Continuity of inverse function at endpoints

[Dave1]

Hello!

*Let $f$ be a strictly increasing continuous function on a closed interval $[a, b]$, let $c = f(a), d = f(b)$, and let $g:[c, d] → [a, b]$ be its inverse. Then $g$ is a strictly increasing continuous function on $[c, d]$.*

How can it be shown that $g$ is continuous at its endpoints $c$ and $d$? I am not familiar with one-sided continuity arguments...

 

[Opalg]

Hello!

*Let $f$ be a strictly increasing continuous function on a closed interval $[a, b]$, let $c = f(a), d = f(b)$, and let $g:[c, d] → [a, b]$ be its inverse. Then $g$ is a strictly increasing continuous function on $[c, d]$.*

How can it be shown that $g$ is continuous at its endpoints $c$ and $d$? I am not familiar with one-sided continuity arguments...

You know that $g$ maps $[c,d]$ bijectively onto $[a,b]$. Given $\varepsilon>0$, there exists $z \in [c,d]$ with $g(z) = a + \varepsilon$ (unless $a + \varepsilon >b$, in which case replace $a + \varepsilon$ by $b$ in what follows). Then $g$ maps $[c,z]$ bijectively onto $[a,a + \varepsilon]$ (because $g$, like $f$, is a strictly increasing function). Let $\delta = z-c$. Then $$c \leqslant y < c + \delta\; \Rightarrow \; y<z\; \Rightarrow \; g(y) < a + \varepsilon\; \Rightarrow \; |g(y) - g( c)| < \varepsilon.$$ That shows that $g$ is continuous (on the right) at $c$. A similar argument shows that $g$ is continuous (on the left) at $d$.