I am studying the proof of the inverse function theorem (see picture) in multivariable analysis. I understand the proof, but I do not understand why we can assume without loss of generality that $x_0=y_0=0$ and $f'(0)=\operatorname{Id}$.

A hint I found online: use $F(x)=f'(x_0)^{-1}(f(x+x_0)-y_0)$

Could somebody please help me? Thank you!

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Suppose that the theorem holds when $x_0=y_0=0$. Then, for a general function $f$, you define $g(x)=f(x+x_0)-y_0$. Since you know that the theorem holds for $g$ at $0$, it also holds for $f$ at $x_0$.

And if the theorem holds for $f$ under the extra assumption that $f'(x_0)=\operatorname{Id}_{\mathbb R^n}$, then, in the general case, let $T=\bigl(f'(x_0)\bigr)^{-1}$. Then, by the chain rule, the derivative of $T\circ f$ at $x_0$ is $\operatorname{Id}_{\mathbb R^n}$. So, the theorem holds for $T\circ f$. But then, since $T$ is a diffeomorphism, the theorem holds for $f$ too.

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