8. Function composition

Function composition

Composition of functions

Let $g:A\to B$ and $f:B\to C$ be functions. The composition of $f$ and $g$ is the function $\boxed{ (f \circ g): A \rightarrow C }$, can also define this for all $a \in A$ as: $$ (f \circ g)(a) = f(g(a)) $$

We only define the composition $f\circ g$ when the codomain of $g$ is equal to the domain of $f$.

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title: Example
Let $X = \{ a,b,c \}$ and $Y = \{ 1,2,3 \}$.
Let the function $g: X \rightarrow X$ be defined by:
 
$$
	g(a) = b, \hspace{12px} g(b) = c, \hspace{12px} g(c) = a
$$
 
Let the function $f: X \rightarrow Y$ be defined by:
 
$$
	f(a) = 3, \hspace{12px} f(b) = 2, \hspace{12px} f(c) = 1
$$
 
Then:
- $(f \circ g)(a) = f(g(a)) = f(b) = 2$
- $(f \circ g)(b) = 1$
- $(f \circ g)(c) = 3$
- $(g \circ g)(a) = c$
- $(g \circ g)(b) = a$
- $(g \circ g)(c) = b$

Properties of composition

• Even if both $f$ and $g$ are defined, $f\circ g$ and $g\circ f$ can differ. Let $f$ and $g$ be both $\mathbb{Z}\to \mathbb{Z}$ functions, defined by $f(x)=2x+3$ and $g(x)=3x+2$, then:
  • $(f\circ g)(x)=6x+7$
  • $(g\circ f)(x)=6x+11$
• $\circ$ is associative: $\boxed{f\circ (g\circ h)=(f\circ g)\circ h}$
• If $f:A\to B$ is a bijection then $\boxed{f^{-1}\circ f={\text{id}}_A}$ and $\boxed{f\circ f^{-1}={\text{id}}_B}$
• For any function $f:A\to B$, $\boxed{f\circ {\text{id}}_A={\text{id}}_B\circ f=f}$