Operations on Functions

Composition. Given $f:X\to Y$ and $g:Y\to Z$, define $h:X\to Z$ by first applying $f$, then $g$: $h(x)=g(f(x))$. This $h$ is the composition of $f$ with $g$, written $g\circ f$ (read "$f$ composite with $g$", applied right-to-left: do $f$ first, then $g$).

More generally, $g\circ f$ can be defined whenever the range of $f$ is contained in the domain of $g$ -- $f$'s co-domain need not literally equal $g$'s domain, as long as it fits inside it.

Worked examples.

• $f=\{(1,2),(3,4),(2,2)\},\ g=\{(2,1),(3,1),(4,2)\}$: range of $f=\{2,4\}\subseteq$ domain of $g=\{2,3,4\}$, so $g\circ f$ exists: $g\circ f=\{(1,1),(2,1),(3,2)\}$. Also range of $g=\{1,2\}\subseteq$ domain of $f=\{1,2,3\}$, so $f\circ g=\{(2,2),(3,2),(4,2)\}$ exists too -- and $f\circ g\ne g\circ f$ as sets, illustrating non-commutativity.
• $f=\{(1,4),(2,5),(3,5)\},\ g=\{(4,1),(5,2),(6,4)\}$: $g\circ f=\{(1,1),(2,2),(3,2)\}$ exists, but $f\circ g$ does not exist, since range of $g=\{1,2,4\}\not\subseteq$ domain of $f=\{1,2,3\}$.
• $f(x)=3x-4,\ g(x)=x^2+3$ on $R$: $(g\circ f)(x)=(3x-4)^2+3=9x^2-24x+19$, while $(f\circ g)(x)=3(x^2+3)-4=3x^2+5$ -- different, confirming composition is generally not commutative.

Theorem. If $f:A\to B$ and $g:B\to C$ are both one-to-one, then $g\circ f$ is one-to-one. Proof: if $x\ne y$ in $A$, injectivity of $f$ gives $f(x)\ne f(y)$, and injectivity of $g$ then gives $g(f(x))\ne g(f(y))$, i.e. $(g\circ f)(x)\ne(g\circ f)(y)$. $\blacksquare$ …