagram commutes: $$ \require{AMScd} \begin{CD} TS^1 @>{\phi}>> S^1\times\mathbb{R}\\ @V{\pi}VV @V{\pi_1}VV \\ S^1 @>{id_{S^1}}>> S^1 \end{CD}$$ where $\pi$ is the associated projection of $TS^1$, and $\pi_1(x,y)=x$ is the standard projection function in the first component.

A hint was given along with the exercise that I should find a nowhere vanishing vector field on $S^1$. However, I don't know how to find one exactly, or what to do subsequent to finding such a vector field. I have seen an analogous example where $\phi$ was given without reason where $S^1$ and $\mathbb{R}$ were both instead $\mathbb{R}^n$. The definition of that $\phi$ was:$$\phi(a^i\frac{\partial}{\partial x^i}(p)) = (p,(a^1,...,a^n)).$$Perhaps the nowhere vanishing vector field on $S^1$ is used in an analogous formula?

Could anyone give some additional hints or a sketch of a proof?

EDIT: Thinking about it, if I get the nowhere vanishing vector field, say, $u$, then because $S^1$ is a 1-manifold, I have that $T_pS^1$ is 1-dimensional as well. So that means that $T_pS^1$ is spanned by $u_p$. So I am thinking we use $\forall v_p\in TS^1$ the unique coefficient given by $\alpha\in\mathbb{R}$ such that $v_p = \alpha u_p$. So perhaps:$$\phi(v_p)=(p,\alpha),$$is our diffeomorphism? In that case, is there a condition that is met by $S^1$ such that it has to have a nowhere vanishing vector field (i.e. I don't have to find an exact formula for one)?

Answer 1

$S^1$ is a Lie group, so the left-multiplication map $L_g h = gh$ induces isomorphisms $DL_g: T_1 S^1 \to T_gS^1$ on each fiber that give a trivialization $DL_* : T_1 S^1 \times S^1 \to TS^1$.

Answer 2

You can embed $S^1=\{(x,y):x^2+y^2=1\}$ into $\mathbb{R}^2$ and then think of $TS^1$ as a subset of $T\mathbb{R}^2=\mathbb{R}^2\times \mathbb{R}^2$ defiend by $\{(x,y)\in \mathbb{R}^2\times \mathbb{R}^2:x\in S^1,y=t(y,-x),t\in \mathbb{R}\} $.

Answer 3

You can parameterize $S^1$ by $(\cos\theta, \sin\theta)$. A unit tangent at each point is given by $\langle -\sin\theta, \cos\theta\rangle$. So, you get a map $S^1\times \mathbb{R}\rightarrow TS^1$ given by $$ ((\cos\theta,\sin\theta),a)\mapsto ((\cos\theta,\sin\theta),\langle -a\sin\theta,a\cos\theta\rangle). $$