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Let $f: S \to T$ be a mapping.

Let $f^{-1} \subseteq T \times S$ be the inverse of $f$, defined as:

$f^{-1} = \set {\tuple {t, s}: \map f s = t}$

Let $Y \subseteq T$.

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.

If no element of $Y$ has a preimage, then $f^{-1} \sqbrk Y = \O$.

Preimage of Subset as Element of Inverse Image Mapping

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 $f$:

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


$\forall Y \subseteq T: f^{-1} \sqbrk Y = \map {f^\gets} Y$

and so the preimage of $Y$ under $f$ is also seen referred to as the inverse image of $Y$ under $f$.

Both approaches to this concept are used in $\mathsf{Pr} \infty \mathsf{fWiki}$.

Also known as

Some sources use counter image or inverse image instead of preimage.

Also see


Related Concepts