I know this is a very basic question. But I am unable to get every detail right.

> 
> I need to show that $TS^1$ is diffeomorphic to $S^1\times \mathbf R$.
> 
> 
> 

(I am using the concept of derivations to define the tangent spaces.)

I asked this of my friends and this is what, in essence, I have learned from them:

Define $f:TS^1\to S^1\times \mathbf R$ as 
$$f
\left(
p;\ \lambda\left(p\_2
\left.\frac{\partial}{\partial x\_1}\right|\_{p}-p\_1\left.\frac{\partial}{\partial x\_2}\right|\_p
\right)
\right)
=
(p,\ \lambda)
$$
for all $p=(p\_1,p\_2)\in S^1$ and all $\lambda\in \mathbf R$, and show that this is a diffeomorphism.

***My problem with this is that:*** 

$\left(p\_2
\left.\frac{\partial}{\partial x\_1}\right|\_{p}-p\_1\left.\frac{\partial}{\partial x\_2}\right|\_p
\right)$ is *not really* a member of $T\_pS\_1$.

It lies in $di\_p(T\_pS^1)\subseteq T\_p\mathbf R^2$, where $i:S^1\to \mathbf R^2$ is the inclusion map.

(Reason: The manifold structure of $S\_1$ is governed by the fact that $i:S^1\to \mathbf R^2$ is a smooth embedding, and thus, a tangent vector $X\_p\in T\_p\mathbf R^2$ is in $di\_p(T\_pS^1)$ if an only if $X\_p\xi=0$ for all $\xi\in \mathcal C^\infty(\mathbf R^2)$ with $\xi|S^1\equiv 0$.)

Keeping this in mind, I attempted the following:

Define $F:T\mathbf R^2\to \mathbf R^2\times \mathbf R^2$ as 
$$
F\left(p,\ a\_1
\left.
\frac{\partial}{\partial x\_1}
\right|\_p+a\_2\left.\frac{\partial}{\partial x\_2}\right|\_p\right)
=
(p,\ a\_1, a\_2)
$$
for all $p\in \mathbf R^2$ and $a\_1, a\_2\in \mathbf R$. We note that $F$ is a diffeomorphism.

Now define $G:\mathbf R^2\times \mathbf R^2\to \mathbf R^2\times \mathbf R$
as
$G(p, a\_1, a\_2)=(p, \sqrt{a\_1^2+a\_2^2})$. We note that $G$ is smooth.

Finally, since $i:S^1\to T\mathbf R^2$ is smooth (it is more than that), we have $di:TS^1\to T\mathbf R^2$ is also smooth.

Now $G\circ F\circ di: TS^1\to S^1\times \mathbf R$ is thus a smooth map, since composition of smooth maps is smooth.

*But I am unable to show that this map is the required diffeomorphism.*

Can somebody help?