T$\mathbb{S}^{n} \times \mathbb{R}$ is diffeomorphic to $\mathbb{S}^{n}\times \mathbb{R}^{n+1}$

My strategy is find a diffeomorphism $\psi: T\mathbb{S}^{n} \times \mathbb{R} \to \mathbb{S}^{n}\times \mathbb{R}^{n+1}$ from other diffeomorphisms. First let $(\mathbb{R},Id_{\mathbb{R}})$ the 1-dimensional euclidean space with canonical chart, which is a global chart. Then $T\mathbb{R} \cong \mathbb{R}\times \mathbb{R}$, thus

$T\mathbb{S}^{n} \times \mathbb{R} \times \mathbb{R} \xrightarrow{(Id)^{-1}} T\mathbb{S}^{n}\times T\mathbb{R}$,

Now, the product of tangent bundles are diffeomorphic to tangent bundle of T($\mathbb{S}^{n}\times \mathbb{R})$, thus

$T\mathbb{S}^{n} \times \mathbb{R} \times \mathbb{R} \xrightarrow{(Id)^{-1}} T\mathbb{S}^{n}\times T\mathbb{R} \xrightarrow{\phi} T(\mathbb{S^{n}\times \mathbb{R})}$.

where $\phi:T(\mathbb{S}^{n} \times \mathbb{R}) \to T\mathbb{S}^{n}\times T\ \mathbb{R}$ is a diffeomorphism mentioned above. My question is there exists a global chart for $\mathbb{S}^{n}\times \mathbb{R}$? If it is true this problem is over. Right?


Solutions Collecting From Web of "T$\mathbb{S}^{n} \times \mathbb{R}$ is diffeomorphic to $\mathbb{S}^{n}\times \mathbb{R}^{n+1}$"

You will not be able to conclude this way, as $TS^n$ is not a trivial bundle. Moreover, I believe there are non diffeomorphic spaces with diffeomorphic tangent bundle (even if I can’t find an example now).

Instead, you can notice that the vector bundle $V = \{ (x,v) : x \in S^n, v \in \mathbb R^{n+1} \}$ is trivial, in particular diffeomorphic to $S^n \times \mathbb R^{n+1}$.

But if $T = \{ (x,v) : x \in S^n, v \in \mathbb R^{n+1}, \langle x, v \rangle = 0 \}$ and $N = \{ (x,v) : x \in S^n \subset \mathbb R^{n+1}, v = \lambda x \}$, we have an isomorphism $T \oplus N \cong V$. Do you see how this solve your problem ?

I think that you were on the right track from the beginning.

Consider the smooth embedding $~\psi:\mathbb{S}^n\times\mathbb{R}\hookrightarrow\mathbb{R}^{n+1}$ given by
where $\alpha:\mathbb{R}\to\left(-\frac{1}{2},\frac{1}{2}\right)$ is any diffeomorphism. Then we obtain an induced diffeomorphism

Then we have a composition of diffeomorphisms:
$$T\mathbb{S}^{n} \times \mathbb{R} \times \mathbb{R} \xrightarrow{\text{id}^{-1}} T\mathbb{S}^{n}\times T\mathbb{R} \xrightarrow{\phi} T(\mathbb{S^{n}\times \mathbb{R})}\xrightarrow{T\psi} \mathbb{S}^n\times\mathbb{R}\times\mathbb{R}^{n+1}$$

We conclude by noting that $T\mathbb{S}^n\times\{0\}\times\mathbb{R}$ maps to $\mathbb{S}^n\times\{0\}\times\mathbb{R}^{n+1}$ under the composition above.