I have a question with its solution. But I dont understand some parts of the solution. I wrote why? near inapparent parts. Please can one explain there? Thank you.
Question:
If $x_0^2+ s_0^2+t_0^2=1$ and $x_0\not =0$
Prove that there exist $r>0$ and a function $g(s,t)$ continuously differentiable on $B_r(s_0,t_0)$ s.t. $x_0=g(s_0,t_0)$ and $x^2+s^2+t^2=1$ for $x=g(s,t)$
Solution:
Define $F(x,s,t)=(x^2+s^2+t^2-1, s,t)$
Let's find $DF(x,s,t)=\begin{pmatrix}2x&2s&2t \\ 0&1&0\\0&0&1\end{pmatrix}$
Then, $\nabla _{F(x,s,t)}= 2x.1.1-2s.0-2t.1.0=2x\not = 0$ for $x\not=0$
At $x=1, s=t=0$ $F(1,0,0)=(0,0,0)$
here, why choose x=1 and s=t=0? Do we want $DF(x,s,t)$ to be identity matrix?
Since $F$ is continuously differentiable and $\nabla_{F(1,0,0)}\not=0$, $F^{-1}$ is continuously differentiable in a nbdhood of $(0,0,0)$
here, since F(1,0,0)=(0,0,0) $\iff$ $F^{-1}(0,0,0)=(1,0,0)$, we wrote "in a nbdhood of (0,0,0). Is it?
Set $F^{-1}(x,y,z)=(\alpha(x,y,z),s,t)$ Since $F^{-1}(0,0,0)$ is continuously diff, $\alpha(x,y,z)$ is continuously differentiable in a nbdhood of $(x,y,z)=(0,0,0)$
Now, define $g(s,t):=\alpha(0,s,t)$
why do we define g(s,t) so? *After that, how can I continue to solve?*
First of all, you should know the big picture: what you're doing is deriving the implicit function theorem (in particular, the implicit function theorem for a single equation) from the inverse function theorem.
The reason they choose $(x,s,t)=(1,0,0)$ is that it is an obvious point on the surface where $x\neq 0$. It has nothing to do with wanting the identity matrix.
One thing that might be confusing you is that you use $(x,y,z)$ in the target space, while you have already used $x$ in the domain. So let's call the target space $(u,v,w)$, and the domain $(x,s,t)$, and we've defined a function as follows:
$$\left\{ \begin{array}{ll}u = x^2 + s^2 + t^2 - 1 \\ v = s \\ w = t\end{array} \right\}.$$
We're obviously interested in the points where $u=0$. We have an inverse function
$$\left\{ \begin{array}{ll}x = \alpha(u,v,w) \\ s=v \\ t=w\end{array} \right\},$$
near a point where $x\neq 0$. So going back to the first equation we can write $u = (\alpha(u,v,w)^2 + s^2 + t^2 - 1)$, in a neighborhood of the point we're looking at. But $u=0$ in the area we're concerned about, so that gives us $0 = (\alpha(0,v,w)^2 + s^2 + t^2 - 1)$. And we may as well ignore the 0 and just consider $\alpha$ as a function of $x$ and $y$! And remember, $u=s$ and $v=t$, so we just have $0 = \alpha(s,t)^2 + s^2 + t^2 - 1$, in a neighborhood of the given point. (So I've written $\alpha$ where you wrote $g$.)