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Something came to mind: if $f:\mathbb R\to\mathbb R$ is a bijection with inverse $g$. So can I say that $g(-x)$ is inverse of $f(-x)$? I think it is false but I have not been able to find a counterexample although visually the answer may seem affirmative since it is only a reflection with respect to the y-axis. Any idea or example? For example if defined $h_1(x)=f(-x)$ and $h_2(x)=g(-x)$ then $h_1(h_2(x))=f(-g(-x))¿=?id_{\mathbb R}$. If $f$ and $g$ were odd I think it would be trivial, but this is not the case.

Zaragosa
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    Let $r(x)=-x$ be the negation. Observe that it is its own inverse, in other words $r^{-1}=r$. Anyway, the function $x\mapsto f(-x)$ is equal to the composition $f\circ r$. By the usual "inverse of the compositon rule" $$(f\circ r)^{-1}=r^{-1}\circ f^{-1}=r\circ f^{-1}.$$ In your case $f^{-1}=g$, so the inverse of $f(-x)$ is $-g(x)$. You can also verfiy this easily enough. – Jyrki Lahtonen Jun 01 '22 at 04:22
  • @JyrkiLahtonen Why are you answering in a comment? – Arthur Jun 01 '22 at 04:26
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    @Arthur I have not had the time to search for a proper duplicate, yet. You may want to question my motives for posting an explanation in a comment. My practice here is to try and make the asker to see the light, and post a summary themself. That process is still in the works. – Jyrki Lahtonen Jun 01 '22 at 04:29
  • @JyrkiLahtonen Wow how fast! It was much more than I expected, I just thought of counterexample $f(x)=x+2$ and $g(x)=x-2$, but your answer is absolute. Thank you – Zaragosa Jun 01 '22 at 04:30
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    As I said in my comment to Arthur, Zaragosa. Unless a suitable duplicate target is found, you are welcome to post a verification of this process as an answer. That way you get more feedback on the details, and any unclear steps are discovered. – Jyrki Lahtonen Jun 01 '22 at 04:33
  • @Arthur, if you think I am violating some principle of the site, do start a meta discussion about it. It is always possible that I have overlooked something, and should revise my approach. – Jyrki Lahtonen Jun 01 '22 at 04:35
  • See here. – Jyrki Lahtonen Jun 01 '22 at 04:44
  • @JyrkiLahtonen Sorry for my little experience but I think you should post your answer, I don't see any weak points in your procedure and also no duplicate questions based on my question approach. Also, I think it's a very good solution. – Zaragosa Jun 01 '22 at 04:48
  • Does this answer your question? How to prove $(f \circ\ g) ^{-1} = g^{-1} \circ\ f^{-1}$? (inverse of composition) – Jyrki Lahtonen Jun 01 '22 at 05:10
  • It doesn't do it directly. – Zaragosa Jun 01 '22 at 05:11

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The function that maps the number $x$ to $f(-x)$ is the composition of the negation $r:x\mapsto -x$ and $f$. By the general result on the inverse of a composition we have $$(f\circ r)^{-1}=r^{-1}\circ f^{-1}.$$ Here $f^{-1}=g$ and $r^{-1}=r$, so the inverse you are looking for is the function $$ r\circ g:x\mapsto -g(x). $$

Jyrki Lahtonen
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