From 9fbccf04f0fee409cb20490a5ebc58e49e397344 Mon Sep 17 00:00:00 2001 From: Luc Date: Wed, 27 Dec 2023 17:37:52 +0100 Subject: [PATCH] Added section to maps in set theory. --- docs/en/mathematics/set-theory/maps.md | 27 ++++++++++++++++++++++++-- 1 file changed, 25 insertions(+), 2 deletions(-) diff --git a/docs/en/mathematics/set-theory/maps.md b/docs/en/mathematics/set-theory/maps.md index 0f0f5ab..66ba566 100644 --- a/docs/en/mathematics/set-theory/maps.md +++ b/docs/en/mathematics/set-theory/maps.md @@ -43,6 +43,29 @@ If $B'$ is a subet of $B$ then the pre-image of $B'$, denoted by $f^{-1}(B') is ??? note "*Proof*:" - Let $a' \in A'$, then $f(a') \in f(A')$ and hence $a' \in f^{-1}(f(A'))$. Thus $A' \subseteq f^{-1}(f(A'))$. + Let $a' \in A'$, then $f(a') \in f(A')$ and hence $a' \in f^{-1}(f(A'))$. Thus $A' \subseteq f^{-1}(f(A'))$. + + Let $a \in f^{-1}(B')$, then $f(a) \in B'$. Thus $f(f^{-1}(B')) \subseteq B'$. + +## Special maps + +> *Definition*: let $f: A \to B$ be a map. +> +> * $f$ is called **surjective**, if for each $b \in B$ there is at least one $a \in A$ with $b = f(a)$. Thus $\text{Im}(f) = B$. +> * $f$ is called **injective** if for each $b \in B$, there is at most one $a$ with $f(a) = b$. +> * $f$ is called **bijective** if it is both surjective and injective. So, if for each $b \in B$ there is a unique $a \in A$ with $f(a) = b$. + +For example the map $\sin: \mathbb{R} \to \mathbb{R}$ is not surjective nor injective. The map $\sin: [-\frac{\pi}{2},\frac{\pi}{2}] \to \mathbb{R}$ is injective but not surjective and the map $\sin: \mathbb{R} \to [-1,1]$ is surjective but not injective. To conclude the map $\sin: [-\frac{\pi}{2},\frac{\pi}{2}] \to [-1,1]$ is a bijective map. + +
+ +> *Theorem*: let $A$ be a set of size $n$ and $B$ a set of size $m$. Let $f: A \to B$ be a map between the sets $A$ and $B$. +> +> * If $n < m$ then $f$ can not be surjective. +> * If $n > m$ then $f$ can not be injective. +> * If $n = m$ then $f$ is injective if and only if it is surjective. + +??? note "*Proof*:" + + Think of pigeonholes. (Not really a proof). - Let $a \in f^{-1}(B')$, then $f(a) \in B'$. Thus $f(f^{-1}(B')) \subseteq B'$. \ No newline at end of file