I was asked for a homework question to prove that $\mathbb{S}^1 \times \mathbb{R}$ is diffemorphic to $T\mathbb{S}^1$. Intuitively, it was clear to me how this could be the case, but when i attempted to prove it, I came up with a proof which seems to work for every smooth manifold $M.$ Since this conclusions is false, I was curious if anyone could tell me where I went wrong in my proof. I wrote:
We will find a diffeomorphism between $T\mathbb{S}^1$ and $\mathbb{S}^1 \times \mathbb{R}.$ Let $(a,b) \in \mathbb{S}^1$ and let $v\in T_{(a,b)}\mathbb{S}^1$ be a tangent vector. By Proposition 3.15, $v$ can be written as $x\partial/\partial \theta \mid_{(a,b)},$ for $x\in\mathbb{R}$ and where $\theta$ is the pre-image of the real coordinate $\theta\in \mathbb{R}.$ In other words, if $(U, \phi)$ is a smooth chart containing $(a,b),$ then \begin{align*} \frac{\partial}{\partial \theta} \biggr|_{(a,b)} = (d\phi)^{-1}_{\phi(a,b)} \left( \frac{\partial}{\partial \theta} \biggr|_{\phi(a,b)}\right). \end{align*} So every $v\in T_{(a,b)}\mathbb{S}^1$ has a unique real "coordinate" $x.$ Now to construct a diffeomorphism. First, let $\pi : T\mathbb{S}^1 \to \mathbb{S}^1$ be the projection. Let $((a,b), x \partial/\partial \theta |_{(a,b)})$ be an element of $T\mathbb{S}^1.$ Consider a smooth coordinate chart $(U, \phi)$ containing $(a,b).$ As per the proof of Proposition 3.18, $\pi^{-1}(U)$ forms the domain of a smooth chart, together with the diffeomorphism $\overline{\phi},$ which is defined as \begin{align*} \overline{\phi}\left( (c, d), x\frac{\partial}{\partial \theta} \biggr|_{(c,d)}\right) = (\phi(c,d), x). \end{align*} Now let \begin{align*} F_U: \pi^{-1} (U) \to \mathbb{S}^1 \times \mathbb{R}: \left( (c, d), x\frac{\partial}{\partial \theta} \biggr|_{(c,d)}\right) \mapsto \left(\phi^{-1}\left(\phi(c,d)\right), x\right). \end{align*} In other words $F_U = (\phi^{-1}, Id_\mathbb{R}) \circ \overline{\phi}.$ Since both of these maps are diffeomorphism, $F_U$ is a diffeomorphism as well. We know that the domains of the charts cover $\mathbb{S}^1.$ Furthermore, it is clear that $F_U$ and $F_V$ agree on $\pi^{-1}(U) \cap \pi^{-1}(V),$ since for a point $((a,b), x \partial/\partial \theta |_{(a,b)})$ both the coordinate $(a,b)$ and the "coordinate" $x$ are independent of the choice of chart containing $((a,b), x \partial/\partial \theta |_{(a,b)})$ and $F_U$ and $F_V$ only depend on these two things. So $F_U$ can be extended to a smooth function $F$ on $T\mathbb{S}^1$ that has a smooth inverse. $F$ is injective, because $F$ once again only depends on the coordinate $(a,b)$ and the "coordinate" $x,$ which are unique for every element of $T\mathbb{S}^1$ by definition and because $T_{(a,b)}\mathbb{S}^1$ is a vector space. Notice that \begin{align*} \text{Im } F &= \bigcup_{\substack{\pi^{-1}(U) \subseteq T\mathbb{S}^1 \\ (\pi^{-1}(U), \overline{\phi}) \text{ is a chart}}} \text{ Im }F_U \\ &= \bigcup_{\substack{\pi^{-1}(U) \subseteq T\mathbb{S}^1 \\ (\pi^{-1}(U), \overline{\phi}) \text{ is a chart}}} U \times \mathbb{R} \\ &= \mathbb{S}^1 \times \mathbb{R}. \end{align*} The second equality holds, since $F_U$ is a diffeomorphism and the third equality holds since the domains charts $(U, \phi)$ of $\mathbb{S}^1$ cover $\mathbb{S}^1.$ So $F$ is surjective. This means $F$ is a diffeomorphism between $T\mathbb{S}^1$ and $\mathbb{S}^1 \times \mathbb{R}.$
Proposition 3.15 states that tangent spaces are vector spaces and specifies their basis. The proof of Proposition 3.18 constructs the smooth charts for the tangent bundle of a manifold in terms of the smooth charts of the manifold.