# Proof of the inverse function theorem

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 $$F(x)=f'(x_0)^{-1}(f(x+x_0)-y_0)$$

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.