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So first of all I am sorry if this is breaking any posting rules. Yesterday i asked a question Example of function that is differentiable at $0$, and has inverse function that is not continous at $0$? , and since people asked for more explanation and it seems i cant edit the question i decided to post a new one.

So the first thing i stated wrong in my question is i need inverse function that is not continuous at $f(0)\neq 0$... And also that discontinuity has to be jump discontinuity.

I would also like to thank everyone who helped me when i first asked the question, and since this is getting longer than it should i will sum it up below by repeating the title of the question:

I need to find example of function $f$, that is Differentiable at $0$, and has inverse function $g$, that has jump discontinuity at $g(f(0))$.

Thanks for reading :)

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You can edit your previous question if you log in with the same account details you used to write it. That would be better than opening a new question. –  Chris Eagle Dec 11 '12 at 19:18
    
If you wish to merge your different accounts, please flag a moderator. –  robjohn Dec 11 '12 at 19:58
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1 Answer

This is not possible, as stated. For $g$ to be the inverse function of $f$ on some neighborhood of $x=0$, by definition $g(f(x)) = f(g(x)) = x$ on that neighborhood. The map $x \mapsto x$ does not have a jump discontinuity.

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