Understanding submersions in differential topology I'm having trouble understanding an example in Guillemin and Pollack. If $f:R^k\rightarrow R$ be defined by  
$f(x) = |x|^2 = x_1^2+...+x_k^2$
The derivative $df_x$ at the point $a = (a_1,..,a_k)$ has matrix $(2a_1,..,2a_k)$. Thus
$df_a:R^k\rightarrow R$ is surjective unless $f(a)=0$, so every nonzero real number is a regular value of $f$. I can't understand how to check $df_a$ is surjective and why it is obvious that every nonzero real number is a regular value of $f$? A hint is appreciated. Thanks.
 A: We know that the differential of $f(x^1,\dots,x^k) = (x^1)^2 + \cdots + (x^k)^2$ at $a=(a_1,\dots,a_k) \in \mathbb{R}^k$ is $df_a = \sum 2a_i \, \text{d}x^i $. It is clear that if $f(a)=a_1^2+\cdots+a_k^2 = 0$, then $a=0$ which implies $df_a = (0,\dots,0)$ is the zero map (obviously not surjective). 
For $f(a)\neq 0$ happen exactly at $a\neq 0$, so at least there is one component of $a$ which is nonzero. W.l.o.g assume $a_1 \neq 0$. For any $\lambda \in \Bbb{R}$, 
$$
df_a\Big(\frac{\lambda}{2a_1} \frac{\partial}{\partial x^1}\Big)  =\frac{\lambda}{2a_1}\,  df_a\Big( \frac{\partial}{\partial x^1} \Big) =    \frac{\lambda}{2a_1} \cdot 2a_1= \lambda.
$$
So $df_a : T_a \mathbb{R}^k \to \mathbb{R}$ is onto. 
A: Hint:  What is the rank of the linear map $2(a_1,\dots, a_k):\Bbb R^k\rightarrow \Bbb R$?  (And, what is the dimension of $\Bbb R$?)
A: For $a \neq 0$, the reason why $df_a$ is surjective because $df_a$ is a linear map from $T_p \mathbb{R}^k$ to $T_{f(p)} \mathbb{R}$ and since $\operatorname{rank} df_a = 1 = \operatorname{dim} T_{f(p)}\mathbb{R}$ it follows that $df_a$ is surjective.
To see that $\operatorname{rank} df_a = 1$, note that the matrix $[ 2a_1 \dots 2a_k]$ will always have some entry to be non-zero (since we assumed that $a \neq 0$ so some $a_i$ must be non-zero for some $0 \leq i \leq k$) and having one non-zero entry is all we need to conclude that the rank of this matrix is $1$.
Recall that the for a smooth map $F : M \to N$ between smooth manifolds, a point $c \in N$ is a regular value of $F$ if and only if either $c$ is not in the image of $F$ or at every point $p \in F^{-1}(c)$ we have $dF_p : T_pM \to T_{F(p)}M$ to be surjective.
So choose some non-zero real number $y$, if $y \not\in f[\mathbb{R^k}] $ then we're done otherwise suppose that $y \in f[\mathbb{R^k}]$, then $f^{-1}(y)$ is nonempty. Choose any $a \in f^{-1}(y)$, and then note that $a$ must clearly be non-zero in $\mathbb{R}^k$, thus by the above we have $df_a$ to be surjective and since $a$ was chosen arbitrarily here, this holds for all $a \in f^{-1}(y)$, and so it follows that every non-zero real number is a regular value for $f$.
