If $f:R^3\rightarrow R$ be a.t. $f(x) = x_1^2+x_2^2+x_3^2$, then I was explained how $df_a:R^3\rightarrow R$ is surjective except except when $f(a)=0$ here Understanding submersions in differential topology
Now, for $f(x) = x_1^2+x_2^2-x_3^2$, $df_a$ is the matrix $(2a_1,2a_2,-2a_3)$. Why is it that for $f(a)=0$, $df_a$ is not surjective in this case also?
If $f(a) = a_1^2+a_2^2-a_3^2=0$, can we not take the derivative map to be the matrix $(2a_1,2a_2,-2\sqrt{a_1^2+a_2^2})$ so that $\forall x\in R$, we can find a point whose image is $x$ under $df_a$ and the pre-image of $x$ is:
$df_a([1.5x/2a_1,1.5x/2a_2,x/\sqrt{a_1^2+a_2^2}) = 2a_1.\dfrac{1.5x}{2a_1}+2a_2.\dfrac{1.5x}{2a_2}- 2\sqrt{a_1^2+a_2^2}.\dfrac{x}{\sqrt{a_1^2+a_2^2}} = x?$ I know that I'm wrong. Can someone please point out where I went wrong? Thanks.
If $f(x_1,x_2,x_3)=x_1^2+x_2^2-x_3^2$, $df_a$ is surjective for every $a\in\mathbb{R}^3-\{0\}$. So it is not surjective on every point of $\{a:f(a)=0\}$.