Injectivity and surjectivity under composition

For $f:A\to B$ and $g:B\to C$, the following implications hold. If both are injective, equality of two composite images can be cancelled first through $g$ and then through $f$, proving that $g\circ f$ is injective. Conversely, if $g\circ f$ is injective, equality under $f$ implies equality under the composite, so $f$ is injective. If $g\circ f$ is surjective, every $c\in C$ is $g(f(a))$, so $g$ is surjective.

There are two useful strengthened cancellation rules. If $g\circ f$ is injective and $f$ is surjective, lift any $b,b^{\prime }\in B$ with $g(b)=g(b^{\prime })$ to preimages under $f$; composite injectivity then proves $b=b^{\prime }$. If $g\circ f$ is surjective and $g$ is injective, apply composite surjectivity to $g(b)$ and cancel $g$ to find a preimage of every $b$. Neither strengthened conclusion follows from the composite property alone.