If A and B are sets, then a function f from A to B, written $f:A\to B$, is a rule that associates to each element x in A a unique element denoted f(x) in B. The element f(x) is called the image of x (under f), and x is called a preimage of f(x) (under f). If $f:A\to B$, then A is called the domain of f, B is called the codomain of f, and the set $\{f(x):x\in A\}$ is called the range of f. Note that the range of f is a subset of B. If $S\subseteq A$, we denote by f(S) the set $\{f(x):x\in S\}$ of all images of elements of S. Likewise, if $T\subseteq B$, we denote by $f^{-1}(T)$ the set $\{x\in A:f(x)\in T\}$ of all preimages of elements in T. Finally, two functions $f:A\to B$ and $g:A\to B$ are equal, written $f=g$, if $f(x)=g(x)$ for all $x\in A$.

As Example 1 shows, the preimage of an element in the range need not be unique. Functions such that each element of the range has a unique preimage are called one-to-one; that is $f:A\to B$ is one-to-one if $f(x)=f(y)$ implies $x=y$ or, equivalently, if $x\ne y$ implies $f(x)\ne f(y)$.

If $f:A\to B$ is a function with range B, that is, if $f(A)=B$, then f is called onto. So f is onto if and only if the range of f equals the codomain of f.

Let $f:A\to B$ be a function and $S\subseteq A$. Then a function $f_S:S\to B$, called the restriction of f to S, can be formed by defining $f_S(x)=f(x)$ for each $x\in S$.

The next example illustrates these concepts.

Let A, B, and C be sets and $f:A\to B$ and $g:B\to C$ be functions. By following f with g, we obtain a function $g\circ f:A\to C$ called the composite of g and f. Thus $(g\circ f)(x)=g(f(x))$ for all $x\in A$. For example, let $A=B=C=R,f(x)=\sin x$, and $g(x)=x^2+3$. Then $(g\circ f)(x)=(g(f(x))={\sin }^{2}x+3$, whereas $(f\circ g)(x)=f(g(x))=\sin (x^2+3)$. Hence, $g\circ f\ne f\circ g$. Functional composition is associative, however; that is, if $h:C\to D$ is another function, then $h\circ (g\circ f)=(h\circ g)\circ f$.

A function $f:A\to B$ is said to be invertible if there exists a function $g:B\to A$ such that $(f\circ g)(y)=y$ for all $y\in B$ and $(g\circ f)(x)=x$ for all $x\in A$. If such a function g exists, then it is unique and is called the inverse of f. We denote the inverse of f (when it exists) by $f^{-1}$. It can be shown that f is invertible if and only if f is both one-to-one and onto.

The following facts about invertible functions are easily proved.

1. If $f:A\to B$ is invertible, then $f^{-1}$ is invertible, and ${(f^{-1})}^{-1}=f$.

2. If $f:A\to B$ and $g:B\to C$ are invertible, then $g\circ f$ is invertible, and ${(g\circ f)}^{-1}=f^{-1}\circ g^{-1}$.