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Let $f: A \to B$ be given by $f(x) = |x|$, such that $A = [-1,0] \cup [1,2]$ and $B = [0,2]$.

I understand that $f$ does not have an inverse, since $f(-1) = f(1)$.

However, albeit my textbook is saying $f$ does not have an inverse $f^{-1}: B \to A$, it is also saying $f^{-1}$ is not continuous on $B$. How can that second sentence be made since $f^{-1}$ does not exist?

Lenora
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    Perhaps the textbook intended to take $A = (-1,0] \cup [1,2]$. – Ben Grossmann Dec 21 '21 at 18:28
  • But in that case $f$ would have an inverse. Unless the author was thinking about the interval $[-1,0]$ at the first statement, and the interval $(-1,0]$ at the second. – Lenora Dec 21 '21 at 18:32
  • Did the author state that $f$ has no inverse and that $f^{-1}$ is discontinuous? – Ben Grossmann Dec 21 '21 at 18:33
  • Yes. In the exact words, a) $f$ does not have inverse $f^{-1}: B \to A$; and b) $f^{-1}$ is not continuous on $B$. – Lenora Dec 21 '21 at 18:36
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    It is impossible for $f$ to not be invertible and for $f^{-1}$ to be a discontinuous function anywhere, because if $f$ is not invertible, then by definition, $f^{-1}$ does not exist. It is nonsensical to use the symbol $f^{-1}$ when talking about functions that are not invertible. The very usage of the symbol assumes $f$ is invertible. So I think the author of the book likely made a mistake. – Angel Dec 21 '21 at 18:40
  • For clarification, can you tell us what book we are using here? Perhaps we can check for context what the author meant. – Angel Dec 21 '21 at 18:41
  • It's a portuguese book. It was a comment on some exercise that appeared couple years ago in an exam. – Lenora Dec 21 '21 at 18:45
  • @Lenora Interesting. – Angel Dec 21 '21 at 19:02
  • @Lenora Well, in any case, let us hypothetically suppose that the author did mean $A=(-1,0]\cup[1,2],$ which makes $f$ invertible. Then it makes sense to discuss whether $f^{-1}$ is continuous or not. And we can conclude it is not continuous at $1$, but continuous everywhere else. If you graph it, I think it will be clear why. – Angel Dec 21 '21 at 19:07
  • Sometimes $f^{-1}$ is also used to refer to the map $f^{-1}(y) = {x \in A : f(x) = y}$ – dfnu Dec 21 '21 at 19:17
  • Comments often undergo far less scrutiny than the main text. My suspicion is the author meant that if the domain of $f$ were restricted so that $f$ becomes invertible, that inverse would not be continuous. – Paul Sinclair Dec 23 '21 at 00:57

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