I'm working on the following problem from Lee's Introduction to Smooth Manifolds:
Show that $T\mathbb{S}^1$ is diffeomorphic to $\mathbb{S}^1\times\mathbb{R}$
I am trying to apply a procedure that I saw e.g. in the computation of the Lie Algebra of $SO(3)$:
For some point $p\in\mathbb{S}^1$ take a curve $\gamma(t)\in\mathbb{S}^1$ with $\gamma(0)=p$ and $\gamma'(0)=v$ where $v$ will be the tangent vector. Differentiating the defining property of the sphere ($x^2+y^2=1$) leads to $2x(0)\cdot x'(0)+2y(0)\cdot y'(0)=0 \Leftrightarrow p^T v=0$
Hence, $T_p\mathbb{S}^1=\left\{v\in\mathbb{R}^2:p^Tv = 0\right\}$. Since $p$ is not zero, this is a 1-dimensional subspace of $\mathbb{R}^2$ which is clearly diffeomorphic to $\mathbb{R}$. Then, $T\mathbb{S}^1$ is the union over all $T_p\mathbb{S}^1$ and thus diffeomorphic to $\mathbb{S}^1\times\mathbb{R}$ (for each point $p\in\mathbb{S}^1$ we have a tangent space at p diffeomorphic to $\mathbb{R}$).
Is my reasoning correct? What would be the easiest way to show this?
As pointed out in the comment, your argument is not correct. And the mistake comes from the "thus" in the quoted comment. All you know in general is that the tangent bundle is locally a product:
$$TM|_{U} \cong U \times \mathbb R^n$$
for all small enough open set in a manifold $M$. There are lots of manifolds with nontrivial tangent bundle, as pointed out by yourself.
So in order to show $T\mathbb S^1 \cong \mathbb S^1 \times \mathbb R$, you need more inputs: for each $p\in \mathbb S^1$, from your description, there is a canonical choice of basis of $T_p\mathbb S^1$: for example, $\sqrt{-1} p$ is such a choice. So every vector $v\in T_p\mathbb S^1$ is given by $v = t \sqrt{-1} p$ for some $t\in \mathbb R$. Try to use this $t$ to write down an explicit isomorphism
$$T\mathbb S^1 \to \mathbb S^1 \times \mathbb R, \ \ \ (p, v)\mapsto ??.$$