# Proof explanation of an inverse function theorem

Let $$I\neq\emptyset$$ be an open interval and $$f:I\to\mathbb{R}$$ a continuous and injective function. If $$f$$ is differentiable at $$x_0\in I$$ and $$f'(x_0)\neq 0$$, then $$f^{-1}$$ is differentiable at $$f'(x_0):=y_0$$ and we have that $$(f^{-1})'(y_0)=\frac{1}{f'(x_0)}=\frac{1}{f'(f^{-1}(y_0))}.$$

Proof:

Note that $$f(I)$$ is an interval, $$f:I\to\mathbb{R}$$ is an homomorphism, and $$f^{-1}:J\to I$$ is continuous. Now, $$f$$ and $$f^{-1}$$ are strictly monotonic (by a a lemma we previously proved).

Let $$(y_n)\in J\setminus\{y_0\}$$ be any sequence such that $$y_n\to y_0$$, and let $$x_n=f^{-1}(y_n)$$.

The continuity and injectivity of $$f^{-1}$$ implies that $$(x_n)$$ is a sequence in $$I\setminus\{x_0\}$$ such that $$x_n\to x_0$$. Since $$f$$ is differentiable at $$x_0$$ and $$f^{-1}(x_0)\neq 0$$, then we have $$\displaystyle\lim_{n\to\infty}\frac{f^{-1}(x_n)-f^{-1}(x_0)}{y_n-y_0}=\displaystyle\lim_{n\to\infty}\frac{x_n-x_0}{f(x_n)-f(x_0)}=\frac{1}{f'(x_0)}.$$

What I don't understand is how the continuity and injectivity of $$f^{-1}$$ implies that $$(x_n)$$ is a sequence in $$I\setminus\{x_0\}$$ such that $$x_n\to x_0$$, and how to go from $$\displaystyle\lim_{n\to\infty}\frac{x_n-x_0}{f(x_n)-f(x_0)}$$ to $$\frac{1}{f'(x_0)}$$. Any help is welcome

Well, each $$x_n = f^{-1}(y_n)$$. So since $$f^{-1} : J \to I$$, we at least have $$x_n \in I$$. Also, each $$y_n \neq y_0$$ (because $$y_n$$ was chosen from $$J \setminus \{y_0\}$$). So, by definition of injectivity, $$x_n = f^{-1}(y_n) \neq f^{-1}(y_0) = x_0$$ Thus, $$x_n \neq x_0$$ and so $$x_n \in I \setminus \{x_0\}$$. Finally, because $$f^{-1}$$ is continuous, it commutes with limits: $$x_0 = f^{-1}(y_0) = f^{-1}(\lim_{n \to \infty}y_n) = \lim_{n \to \infty}f^{-1}(y_n) = \lim_{n \to \infty}x_n$$ So $$x_n \to x_0$$.
Next, let $$L := f'(x_0)$$ and let $$D(x) := \frac{f(x) - f(x_0)}{x - x_0}$$. This is a function $$D : I \setminus \{x_0\} \to \mathbb{R}$$. $$x_0$$ is not in the domain of $$D$$ but it is still an adherent point of that domain. So $$\lim\limits_{x \to x_0; x \in I \setminus \{x_0\}}D(x)$$ makes sense and, in fact, this limit is just $$L$$. In other words, $$D(x)$$ converges to $$L$$ at $$x_0$$ in $$I \setminus \{x_0\}$$. Or equivalently, by the sequential criterion for convergence:
If $$x_n \to x_0$$ where $$x_n$$ is in the domain of $$D$$ (i.e. $$I \setminus \{x_0\}$$), then $$D(x_n) \to L$$.
As a result, $$\lim\limits_{n \to \infty} D(x_n) = L$$. So, as long as $$D(x_n) \neq 0$$ and $$L \neq 0$$, we also have $$\lim_{n \to \infty} \frac{x_n - x_0}{f(x_n) - f(x_0)} = \lim_{n \to \infty} \frac{1}{D(x_n)} = \frac{1}{L} = \frac{1}{f'(x_0)}$$ by elementary limit laws for sequences. But lo and behold, we indeed have $$L = f'(x_0) \neq 0$$ and $$D(x_n) = \frac{f(x_n) - f(x_0)}{x_n - x_0} \neq 0$$ because the numerator is non-zero: $$f(x_n) = y_n \neq y_0 = f(x_0)$$.