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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!

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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.

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  • $\begingroup$ So if we have that our implicit function is $g(x)$ then $F\big(x,g(x)\big)=f\big(g(x)\big)-x=0 \Longrightarrow f\big(g(x)\big)=x $. So is the idea that if we have that $g(x)$ is an implicit function that satisfies $F\big(x,g(x)\big)=0$, then by defining $F$ the way they did shows that $f$ will be the inverse function? $\endgroup$ – john fowles Jul 5 '18 at 10:39
  • $\begingroup$ No, it shows that $g$ is the candidate for the (local) inverse function of $f$. Of course, to have an inverse function, you must check both $f(g(x))=x$ and $g(f(y))=y$ ... $\endgroup$ – Ted Shifrin Jul 5 '18 at 15:20
  • $\begingroup$ I'm tempted to use the same $g$ and write $F\big(g(y),y\big)$, but am not sure if this is fine? If I can then I get $f(y)-g(y)=0 \Longrightarrow f(y)=g(y)$ and from that I would write $g\big(f(y)\big)=g\big(g(y)\big)$, and I'm not sure how to show that this would equal $y$. $\endgroup$ – john fowles Jul 5 '18 at 17:14
  • $\begingroup$ This doesn't make sense. Remember that $y$ is your variable in the domain $\Bbb R^n$ and $x$ is your variable in the range $\Bbb R^n$. Whereas $f$ maps $y$ to $x$, the inverse function maps $x$ to $y$. The domains of $f$ and $g$ are totally different. ... See my edit for more clarification. $\endgroup$ – Ted Shifrin Jul 5 '18 at 20:10
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    $\begingroup$ Well, we're working only at points $(x_0,y_0)$ for which $Df(y_0)$ is nonsingular (and $F(x_0,y_0)=0$). The theorem gives a neighborhood on which my statement applies. This is what I meant by "locally." $\endgroup$ – Ted Shifrin Jul 6 '18 at 13:34

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