Definition:Preimage/Mapping/Subset

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Definition

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 $Y \subseteq T$.


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:

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


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


Also known as

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


Examples

Subset of Image of Square Root Function

Let $f: \R \to \R$ be the real function defined as:

$\forall x \in \R: \map f x = x^2$


Let $A \subseteq \R$ be defined as:

$A := \closedint 4 9 = \set {x \in \R: 4 \le x \le 9}$

Then the preimage of $A$ under $f$ is:

$f^{-1} \sqbrk A = \closedint {-3} {-2} \cup \closedint 2 3$


Let $B \subseteq \R$ be defined as:

$B := \closedint {-9} {-4} = \set {x \in \R: -9 \le x \le -4}$

Then the preimage of $B$ under $f$ is:

$f^{-1} \sqbrk B = \O$


Identity Function with Discontinuity

Let $f: \R \to \R$ be the real function defined as:

$\forall x \in \R: \map f x = \begin {cases} x & : x \ne 0 \\ 1 & : x = 0 \end {cases}$


Let the following subsets of $\R$ be defined:

\(\ds A\) \(=\) \(\ds \openint \gets 0\)
\(\ds B\) \(=\) \(\ds \openint 0 \to\)
\(\ds C\) \(=\) \(\ds \openint 0 1\)
\(\ds D\) \(=\) \(\ds \openint {-1} 1\)
\(\ds E\) \(=\) \(\ds \openint {-2} {-1}\)
\(\ds F\) \(=\) \(\ds \openint {\dfrac 1 2} 2\)

Then the preimages under $f$ of these sets is:

$f^{-1} \sqbrk A = \set {x \in \R: \map f x < 0} = \openint \gets 0$
$f^{-1} \sqbrk B = \set {x \in \R: \map f x > 0} = \openint 0 \to$
$f^{-1} \sqbrk C = \set {x \in \R: 0 < \map f x < 1} = \openint 0 1$
$f^{-1} \sqbrk D = \set {x \in \R: -1 < \map f x < 1} = \openint {-1} 0 \cup \openint 0 1$
$f^{-1} \sqbrk {E \cup F} = \openint {-2} {-1} \cup \openint {\dfrac 1 2} 2 \cup \set 0$


Also see


Generalizations


Related Concepts