I want to prove that the implicit function theorem implies the inverse function theorem. I saw in another post what's written below as the proof for this but I don't understand what they've done.
$$ \text{ For } f : \mathbb{R}^n \to \mathbb{R}^n \text{, consider } F:\mathbb{R}^n\times\mathbb{R}^n \to \mathbb{R}^n \text{ given by } F({\bf x}, {\bf y}) = f({\bf y}) - {\bf x}$$
Do we consider $f(x)$ to be the implicit function satisfying $F\big(x,f(x)\big)=0$ , and by the definition of $F$ we get $F\big(x,f(x)\big)=0=f\big(f(x)\big)-x \Longrightarrow f\big(f(x)\big)=x$. It seems I was wrong by assuming defining $f$ as the implicit function and I should have just let it be $g$. So if we instead get the final implcation being $f\big(g(x)\big)=x$ is that proof of the inverse function theorem?
Thanks!
No, the notation is messing you up. You want to solve for $y=g(x)$ so that $F(x,g(x))=0$. Can you check that this works?
EDIT: So locally we get $y=g(x)$ if and only if $F(x,y) = f(y)-x = 0$. This says that $f(g(x))-x=0$, and so $f(g(x))=x$.
On the other hand, using the if part of this statement, we note that $F(f(y),y)=f(y)-f(y)=0$, and so we must have $y=g(f(y))$, as desired.