Show that a map is a submersion Please note: I am still a beginner in differential geometry.
Let $\psi: \mathbb{R}^{n+1}\backslash\{0\} \mapsto S^n, x \mapsto \frac{x}{||x||}$. Show that $\psi$ is a submersion.
The definition I have of submersion is the following: if $M, N$ are smooth manifolds, a map $F:M\mapsto N$  is a submersion if and only if its pushforward $F_{*_p}$ is surjective at every point $p \in M$, that is, $rank(F) = dim(N)$. The rank of a map is the rank of the jacobian matrix of said map.
It's not difficult to compute the jacobian matrix. One has
\begin{equation*}
\frac{\partial}{\partial x_i} \frac{x_i}{||x||}= \frac{x_1^2+\dots+x_{i-1}^2+x_{i+1}^2+\dots+x_n^2}{(x_1^2+\dots+x_n^2)^{\frac{3}{2}}}
\end{equation*}
and
\begin{equation*}
\frac{\partial}{\partial x_i} \frac{x_j}{||x||}= \frac{- x_i x_j}{(x_1^2+\dots+x_n^2)^{\frac{3}{2}}}
\end{equation*}
How do I go on from here? How do I compute the rank of the resulting matrix? Is this the correct method at all?
 A: Hint: Given a curve $\gamma : (-1,1) \to S^{n}$, can you find a lift $\tilde{\gamma}:(-1,1) \to \mathbb{R}^{n+1} \backslash\{0\}$ such that $\psi \circ \tilde{\gamma} = \gamma$? What does this tell you about $\psi_*$? How does the point $p$ come into play here?
Edit: I'll flesh this out to a complete answer. Set $M$=$\mathbb{R}^{n+1} \backslash \{0\}$, $N=S^n$. Recall that the push forward map $\psi_{*p} : T_pM \to T_{\psi(p)}N$ is defined as follows: for any $v \in T_pM$, if $\gamma: (-1,1) \to M$ is a path satisfying $\gamma(0)=p$ and $\gamma'(0)=v$, then $\psi_{*p}(v)=(\psi \circ \gamma)'(0)$. 
Now, if we fix $p \in M$, then for any $v \in T_{\psi(p)}N$ with representative path $\gamma$ in $N$ (so $\gamma(0)=\psi(p) = \frac{p}{\|p\|}$, $\gamma'(0)=v$, and $\|\gamma(t)\|=1$), we may define the path $\tilde{\gamma}$ in $M$ by $\tilde{\gamma}(t)=\|p\| \cdot \gamma(t)$, so that $\psi \circ \tilde{\gamma} = \gamma$ and $\tilde{\gamma}(0)=p$. By definition of the push forward, then,
$$\psi_{*p}(\tilde{\gamma}'(0)) = (\psi \circ \tilde{\gamma})'(0) = \gamma'(0) = v,$$
showing that $\psi_{*p}$ is surjective.
This is an argument that, once you've seen once, you shouldn't need to write out again: it's clear that surjectivity of the pushforward is equivalent to being able to find a lift $\tilde{\gamma}$ through $p$ of any (sufficiently short) path $\gamma$ through $\psi(p)$.
A: You also can prove this without any calculation. Let $\Phi : S\times R^+ \to \bf R^n$ defined by $\Phi(y,t)=t.y$, and $\tilde \Psi (x)= (\Psi(x),{x\over \vert x \vert}) $so that $\tilde \Psi \circ \tilde \Phi  = Id$ and $\tilde \Phi \circ \Psi = Id$. It follows that $\tilde \Phi$ is a diffeormophism, but the projection $S\times R^+\to S$ is a submersion, and therefore $\Phi = p\circ \tilde \Psi$ is a submersion.
