Definition:Mapping/Diagrammatic Presentations/Finite

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Diagrammatic Presentation of Mapping on Finite Set

The following diagram illustrates the mapping:

$f: S \to T$

where $S$ and $T$ are the finite sets:

\(\displaystyle S\) \(=\) \(\displaystyle \set {a, b, c, i, j, k}\)
\(\displaystyle T\) \(=\) \(\displaystyle \set {p, q, r, s}\)

and $f$ is defined as:

$f = \set {\tuple {a, p}, \tuple {b, p}, \tuple {c, p}, \tuple {i, r}, \tuple {j, s}, \tuple {k, s} }$


Thus the images of each of the elements of $S$ under $f$ are:

\(\displaystyle \map f a\) \(=\) \(\displaystyle \map f b = \map f c = p\)
\(\displaystyle \map f i\) \(=\) \(\displaystyle r\)
\(\displaystyle \map f j\) \(=\) \(\displaystyle \map f k = s\)
MappingFinite.png
$S$ is the domain of $f$.
$T$ is the codomain of $f$.
$\set {p, r, s}$ is the image of $f$.


The preimages of each of the elements of $T$ under $f$ are:

\(\displaystyle \map {f^{-1} } p\) \(=\) \(\displaystyle \set {a, b, c}\)
\(\displaystyle \map {f^{-1} } q\) \(=\) \(\displaystyle \O\)
\(\displaystyle \map {f^{-1} } r\) \(=\) \(\displaystyle \set i\)
\(\displaystyle \map {f^{-1} } s\) \(=\) \(\displaystyle \set {j, k}\)


Note that $f$ is neither injective nor surjective:

$\map {f^{-1} } p$ is not a singleton: $\map f a = \map f b = \map f c$
$\map {f^{-1} } q = \O$


Sources