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 $0$?

  • $\begingroup$ Well, $x_0$ can be $0$, why not? What would be the problem then? $\endgroup$
    – Berci
    Commented Feb 21, 2018 at 1:16
  • $\begingroup$ How can I get $f_i(x_0) = y_i$? $\endgroup$ Commented Feb 21, 2018 at 1:35
  • $\begingroup$ 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 $Df(0)\neq0$. $\endgroup$
    – mr_e_man
    Commented Jun 28, 2022 at 15:48

1 Answer 1


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 $z=0$.


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