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Let $\mathcal R \subseteq S \times T$ be a relation.

Let $\mathcal R^{-1} \subseteq T \times S$ be the inverse relation to $\mathcal R$, defined as:

$\mathcal R^{-1} = \left\{{\left({t, s}\right): \left({s, t}\right) \in \mathcal R}\right\}$

Preimage of Element

Every $s \in S$ such that $\left({s, t}\right) \in \mathcal R$ is called a preimage of $t$.

In some contexts, it is not individual elements that are important, but all elements of $S$ which are of interest.

Thus the preimage of $t \in T$ is defined as:

$\mathcal R^{-1} \left ({t}\right) := \left\{{s \in S: \left({s, t}\right) \in \mathcal R}\right\}$

This can also be written:

$\mathcal R^{-1} \left ({t}\right) := \left\{{s \in \operatorname{Im} \left({\mathcal R^{-1}}\right): \left({t, s}\right) \in \mathcal R^{-1}}\right\}$

That is, the preimage of $t$ under $\mathcal R$ is the image of $t$ under $\mathcal R^{-1}$.

Preimage of Subset

Let $Y \subseteq T$.

The preimage of $Y$ under $\mathcal R$ is defined as:

$\mathcal R^{-1} \sqbrk Y := \set {s \in S: \exists t \in Y: \tuple {s, t} \in \mathcal R}$

That is, the preimage of $Y$ under $\mathcal R$ is the image of $Y$ under $\mathcal R^{-1}$:

$\mathcal R^{-1} \sqbrk Y := \set {s \in S: \exists t \in Y: \tuple {t, s} \in \mathcal R^{-1} }$

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

Preimage of Relation

The preimage of $\mathcal R \subseteq S \times T$ is:

$\Preimg {\mathcal R} := \mathcal R^{-1} \sqbrk T = \set {s \in S: \exists t \in T: \tuple {s, t} \in \mathcal R}$

Also known as

A preimage is also known as an inverse image.

Also see

  • Results about preimages under relations can be found here.

Technical Note

The $\LaTeX$ code for \(\Preimg {f}\) is \Preimg {f} .

When the argument is a single character, it is usual to omit the braces:

\Preimg f