I'm working on a problem where I'm given a relation "~" on functions from some non-empty set X mapping to that same set X such that f1~f2 if there is some function g such that f1(g(x)) = g(f2(x)).
I'm to show by counterexample that ~ isn't necessarily an equivalence relation and that it would be an equivalence relation if I stipulated that g must be bijective.
So I know I need to check reflexivity, symmetry, and transitivity, and that the counterexample most likely will come from the fact the g is not bijective. The inverse of g will probably be involved in proving in the second portion that ~ is an equivalence relation. But the concept of finding some function such that composing with with two other functions in two different orders is so abstract that I don't know where to start.
HINT: For the counterexample, assume that $X$ has at least two elements $a$ and $b$, and let $f_2$ and $g$ be the constant function $g(x)=f_2(x)=a$ for all $x\in X$. Now find an $f_1:X\to X$ such that $f_1\circ g\ne g$.
If you require $g$ to be a bijection, note that $f_1\circ g=g\circ f_2$ if and only if $f_2=g^{-1}\circ f_1\circ g$ if and only if $g^{-1}\circ f_1=f_2\circ g^{-1}$. This already proves that $\sim$ is symmetric, since $g^{-1}$ is a bijection whenever $g$ is. There is an obvious choice of $g$ for showing reflexivity. Finally, the fact that $f_1\sim f_2$ if and only if there is a bijection $g:X\to X$ such that $f_2=g^{-1}\circ f_1\circ g$ should help you show that $\sim$ is transitive.