Past exam of the mathematics course of the University of Oxford / 2017 / Prelims / Mathematics II / 4 / a

If $x\ne y$, one of $x<y$ and $y<x$ holds. Since $f$ is strictly increasing, the corresponding values are strictly ordered, so $f(x)\ne f(y)$. Thus $f$ is an injection. Every element of the image of a function $f([a,b])$ equals $f(x)$ for some $x\in [a,b]$, which is precisely surjectivity onto that image of a function. Hence $f:[a,b]\to f([a,b])$ is a bijection.

Let $u<v$ lie in $f([a,b])$, and put $x=g(u)$ and $y=g(v)$. Equality $x=y$ would imply $u=v$. If $x>y$, strict monotonicity of $f$ would give $u=f(x)>f(y)=v$. Both are impossible, so $x<y$. Therefore the inverse function $g$ is strictly increasing.