Past exam of the year 1 of the computer science course of the University of Oxford / 2023 / Discrete mathematics, probability and continuous mathematics / 2 / d / iv

Suppose $g(b)=g(b^{\prime })$. Since $f$ is surjective, choose $a,a^{\prime }\in A$ with $f(a)=b$ and $f(a^{\prime })=b^{\prime }$. Then $(g\circ f)(a)=(g\circ f)(a^{\prime })$, so injectivity gives $a=a^{\prime }$ and therefore $b=b^{\prime }$. Thus $g$ is injective. The use of surjectivity is what makes every point of $B$ accessible to this argument.