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We have seen in topology, that if a function is continuous and has an inverse, its inverse is not necessarily continuous. However the examples we had were from higher dimensions. So I've been thinking if this is also true for a function from the reals to reals. I couldn't find a counter example, and when I think geometrically, it's just a reflection on the line y=x. So I don't find any reason why the inverse wouldn't be continuous. So I just wanted to verify if this claim is true or not. If it is I will try to prove, and if it isn't I would appreciate if someone can give counter example. Thanks!

Charles Carmichael
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  • Those functions exist, indeed: http://math.stackexchange.com/questions/68800/functions-which-are-continuous-but-not-bicontinuous – Daniel Mar 25 '15 at 12:15
  • Ok, I've got it, you need the domain to be $\mathbb{R}$. – Daniel Mar 25 '15 at 12:16
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    If $f : \mathbb R \to \mathbb R$ is continuous and injective, then the inverse map is also continuous. For more advanced readers: Use the fact that $\mathbb R$ is locally compact. – GEdgar Mar 25 '15 at 12:16
  • Awesome, Thanks! – Charles Carmichael Mar 25 '15 at 12:17
  • @DanielEscudero , then if I define f from the reals to the reals (In other words Rangef=R) that is differentiable and has an inverse,then What I am saying is actually correct, right? – Charles Carmichael Mar 25 '15 at 12:25
  • That's right. The inverse function theorem guarantee's that (always that $f$ is of class $C^1$ and the other hypothesis of the theorem hold). However, the theorem is still true with weaker hypothesis, as GEdgar noticed. – Daniel Mar 25 '15 at 12:32

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