differential structure of $S^n$ and diffeomorphism map Let $S^n$ be the unit sphere in $\mathbb{R}^{n+1}$, and $TS^n=\{(x,v): x\in S^n,v\in T_x S^n\}$.
Show that $F: TS^n\times\mathbb{R}\to S^n\times\mathbb{R}^{n+1}$ given by $F((z,v),\lambda)=(z,v+\lambda z)$ is a diffeomorphism.
I started with…
Let $M=S^n$ be the unit sphere in $\mathbb{R}^{n+1}$. For every $x\in S^n$, the tangent space $T_x S^n$ can be identified with the subspace of vectors $v$ orthogonal to $x$. This yields a bijection $F$ from $TS^n$ onto the subset
        \begin{align*}
   \{(x,v)\in S^n\times\mathbb{R}^{n+1},\,\,v\in T_x S^n\}
  \end{align*}
    of $\mathbb{R}^{2(n+1)}$. This is the level set of smooth map
    \begin{align*}
  F: \mathbb{R}^{2(n+1)}\to\mathbb{R}^{n+1},\quad F(x,v):=(\|x\|^{n+1},x\cdot v)
 \end{align*}
    at some suitable regular value.
I want to first identify the level set G, then compute the Derivative at a point $DG|_x$, but i don't know how to find the level set of $S^n$. And 
I don't know how does this help to show the map $F((z,v),\lambda)=(z,v+\lambda z)$ is a diffeomorphism.  
Appreciate for any help.
 A: The easiest approach to me just constructs an inverse and then we observe that both of these functions are smooth so $F$ is a diffeomorphism.
To this end, define $F^{-1}(z,w)=((z,w-\langle z,w\rangle z),\langle z,w\rangle)$. Now we can check that $F\circ F^{-1}$ and $F^{-1}\circ F$ are the identity on their respective domains:
$F^{-1}\circ F((z,v),\lambda)=F^{-1}(z,v+\lambda z)=((z,v+\lambda z-\langle z,v+\lambda z\rangle z),\langle z,v+\lambda z\rangle)=(z,v+\lambda z-\lambda z)=(z,v)$
I used that the norm of $z$ is $1$ and that $z$ and $v$ are perpendicular by definition.
$F\circ F^{-1}(z,w)=F((z,w-\langle z,w\rangle z),\langle z,w\rangle)=(z,w-\langle z,w\rangle z+\langle z,w\rangle z)=(z,w)$
Finally, every operation we're using is smooth, like addition, multiplication  of scalars, and dot product, so both maps are smooth from which we conclude that $F$ is a diffeomorphism.
I can't figure out how to typeset a matrix, but I'll describe $dF$, as per request (see comments). It has $2n+2$ rows and $2n+3$ columns. The top left corner is an $(n+1)\times (n+1)$ identity matrix and the bottom left corner is an $(n+1)\times (n+1)$ identity matrix times $\lambda$. To the right of this $\lambda I$ block is another $(n+1)\times (n+1)$ identity matrix, and directly above this is an $(n+1)\times (n+1)$ zero matrix. In the final column, the top $n+1$ entries are $0$, and the last $n+1$ are the vector $z$.
I had to review a couple definitions, but it seems that the import for $K$-theory is that the tangent bundle over $S^{n}$ and the $n$-dimensional trivial bundle over $S^{n}$ are stably isomorphic and thus represent the same element in the real $K$-theory of $S^n$
