Function composition ($\circ$)

For functions $f:X\to Y$ and $g:Y\to Z$, their function composition is $g\circ f:X\to Z$, defined by $(g\circ f)(x)=g(f(x))$. The rightmost function is applied first. Matching the intermediate codomain and domain of a function makes the expression well-defined.

For self-maps of the real numbers, take $f(x)=x+1$ and $g(x)=2x$. Then $(g\circ f)(x)=2x+2$, whereas $(f\circ g)(x)=2x+1$. Thus function composition need not be commutative. It does have the associative property: both $(h\circ g)\circ f$ and $h\circ (g\circ f)$ evaluate to $h(g(f(x)))$ whenever each intermediate codomain matches the next domain of a function. The identity function leaves composition unchanged, making self-maps particularly useful as a monoid of transformations.

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.