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I have the following statement:

Determine if is true that if $g: \mathbb{R} \to \mathbb{R}, f: \mathbb{R} \to \mathbb{R}$ and $ (g\circ f)(x) = x$ therefore $g = f^{-1}$

My attempt was:

$i)$ Since $g \circ f$ is injective then $f$ is injective.

$ii)$ Since $g\circ f$ is surjective then $g$ is surjective.

I couldn't found a counterexample function, so I think this is true. So I will try to show that $f$ is bijective and then use the property $(f^{-1}\circ f)(x)=x$ that is $x = (g \circ f)(x)$

But i could not prove that $f$ will be surjective to determine that $f$ is bijective. So any hint is appreciated.

Thanks in advanced.

ESCM
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  • What is your definition of the concept of "$f^{-1}$" for a given function $f\colon A\to B$? A function $B\to A$ such that $f^{-1}\circ f=\operatorname{id}_A$ and $f\circ f^{-1}=\operatorname{id}_B$? -- If so, you are apparently only given half of the requirements – Hagen von Eitzen Jun 07 '20 at 20:44
  • yes @HagenvonEitzen, that is my definition of the inverse function $f^{-1}$ and iff$f$ is bijective. – ESCM Jun 07 '20 at 20:46

1 Answers1

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Hint: $f$ must be injective (if $f(x)$ lost information about $x$, then $g$ could not recover $x$ from $f(x)$). However, the value of $g(x)$ is irrelevant for $x$ outside the range of $f$. So try taking $f$ to be an injection of $\Bbb{R}$ into a proper subset of $\Bbb{R}$. E.g., take $f(x) = \arctan(x)$.

Rob Arthan
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  • $arctan(x)$ is injective and with domain $(-\frac{\pi}{2}, \frac{\pi}{2})$, by other side what about of $g(x)$ ? I know that it must be surjective since $gof$ is surjective implies that $g$ is. So possible i can select $g$ as a piecewise function such that $tanx$ when $x \neq (2k-1)\cdot \frac{\pi}{2}, $ with $k \in \mathbb{Z}$ and $0$ in the other case. We have that $f$ is injective but not surjective therefore not bijective, $g$ is surjective and $(g \circ f)(x) = tan(arctan(x)) = x \forall x \in \mathbb{R}$ but $g$ can't be $f^-1$ since f is not surjective. What do you think? – ESCM Jul 07 '20 at 03:20
  • $(\tan \circ \arctan)(x) = x$ But $\tan \neq \arctan^{-1}$. In fact $\arctan^{-1}$ does not exist, because $\arctan$ is not surjective. – Rob Arthan Jul 07 '20 at 12:21
  • Yes, that i said. So is all correct according to you thought? – ESCM Jul 07 '20 at 15:11
  • As your problem requires $g$ to be a total function, yes you do need to make $\tan$ total somehow - it doesn't matter how. Aside: $(-{\pi \over 2}, {\pi \over 2})$ is the range of $\arctan$ not the domain. – Rob Arthan Jul 07 '20 at 15:35