If $F$ is injective, and so is $F \circ G$, does $G$ have to be?

Question

If $F$ is an injective (one-to-one) function, and the composite of the two $(F \circ G)$ is injective, is it possible for $G$ to not be injective?

Answer 1

No. If $G(x)=G(y)$ with $x\neq y$ then $$ (F\circ G)(x)=F(G(x))=F(G(y))=(F\circ G)(y), $$ i.e. $F\circ G$ is not injective

Answer 2

If $f\circ g$ is one to one, then so is $g$

Regardless if $f$ is or not.