= 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 $b,b'\in B$ with $g(b)=g(b')$ to preimages under $f$; composite <injectivity> then proves $b=b'$. 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.
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