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Suppose A is a non-empty set and f is a function on A. Suppose for all g(which is also a function on A), composition of functions f and g is f, then f is a constant function.

I try to prove it but the attempt turned out to be futile. First I have assumed g to be arbitrary and try to work on it, but it leads to nowhere. The hint says that what happens if g is a constant function. I think it is trying to tell me that we should define g to be a constant function. But isn't g should be arbitrary and we shall assume nothing about g? Below is the link for the same question, but I can't understand the solution: http://mathhelpforum.com/discrete-math/188033-problem-involving-composite-functions.html

Please explain, thanks in advance.

Dave Clifford
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There exists at least one constant function $g$ defined on $A$. For every $a\ne b$ in $A$, $g(a)=g(b)$, hence $f\circ g=f$ implies that $f(a)=f(g(a))=f(g(b))=f(b)$. Thus, $f$ is constant.

Did
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    Did not check the mathhelp solution. – Did Feb 06 '14 at 09:49
  • I agree with your answer. But the question states that g is arbitrary; it doesn't ask that there exists some function, rather it says that for all functions g. Even g is an arbitrary function I know that the statement still holds, I am just dubious about the hint the book gives. – Dave Clifford Feb 06 '14 at 09:54
  • In the answer you assume g to be a constant function, so it is not arbitrary, am I right? – Dave Clifford Feb 06 '14 at 09:54
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    Something is assumed to hold for every function g. Hence it holds for every constant function g, as well as the consequences of the fact that it holds for these specific functions. – Did Feb 06 '14 at 10:48