= Solution
A <function> is <injective> when $f(x)=f(x')$ implies $x=x'$. 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 $f(x)=f(x')$ implies
$$
(g\circ f)(x)=(g\circ f)(x'),
$$
hence $x=x'$. Thus \b[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 \b[surjectivity of the composition implies surjectivity of the second map]. The arguments include empty domains and codomains whenever the specified <functions> exist.
Back to article page
