Let \(X\) and \(Y\) be two sets. A function from \(X\) to \(Y\), denoted \(f\colon X\rightarrow Y\text{,}\) is a rule which, given any input \(x\in X\text{,}\) returns an output \(y\in Y\text{.}\) In this case we write \(y=f(x)\) and call \(y\) the image of \(x\) under \(f\), or the value of \(f\) at \(x\). Similarly, we say \(f\) maps (or sends) the input \(x\) to the output \(y\text{.}\)
Invoking the notion of a rule in the definition of a function is admittedly somewhat vague. A more precise definition can be given using the Cartesian product. Namely, given sets \(X\) and \(Y\text{,}\) we define a function \(f\colon X\rightarrow Y\) to be a subset \(f\subseteq X\times Y\) satisfying the following property: for all \(x\in X\) there is a unique element \((x,y)\in f\text{.}\) The uniqueness of the pair \((x,y)\) then allows us to define the value \(f(x)\) of \(f\) at \(x\) as \(y\text{,}\) denoted \(f(x)=y\text{.}\)
As with sets and tuples, we are obliged to specify what we mean for two functions to be equal. The definition below makes clear how the the domain and codomain, as well as the rule that takes inputs to outputs, are all essential ingredients of a function.
Given a function \(f\colon X\rightarrow Y\) and a subset \(A \subseteq X\text{,}\) the image of \(A\) under \(f\), denoted \(f(A)\text{,}\) is defined as
\begin{equation*}
f(A)=\left\{ y \in Y \colon f(a)=y \text{ for some } a \in A \right\}\text{.}
\end{equation*}
In other words, \(f(A)\) is the set of all outputs \(f(a)\text{,}\) where \(a\in X\text{.}\)
The function \(f\) is injective (or one-to-one) if for all \(x,x'\in X\text{,}\) if \(f(x)=f(x')\text{,}\) then \(x=x'\text{:}\) equivalently, if \(x\ne x'\text{,}\) then \(f(x)\ne f(x')\text{.}\)
The function \(f\) is surjective (or onto) if for all \(y\in Y\text{,}\) there is an \(x\in X\) such that \(f(x)=y\text{:}\) equivalently, \(\im f=Y\text{.}\)
Example0.2.8.Role of domain and codomain in injectivity and surjectivity.
Consider the following three functions
\begin{align*}
f\colon \R \amp\rightarrow \R \amp g\colon [0,\infty)\amp\rightarrow \R \amp h\colon [0,\infty)\amp \rightarrow [0,\infty) \\
x \amp\mapsto x^2 \amp x \amp\mapsto x^2 \amp x \amp\mapsto x^2 \amp \text{.}
\end{align*}
Although all three functions are defined using the same rule (\(x\mapsto x^2\)), they are not equal thanks to their differing domains and/or codomains. The choice of domain and codomain in these examples also plays a crucial role in determining whether the function is injective and/or surjective. The function \(f\) is neither injective (\(f(-2)=f(2)=4\)) nor surjective (\(f(X)=[0,\infty)\ne \R\)); the function \(g\) is injective but not surjective; the function \(h\) is both injective and surjective, hence bijective.
Suppose \(f\colon Z\rightarrow W\) and \(g\colon X\rightarrow Y\) are functions satisfying \(Y\subseteq Z\text{.}\) The composition of \(f\) and \(g\) is the function \(f\circ g\colon X\rightarrow W\) defined as \(f\circ g\, (x)=f(g(x))
\text{,}\) for all \(x\in X\text{.}\)
A function \(f\colon X\rightarrow Y\) is invertible if there is a function \(g\colon Y\rightarrow X\) satisfying \(g\circ f=\id_X\) and \(f\circ g=\id_Y\text{:}\) i.e.,
\begin{align*}
g(f(x))\amp =x \text{ for all } x\in X, \\
f(g(y))\amp=y \text{ for all } y\in Y \text{.}
\end{align*}
For each of the following functions, \(f:\textbf{R}\longrightarrow\textbf{R}\) state whether they are one-to-one, onto, both or neither. Use the following key:
