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Let's suppose there are two functions $f:A\longrightarrow B$ and $g:B\longrightarrow B$, such that: $f(x)=f(g(f(x)))$ and $g(x)=g(f(g(x)))$.

I know this should be a trivial question, but what does this mean exactly? Is it possible that this is equivalent to saying that $g$ is an inverse for $f$?

Thanks for any kind of help!

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    Consider $f : x \mapsto 3$ and $g : x \mapsto 6$, then $f(x) = f(g(f(x)))$ and $g(x) = g(f(g(x)))$, but is $g$ an "inverse" for $f$? – 2017-01-16
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    Your second example has a type error. – 2017-01-16
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    @QthePlatypus What do you mean by type error? – 2017-01-16
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    @Raito No, it's not. So stating those two properties means nothing, doesn't it? – 2017-01-16
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    @any_one I wouldn't say that these two properties mean nothing, at least, they won't be the characterization of inverse functions. – 2017-01-16
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    @Ratio Is it possible that these two properties are a characterization for a sort of Galois connection? – 2017-01-16
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    @any_one You'd need more conditions on these functions to get something interesting, I believe. – 2017-01-16
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    @Ratio mmmh what a pity...thanks anyway! – 2017-01-16
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    @raito in any_one's example they compose g which has a codomain of B with f which has a domain of A. $f(g(x))$ doesn't work. – 2017-01-16
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    @QthePlatypus Right, I think $g : x \mapsto 3$ would suffice to show that $f \circ g \neq \mathrm{Id_{A}}$. – 2017-01-16

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