True or False? Given any set $X$ and given any function $f:X\rightarrow X$ , $g:X\rightarrow X$ and $h:X\rightarrow X$, if $h$ is one-to-one function and $f(h(x))= g(h(x))$, then $f=g$. Justify your answer.
I know that this is false as $f$ and $g$ may not be injective and hence $f$ not equals to $g$ But how do I prove it more rigorously ? Like what counterexamples can I give ?
HINT: If $h=e^x$, then $h$ is one-to-one. But the range of $h$ is the positive real numbers. Can you construct $f$ and $g$ so that they agree on the positive real numbers, but not the negatives?
Actually I think this is true? (correct me if I'm mistaken) because you said h was a one to one function on itself. That means its also a bijection on itself because the cardinality of the sets is same (corollary of Cantor-Bernstein theorem) and by definition also a permutation.
Thus, we can conclude that the range of h must also be the codomain which here is defined to be the domain
Since f and g share this domain, namely X, we know for certain that they must agree on every point in their agreeing domains, thus satisfying the definition for function equality
