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It is proved that if $(X,d)$ is a metric space and $f: X\to X$ is surjective, there exists $f^\ast:X\to X$ such that $f\circ f^\ast x=x$ for all $x \in X$. Here, $f^\ast$ is called right inverse.

I claim that if $f$ is bijective and continuous then $f^\ast$ is continuous also. Am I right?

In other words, what are the necessary conditions on $f$ ensuring the continuity of $f^\ast$.

Rere
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  • Nope. "$f^*$", which I'd rather call $f^{-1}$ for the occasion, since $f$ is assumed to be bijective, is continuous if and only if $f$ is open (id est, if $f(A)$ is open for every open set $A$). Examples of non-open continuous bijections between metric spaces can be found here. It is however true that continuous bijections $\Bbb R^n\to\Bbb R^n$ are always open. –  Aug 20 '17 at 07:23
  • FYI your claim is true if $X$ is complete and $f$ is linear. This is a special case of what's called inverse mapping thereom (or its two equivalents, open mapping theorem/closed graph theorem) – user160738 Aug 20 '17 at 07:44
  • But I am talking about the right inverse, not the usual inverse.! – Rere Aug 20 '17 at 15:00
  • I deleted the following: Observe that, for the continuity of the inverse mapping $f^{-1}$, the compactness of $X$ is necessary (besides the continuity of $f$ and bijectiveness). – Rere Aug 20 '17 at 15:07
  • @Rere If $f:X\to Y$ is bijective, $f^{-1}$ is not only the sole function $g:Y\to X$ such that $f\circ g=id_Y$ and $g\circ f=id_X$, but also: 1) the only function $g$ such that $f\circ g=id_Y$; 2) the only function $g$ such that $g\circ f=id_X$. This is an easy set-theoretic lemma. –  Aug 20 '17 at 16:40
  • Dear @G.Sassatelli, I am not interested in the inverse function $f^{-1}$. I need the only right inverse and in such situation, $f$ need to be surjective to ensure the existence of $^\ast$. I want the necessary conditions on $f$ such that $f^\ast$ is continuous. – Rere Aug 21 '17 at 03:41
  • If there is a the right inverse then $f$ necessarily bijective – Hagen von Eitzen Aug 21 '17 at 04:07
  • @HagenvonEitzen No, Surjective is enogh. see: http://mathworld.wolfram.com/RightInverse.html – Rere Aug 21 '17 at 04:44
  • Dear @Rere, I'll gladly leave you stuck in the misconception you refuse to crawl out of. –  Aug 21 '17 at 06:29
  • Thanks, dear @G.Sassatelli I appreciate your effort. – Rere Aug 21 '17 at 13:57
  • Does this answer your question? Functions which are Continuous, but not Bicontinuous – GEdgar Feb 16 '24 at 16:10

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We know that, if $f$ is bijective then the right inverse, as well as the left inverse, are the same and equivalent to $f^{-1}$.

I found in (Q.H. Ansari, Metric spaces ... ) that "If $f$ is bijective and continuous and $X$ is compact, then $f^{-1}$ is continuous". Therefore, $f^*$ is continuous under the same situation.

Rere
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  • Since you've answered (and answers fall under a more demanding quality level than questions and comments): if $X$ is not empty, $f:X\to Y$, $g: Y\to X$ are two functions such that $f\circ g=id_X$ and $f$ is not injective, then there is a function $h:Y\to X$ such that $h\ne g$ and $f\circ h=f\circ g=id_X$. Which one is your $f^*$? In fact, there are plenty of discontinuous right-inverses of the map $\pi:[0,1]^2\to[0,1]$, $\pi(x,y)=x$. –  Aug 21 '17 at 14:47
  • Firstly, the mapping in the question is self-mapping. Secondly, $f$ should be surjective to ensure the right inverse. Thirdly, I didn't mention that $f^\ast$ is unique. – Rere Aug 22 '17 at 05:43
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    Ok, then, $\sin(36\pi x)$ from $[-1,1]$ to $[-1,1]$, which has a considerable number of discontinuous sections: say, assign to some $y_j$ the number $\frac1{36}\arcsin y_j$, to some other $y_h$ assign $\frac{\pi-\arcsin y_h}{36}$, to some other $\frac\pi{18}+\frac1{36}\arcsin y_u$ et cetera. –  Aug 22 '17 at 06:51
  • Thanks Dear @G.Sassatelli. Now, I see the point. – Rere Aug 26 '17 at 05:49