Equivalence of Definitions of Preimage of Subset under Mapping
Theorem
Let $f: S \to T$ be a mapping from a set $S$ to a set $T$.
Let $Y \subseteq T$ be a subset of $T$.
The following definitions of the concept of Preimage of Subset under Mapping are equivalent:
Definition 1
The preimage of $Y$ under $f$ is defined as:
- $f^{-1} \sqbrk Y := \set {s \in S: \exists t \in Y: \map f s = t}$
That is, the preimage of $Y$ under $f$ is the image of $Y$ under $f^{-1}$, where $f^{-1}$ can be considered as a relation.
Definition 2
The preimage of $Y$ under $f$ can be seen to be an element of the codomain of the inverse image mapping $f^\gets: \powerset T \to \powerset S$ of $f$:
- $\forall Y \in \powerset T: \map {f^\gets} Y := \set {s \in S: \exists t \in Y: \map f s = t}$
Thus:
- $\forall Y \subseteq T: f^{-1} \sqbrk Y = \map {f^\gets} Y$
Proof
The difference in definitions is no more than a difference in notations.
Let $X$ be a preimage by definition $2$.
Then by definition:
- $X := \map {f^\gets} Y$
By definition of inverse image mapping of mapping:
- $\forall Y \in \powerset T: \map {f^\gets} Y = \set {s \in S: \exists t \in Y: \map f s = t}$
Thus $X$ is a preimage by definition $1$.
$\blacksquare$