Question on the proof of the inverse function theorem

Question

I was reading the proof here for the inverse function theorem. I understand everything except in the middle of page 2, the proof says

$$\sum^n_{i=1}-2(f_i(x_0) - y_i) \frac{\partial{f_i}}{\partial{x_j}}(x_0) = 0$$

Since we know that $det(Df(x_0))$ is not zero and thus invertible, $f_i(x_0) - y_i = 0$.

This is used to show that $f$ maps an open set to an open set. My question: since $Df(x_0)$ is invertible so it only maps from $0$ to $0$. Then what prevents $x_0$ from being $0$?

Well, $x_0$ can be $0$, why not? What would be the problem then? — Berci, Feb 21 '18 at 01:16
I think there's confusion about $Df(x_0)$ being a linear transformation. $Df(x_0)(y)$ is linear in $y$, not linear in $x_0$. So $Df(0)\neq0$. — mr_e_man, Jun 28 '22 at 15:48

score 0 · Answer 1 · answered Feb 17 '19 at 15:12

0

Regard ${z_i}:=-2(f_i(x_0)-y_i)$ and let $\vec{z}=(z_1,z_2.\cdots,z_n)$ as a vector. Now, that expression above is the matrix $Df(x_0)$ multiplied into $z$. Since this product is zero and derivative matrix is nonsingular, we get $z=0$.

answered Feb 17 '19 at 15:12

Behnam Esmayli

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Question on the proof of the inverse function theorem

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