MHB Continuity of inverse function at endpoints

Dave1
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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...
 
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Dave said:
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 $d$.
 
A sphere as topological manifold can be defined by gluing together the boundary of two disk. Basically one starts assigning each disk the subspace topology from ##\mathbb R^2## and then taking the quotient topology obtained by gluing their boundaries. Starting from the above definition of 2-sphere as topological manifold, shows that it is homeomorphic to the "embedded" sphere understood as subset of ##\mathbb R^3## in the subspace topology.

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