composition

function composition

$f:A\to B,g:B\to C$
The composition of $f$ and $g$ is the function $g\circ f:A\to C,x\mapsto g(f(x))$
Composition is not commutative.
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See also: Category Theory and Why We Care 1.1

EXAMPLE

$f:\mathbb{R}\to \mathbb{R},x\mapsto x^2,g:\mathbb{R}\to \mathbb{R},x\mapsto \cos (x)$
$g\circ f:\mathbb{R}\to \mathbb{R},x\mapsto \cos (x^2)$
Since domain and codomain are the same, we can also compose $f\circ g:\mathbb{R}\to \mathbb{R},x\mapsto \cos (x{)}^2$ and in this case the result is different.

The composition of two injective or surjective functions is also injective or surjective, respectively.

$f:A\to B,g:B\to C,f,g\text{ injective/surjective}⟹g\circ f\text{ surjective/surjective}$

Proof:
$f,g$ injective
$g(f(x))=g(f(y))⟹f(x)=f(y)⟹x=y$ → $g\circ f$ injective
$f,g$ surjective
$z\in C⟹\exists y\in B:g(y)=z\wedge \exists x\in A:f(x)=y⟹(g\circ f)(x)=g(f(x))=g(y)=z$ → $g\circ f$ surjective
→ $f,g$ bijective $⟹g\circ f$ bijective
→ $(g\circ f{)}^{-1}=f^{-1}\circ g^{-1}$

Properties

$(A^A,\circ )$: associativity
$(f\circ g)\circ h=f\circ (g\circ h)$
Proof: $(f\circ g)\circ h=(x\mapsto f(g(h(x))))=f\circ (g\circ h)$
$(A^A,\circ )$: neutral element
${\text{id}}_A\circ f=f\circ {\text{id}}_A=f$
Proof: $({\text{id}}_A\circ f)(x)={\text{id}}_A(f(x))=f(x)$
$\neg$ commutative: $f\circ g\ne g\circ f$
$\neg$ invertible: $f\circ g$ is not necessarily invertible if $f$ and $g$ are invertible. Only for bijections are invertible, or of $|A|=1$.