# What is the domain of (f@g)(x)?

Jul 24, 2015

If $g : A \to B$ and $f : B \to C$, then the domain of $f \circ g$ is

${\overline{g}}^{- 1} \circ {\overline{f}}^{- 1} \left(C\right)$

using the notation described below...

#### Explanation:

If $g$ is a function that maps some elements of a set $A$ to elements of a set $B$, then the domain of $g$ is the subset of $A$ for which $g \left(a\right)$ is defined.

More formally:

$g \subseteq A \times B :$

$\forall a \in A \forall {b}_{1} , {b}_{2} \in B$

$\left(\left(a , {b}_{1}\right) \in g \wedge \left(a , {b}_{2}\right) \in g\right) \implies {b}_{1} = {b}_{2}$

Use the notation ${2}^{A}$ to represent the set of subsets of $A$ and ${2}^{B}$ the set of subsets of $B$.

Then we can define the pre-image function:

${\overline{g}}^{- 1} : {2}^{B} \to {2}^{A}$ by ${\overline{g}}^{- 1} \left({B}_{1}\right) = \left\{a \in A : g \left(a\right) \in {B}_{1}\right\}$

Then the domain of $g$ is simply ${\overline{g}}^{- 1} \left(B\right)$

If $f$ is a function that maps some elements of set $B$ to elements of a set $C$, then:

${\overline{f}}^{- 1} : {2}^{C} \to {2}^{B}$ is defined by ${\overline{f}}^{- 1} \left({C}_{1}\right) = \left\{b \in B : f \left(b\right) \in {C}_{1}\right\}$

Using this notation, the domain of $f \circ g$ is simply

${\overline{g}}^{- 1} \left({\overline{f}}^{- 1} \left(C\right)\right) = \left({\overline{g}}^{- 1} \circ {\overline{f}}^{- 1}\right) \left(C\right)$