Inverse functions

**espen180** · November 9th 2008, 08:03 AM

Show that $f\left(g^{-1}(x)\right)=g\left(f^{-1}(x)\right) \Leftrightarrow f^{-1}\left(g(x)\right)=g^{-1}\left(f(x)\right)$

If we substitute $x=f(u)$ , we get

$f\left(g^{-1}\left(f(u)\right)\right)=g^{-1}\left(f\left(f^{-1}(u)\right)\right)$
$\Updownarrow$
$f\left(g^{-1}\left(f(u)\right)\right)=g(u)$

This means that $g^{-1}\left(f(u)\right)=f^{-1}\left(g(u)\right)$ , because $f\left(f^{-1}\left(g(u)\right)\right)=g(u)$ .

Now, if the above equality holds for $u$ , it must also hold for $x$ , and we have that

$g^{-1}\left(f(x)\right)=f^{-1}\left(g(x)\right)$

and that

$f\left(g^{-1}(x)\right)=g\left(f^{-1}(x)\right) \Leftrightarrow f^{-1}\left(g(x)\right)=g^{-1}\left(f(x)\right)$

Which was to be shown.

Does this hold, or have I overlooked something?

Thanks in advance.

**ThePerfectHacker** · November 9th 2008, 08:19 AM

Originally Posted by espen180

Does this hold, or have I overlooked something?

That is good. The only issue I have with this is that we do not know much about $f,g$ themselves.
If these were bijections on $\mathbb{R}$ then what you did is okay.

**espen180** · November 9th 2008, 08:29 AM

Yes, I assumed that was the case with f and g. But what if they were bijections on $\mathbb{C}$ ? Would it not hold then?

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