Exercise 3.3.5 (Composition injective, surjective)

Proof. Assume that $g\circ f:X\to Z$ is injective. We show that $f$ is injective.

If $f(x)=f(x^{\prime })$, where $x,x^{\prime }\in X$, then $(g\circ f)(x)=g(f(x))=g(f(x^{\prime }))=(g\circ f)(x^{\prime })$. Since $g\circ f$ is injective, the equality $(g\circ f)(x)=(g\circ f)(x^{\prime })$ implies $x=x^{\prime }$. This shows that $f$ is injective.

It is not true that $g$ must also be injective. To prove this, consider the following counterexample: let $f:\{1,2\}\to \{1,2,3\}$ defined by $f(1)=1,f(2)=2$ and $g:\{1,2,3\}\to \{1,2\}$ defined by $g(1)=1,g(2)=g(3)=2$. Then $g\circ f=1_{\{1,2\}}$ is the identity on $X=\{1,2\}$, thus is injective, but $g$ is not injective, since $g(2)=g(3)$. Another counterexample is given by

Assume now that $g\circ f$ is surjective. If $z$ is any element of $Z$, there exists some $x\in X$ such that $(g\circ f)(x)=z$. Define $y=f(x)$. Then $y\in Y$, and $g(y)=z$. This shows that $g$ is surjective.

$f$ must not be surjective: in the two preceding examples $g\circ f$ is surjective, but $f$ is not surjective. □