Let be $f:I\to\mathbb{R}$ a continuous, bijective function and $I$ an intervall of $\mathbb{R}$. Further, let be $g:J\to I$ the inverse function of $f$ such that for all $y\in J$ we have $f(g(y))=y$ and for all $x\in I$ we have $g(f(x))=x$.
Show that $g$ is continuous.
Let be $\epsilon>0$ and $a\in I$ (both arbitrarily chosen). As $f$ is continuous there exists a $\delta>0$ such that for all $x\in I$ with $|x-a|<\delta \implies |f(x)-f(a)|<\epsilon$.
It follows from the properties of the inverse function $g$ that for $a\in I$ there exists a unique $b\in J$ and for each $x\in I$ a unique $y\in J$ such that $|x-a|=|g(y)-g(b)|<\delta \implies |f(x)-f(a)|=|y-b|<\epsilon$. As $f$ is bijective this implication also holds reversely.
So if I take another arbitrary $\epsilon'>0$ and shrink (if necessary) $\delta$ such that $0<\delta<\epsilon'$ then it holds that for all $y\in J$ with $|y-b|<\epsilon\implies |g(y)-g(b)|<\delta<\epsilon'$ which shows the continuity of the inverse function $g$.
I know that there are already numerous proofs on this issue which use strict monotonicity (also the one in my lecture notes). However, as my approach is so short I have the feeling that it is wrong and that I have overlooked something. Where is it flawed?