Functions

Motivating example. Suppose $A$ is the set of students who wrote a test and $B=\{0,1,\dots,100\}$ is the set of possible marks; relate a student $a$ to a mark $b$ if $a$ scored $b$. Two facts hold: (1) every student got some mark -- every $a\in A$ has some $b$ with $(a,b)\in R$; and (2) no student got two different marks -- if $(a,b),(a,c)\in R$ then $b=c$. Relations with exactly these two properties are functions.

Definition. A relation $f\subseteq A\times B$ is a function from $A$ to $B$ if:

1. for every $a\in A$, there is some $b\in B$ with $(a,b)\in f$ (every domain element has an image), and
2. if $(a,b)\in f$ and $(a,c)\in f$ then $b=c$ (that image is unique). $A$ is the domain, $B$ the co-domain. If $(a,b)\in f$, write $f(a)=b$: $b$ is the image of $a$, $a$ is a pre-image of $b$ (the article changes: an element has exactly one image but can have several pre-images), and $f(a)$ is "the value of $f$ at $a$". The range is $\{b:(a,b)\in f\text{ for some }a\}\subseteq B$. If $B\subseteq R$, $f$ is a real-valued function. Two functions $f,g$ are equal if they share the same domain and $f(a)=g(a)$ for every $a$ in it. We write $f:A\to B$ ("$f$ is from $A$ to $B$", or "$f$ maps $A$ into $B$"). There is no requirement that every co-domain element have a pre-image (that property is studied separately, as "onto"), and there is no restriction on how many pre-images a co-domain element may have (that is what makes "one-to-one" a separate, extra property) -- both of these follow from asymmetry already built into the definition: only the domain side is required to be fully and uniquely covered. Every function is a relation, but not every relation is a function. $f=\{(a,1),(b,2),(c,2),(d,4)\}$ is a function from $\{a,b,c,d\}$ to $\{1,2,4\}$; it is not a function from $\{a,b,c,d,e\}$ to $\{1,2,3,4\}$ (element $e$ has no image), and not a function into $\{1,2,3,5\}$ either (the image $4$ of $d$ is not in that co-domain) -- so both domain and co-domain must always be stated explicitly. …