Is it possible to find a necessary and sufficient condition to conclude when $$f(g(x))=x \implies f(x)=g^{-1}(x) \wedge f^{-1}(x)=g(x),$$ if both functions are well defined?
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If either $f$ is injective or $g$ is surjective, then $f\circ g={\rm id}$ implies $\exists f^{-1},g^{-1}$ then necessarily $f^{-1}=g$ and $g^{-1}=f$.
Berci
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1Is this sufficient condition also necessary? – vadim123 May 14 '13 at 23:44
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$f^{-1}=g$ itself implies $g^{-1}=f$ and $f\circ g={\rm id}$. Also, $\exists f^{-1}$ means $f$ is bijective. So, necessary and sufficient. – Berci May 14 '13 at 23:50
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Hint: If $f,g\colon X\to X$, then $f^{-1}=g$ if and only if $f\circ g=g\circ f=\text{Id}_X$
Asaf Karagila
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