1.2: Functions

Example \(\PageIndex{1}\)

Consider the function \(f: \mathbb{Z} \to \mathbb{R}\) defined by \(f(x)=x^2\text{.}\) The domain of \(f\) is \(\mathbb{Z}\) and the codomain of \(f\) is \(\mathbb{R}\text{;}\) the range of \(f\) is \(\{x^2\,:\,x\in \mathbb{Z}\}=\{0,1,4,9,\ldots\}\text{.}\) The image of \(\{-2,-1,1,2\}\) under \(f\) is the two-element set \(\{1,4\} \subseteq \mathbb{R} \text{,}\) and the preimage of \(\{4,25,\pi\}\) under \(f\) is the set \(\{\pm 2, \pm 5\}\text{.}\) (Do you see why \(\pm \sqrt{\pi}\) are not in this preimage?) What is the preimage of just \(\{\pi\}\) under \(f\text{?}\)

The following definitions will be very important in our future work.

Definition: One-to-one, Onto, and Bijection

Let \(S\) and \(T\) be sets, and \(f:S\to T\text{.}\)

Function \(f\) is one-to-one (1-1) if whenever \(s_1, s_2\in S\) with \(f(s_1)=f(s_2)\text{,}\) we have \(s_1=s_2\text{.}\) Equivalently, \(f\) is one-to-one if whenever \(s_1\neq s_2 \in S\text{,}\) then \(f(s_1)\neq f(s_2) \in T\text{.}\)
Function \(f\) is onto if for every \(t\in T\text{,}\) there exists an element \(s\in S\) such that \(f(s)=t\text{.}\) Equivalently, \(f\) is onto if \(f(S)=T\text{.}\)
Function \(f\) is a bijection if it is both one-to-one and onto.

Remark

We will often have to show functions are one-to-one or onto. Given a function \(f:S\to T\text{,}\) the following methods are recommended.

To prove that \(f\) is one-to-one: Let \(s_1,s_2 \in S\) with \(f(s_1)=f(s_2)\) and prove that then \(s_1=s_2\text{.}\)

Note: It is not sufficient to prove that if \(s_1=s_2\) in \(S\) then \(f(s_1)=f(s_2)\text{;}\) that holds true for ANY function from \(S\) to \(T\text{!}\) Be careful to assume and prove the correct facts.
To prove that \(f\) is not one-to-one: Identify two elements \(s_1 \neq s_2\) of \(S\) such that \(f(s_1)=f(s_2)\text{.}\)
To prove that \(f\) is onto: Let \(t\in T\) and prove that there exists an element \(s\in S\) with \(f(s)=t\text{.}\)

Note: It is not sufficient to prove that if \(s\in S\) then \(f(s)\) is in \(T\text{;}\) that holds true for ANY function from \(S\) to \(T\text{!}\) Again, be careful to assume and prove the correct facts.
To prove that \(f\) is not onto: Identify an element \(t\in T\) for which there is no \(s\in S\) with \(f(s)=t\text{.}\)

Example \(\PageIndex{2}\)

Consider the function \(f: \mathbb{R}^* \to \mathbb{R}^+\) defined by \(f(x)=x^2\text{.}\) Function \(f\) is not one-to-one: indeed, \(-1\) and \(1\) are in \(\mathbb{R}^*\) with \(f(-1)=1=f(1)\) in \(\mathbb{R}^+\text{.}\) However, \(f\) is onto: indeed, let \(t\in \mathbb{R}^+\text{.}\) Then \(\sqrt{t} \in \mathbb{R}^*\) with \(t=f(\sqrt{t})\text{,}\) so we're done.

Example \(\PageIndex{3}\)

Consider the function \(f: \mathbb{Z}^+ \to \mathbb{R}\) defined by \(f(x)=\dfrac{x}{2}\text{.}\) Function \(f\) is one-to-one: indeed, let \(s_1, s_2 \in \mathbb{Z}^+\) with \(f(s_1)=f(s_2)\text{.}\) Then \(\dfrac{s_1}{2}=f(s_1)=f(s_2)=\dfrac{s_2}{2}\text{;}\) multiplying both sides of the equation \(\dfrac{s_1}{2}=\dfrac{s_2}{2}\) by \(2\), we obtain \(s_1=s_2\text{.}\) However, \(f\) is not onto: for example, \(\pi \in \mathbb{R}\) but there is no positive integer \(s\) for which \(f(s)=\dfrac{s}{2}=\pi\text{.}\)

Recall that we can combine certain functions using composition:

Definition: Composition and Identity Function

If \(f:S\to T\) and \(g:T\to U\text{,}\) then the composition of \(f\) and \(g\) is the function \(g\circ f: S\to U\) defined by

\begin{equation*} (g\circ f)(s)=g(f(s)) \end{equation*}

for all \(s\in S\text{.}\) (More generally, you can compose functions \(f:S\to T\) and \(g:R\to U\) to form \(g\circ f:S\to U\text{,}\) as long as \(f(S)\subseteq R\text{.}\)) Also recall that given any set \(S\text{,}\) the identity function on \(S\) is the function \(1_S: S\to S\) defined by \(1_S(s)=s\) for every \(s\in S\text{.}\)

Definition: Inverse

Let \(f\) be a function from \(S\) to \(T\text{.}\) A function \(g\) from \(T\) to \(S\) is an inverse of \(f\) if \(g\circ f\) and \(f\circ g\) are the identity functions on \(S\) and \(T\text{,}\) respectively; that is, if for all \(s\in S\) and \(t\in T\text{,}\) \(g(f(s))=s\) and \(f(g(t))=t\text{.}\)

We say a function is invertible if it has an inverse.

We have the following useful theorems.

Theorem \(\PageIndex{1}\)

Function \(f:S\to T\) is invertible if and only if \(f\) is a bijection.

Proof

Suppose \(f\) has inverse \(g\). Let \(t∈T\). Then \(f(g(t))=t\), so \(f\) is onto. Next, suppose \(s_1,s_2∈S\) with \(f(s_1)=f(s_2)\). Then

\(s_1=g(f(s_1))=g(f(s_2))=s_2\),

so \(f\) is one-to-one.

Conversely, suppose \(f\) is a bijection. Then for every \(t∈T\), there exists a unique element \(s_t∈S\) such that \(f(s_t)=t.\) It is then straighforward to show that the function \(g:T→S\) defined by

\(g(t)=t_s\) for every \(t∈T\)

is an inverse of \(f\).

Theorem \(\PageIndex{2}\)

Let \(f:S\to T\) be invertible. Then the inverse of \(f\) is unique; we denote the unique inverse of \(f\) by \(f^{-1}\text{.}\)

Proof: Suppose \(f\) has inverses \(g\) and \(h\). Then for every \(t∈T\), \(f(g(t))=t=f(h(t))\). But since \(f\) is invertible, it's one-to-one, so we must have \(g(t)=h(t)\) for every \(t∈T\). Thus, \(g=h\), and so \(f\) cannot have two or more distinct inverses.

Theorem \(\PageIndex{3}\)

Let \(f:S\to T\) and \(g:T\to U\) be functions. If \(f\) and \(g\) are both 1-1 [resp., onto], then \(g\circ f: S\to U\) is 1-1 [resp., onto]. It follows that if \(f\) and \(g\) are both bijections, then \(g\circ f: S\to U\) is a bijection.

Proof

Assume \(f\) and \(g\) are 1-1. Let \(s_1,s_2∈S\) with \((g \circ f)(s_1)=(g \circ f)(s_2)\). Then \(g(f(s_1))=g(f(s_2));\) since \(g\) is one-to-one, this shows that \(f(s_1)=f(s_2)\). Then since \(f\) is one-to-one, we must have \(s_1=s_2\). Thus, \(g \circ f\) is one-to-one.

The proof that \(g \circ f\) is onto if \(f\) and \(g\) are onto is left as an exercise for the reader.