I want to disprove:
If g of f is one-to-one, then g is one-to-one.
Thanks for your help
Take $\displaystyle g(x)=x^2$ and $\displaystyle f(x)=e^x$. The key is considering $\displaystyle f$ such that $\displaystyle g$ is injective restricted to the range of $\displaystyle f$, but not in all the domain of $\displaystyle g$. You can easily construct a los of these examples if you understand properly why this one works.