# Equivalence of Definitions of Inverse Image Mapping of Mapping

## Theorem

The following definitions of the concept of Inverse Image Mapping of Mapping are equivalent:

### Definition 1

The inverse image mapping of $f$ is the mapping $f^\gets: \powerset T \to \powerset S$ that sends a subset $Y \subseteq T$ to its preimage $f^{-1} \paren T$ under $f$:

$\forall Y \in \powerset T: \map {f^\gets} Y = \begin {cases} \set {s \in S: \exists t \in Y: \map f s = t} & : \Img f \cap Y \ne \O \\ \O & : \Img f \cap Y = \O \end {cases}$

### Definition 2

The inverse image mapping of $f$ is the direct image mapping of the inverse $f^{-1}$ of $f$:

$f^\gets = \paren {f^{-1} }^\to: \powerset T \to \powerset S$:

That is:

$\forall Y \in \powerset T: \map {f^\gets} Y = \set {s \in S: \exists t \in Y: \map f s = t}$

## Proof

Consider the mapping defined as:

$\forall Y \in \powerset T: \map {f^\gets} Y = \set {s \in S: \exists t \in Y: \map f s = t}$

Let $\Img f \cap Y = \O$.

Then:

$\forall t \in T: \neg \exists t \in \Img f \cap Y$

and so:

$\set {s \in S: \exists t \in Y: \map f s = t} = \O$

and so:

$\forall Y \in \powerset T: \map {f^\gets} Y = \begin {cases} \set {s \in S: \exists t \in Y: \map f s = t} & : \Img f \cap Y \ne \O \\ \O & : \Img f \cap Y = \O \end {cases}$

$\blacksquare$