 Solution

A function is injective when $f(x)=f(x^{\prime })$ implies $x=x^{\prime }$. It is surjective when every element of its codomain equals $f(x)$ for at least one element of its domain.

If $g\circ f$ is injective, then $f(x)=f(x^{\prime })$ implieshence $x=x^{\prime }$. Thus injectivity of the composition implies injectivity of the first map.

If $g\circ f$ is surjective, then for every $z\in Z$ there is an $x\in X$ with $z=g(f(x))$. Taking $y=f(x)$ proves that $z$ lies in the image of $g$. Thus surjectivity of the composition implies surjectivity of the second map. The arguments include empty domains and codomains whenever the specified functions exist.