Let $f,g,h$ be functions from $A$ to $A$. Assume the composition $f \circ g \circ h$ is invertible. Is $g$ injective? Is $g$ on $A$? This seems to be true if $A$ is a finite set, but is it true if $A$ is not a finite set?
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2@Surb "Let f,g,h be functions from A to A", so your example does not match the requirements. – Gyro Gearloose Dec 19 '23 at 20:05
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A function from A to A is invertilbe $\Leftrightarrow$ the function is bijective $\Leftrightarrow$ the function is injective and surjective. Does that help? – Cecilia Dec 19 '23 at 20:16
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@GyroGearloose Good point! Then $h=\tanh$, $f=\tanh^{-1}$, $g=\sin(\pi x)$ could be an example bijective composition and $g(x)$ could be redefined arbitrarily for $|x|>1$. – Surb Dec 19 '23 at 20:18
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Some related posts which might help you: Does $f\circ g$ injective imply $f$ injective for functions $f,g:A\to A$? or Injective and Surjective Functions and som of the questions linked there. – Martin Sleziak Dec 20 '23 at 10:37
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Let $A=\{(a_1, a_2, \ldots ) \, | \, a_i \in \mathbb{R}\}$ be the set of sequences. Consider \begin{align*} f(a_1, a_2, \ldots )&=(a_2, a_3, \ldots )\\ g(a_1, a_2, \ldots )&=(0,a_1, a_2, \ldots )\\ h(a_1, a_2, \ldots )&=(a_1, a_2, \ldots ) \end{align*} Then $f\circ g\circ h$ is the identity function, hence invertible. But $g$ (not onto) and $f$ (not one-one) are not.
Note: When $|A|< \infty$, then if a function $k:A \longrightarrow A$ has one of the two among injectivitiy and surjectivity, then it has the other as well. So, it works in the finite case.
Anurag A
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@GyroGearloose $(1,2,3, \ldots) \xrightarrow{h} (1,2,3, \ldots) \xrightarrow{g} (0,1,2,3, \ldots) \xrightarrow{f} (1,2,3, \ldots)$. Another example, $(0,1,2,3, \ldots) \xrightarrow{h} (0,1,2,3, \ldots) \xrightarrow{g} (0,0,1,2,3, \ldots) \xrightarrow{f} (0,1,2,3, \ldots)$. – Anurag A Dec 19 '23 at 20:56
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Nice! This means $g$ is not necessarily on $A$, but is it necessarily injective? – Adi Hendel Dec 19 '23 at 22:38
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by replacing $f$ with $h$, $g$ with $f$ and $h$ with $g$ we get an example in which $g$ isn't injective. – Adi Hendel Dec 19 '23 at 22:54