Answer
$g$ must be one-to-one if $f\circ g$ is one-to-one.
(see proof below)
Work Step by Step
Let us suppose that g is not one-to-one and $f\circ g $ is.
Not being one-to-one means that there exist two distinct values $x_{1}\neq x_{2}$ for which $g(x_{1})=g(x_{2})=c$.
But then
$f[g(x_{1})]=f(c)$ and
$f[g(x_{2})]=f(c)$, meaning that
$(f\circ g)(x_{1})= (f\circ g)(x_{2})$, even though $x_{1}\neq x_{2}$.
This is a contradiction to $f\circ g$ being one-to-one.
So, assuming that g is not one-to-one has lead us to a contradiction. The assumption is wrong, and it must be that the opposite is true.
$g$ is one-to-one.
