Please tell me if the following is correct.
We have a continuous, strictly monotonic, increasing function on some closed and bounded interval $I$, and $x_0\in I$. Let $g(y)$ be its inverse, and $f(x_0)=y_0$. I want to show that $|g(y)-g(y_0)|<\epsilon\implies|y-y_0|<\delta$.
$$\begin{align*}
|g(y)-g(y_0)|<\epsilon&\Leftrightarrow x_0-\epsilon<g(y)<x_0+\epsilon\\
&\Leftrightarrow f(x_0-\epsilon)<y<f(x_0+\epsilon)\\
&\Leftrightarrow f(x_0-\epsilon)-f(x_0)<y-f(x_0)<f(x_0+\epsilon)-f(x_0)\\
&\Leftrightarrow -(y_0-f(x_0-\epsilon))<y-y_0<f(x_0+\epsilon)-y_0\\
\end{align*}$$
If I consider only the $y$s that are extremely close to $y_0$, then I think that I can set $\delta = \min(y_0-f(x_0-\epsilon), f(x_0+\epsilon)-y_0)$. For the case where the function is decreasing, I just flip the inequality symbols from line one to two, and take $\delta=\min(f(x_0-\epsilon)-y_0,y_0-f(x_0+\epsilon))$.
Best Answer
I don't prove that $f$ is bijective, as it's very easy. Just injectivity requires the monotonicity and surjectivity requires IVT. We know since $f$ is continuous on $[a,b]$ then for any $\varepsilon >0$ one can find $\delta >0$ such that $|x-c|<\delta$ implies $|f(x)-f(c)|<\varepsilon$. Now there is a unique $y,y_0$ such that $|x-c|=|g(y)-g(y_0)|$ and this solves the problem. (Just change the job of epsilon and delta in your approach)