A function $f\colon A\to B$ is bijective (or $f$ is a bijection) if each $b\in B$ has exactly one preimage. Since "at least one'' + "at most one'' = "exactly one'', $f$ is a bijection if and only if it is both an injection and a surjection. A bijection is also called a one-to-one correspondence.

Example 4.6.1 If $A=\{1,2,3,4\}$ and $B=\{r,s,t,u\}$, then

$$ \begin{array}{} f(1)=u&f(3)=t\\ f(2)=r&f(4)=s\\ \end{array} $$

is a bijection. $\square$

Example 4.6.2 The functions $f\colon \R\to \R$ and $g\colon \R\to \R^+$ (where $\R^+$ denotes the positive real numbers) given by $f(x)=x^5$ and $g(x)=5^x$ are bijections. $\square$

Example 4.6.3 For any set $A$, the identity function $i_A$ is a bijection. $\square$

Definition 4.6.4 If $f\colon A\to B$ and $g\colon B\to A$ are functions, we say $g$ is an inverse to $f$ (and $f$ is an inverse to $g$) if and only if $f\circ g=i_B$ and $g\circ f=i_A$. $\square$

Example 4.6.5 If $f$ is the function from example 4.6.1 and

$$ \begin{array}{} g(r)=2&g(t)=3\\ g(s)=4&g(u)=1\\ \end{array} $$

then $f$ and $g$ are inverses. For example, $f(g(r))=f(2)=r$ and $g(f(3))=g(t)=3$. $\square$

Example 4.6.6 An inverse to $x^5$ is $\root 5 \of x$: $$ (\root 5 \of x\,)^5 = x, \quad \root 5 \of {x^5} = x. $$ $\square$

Example 4.6.7 If we think of the exponential function $e^x$ as having domain $\R$ and codomain $\R^{>0}$ (the positive real numbers), and $\ln x$ as having domain $\R^{>0}$ and codomain $\R$, then they are inverses: and $$ \ln e^x = x, \quad e^{\ln x}=x. $$ $\square$

Example 4.6.8 The identity function $i_A\colon A\to A$ is its own inverse. $\square$

If you understand these examples, the following should come as no surprise.

Theorem 4.6.9 A function $f\colon A\to B$ has an inverse if and only if it is bijective.

Proof. Suppose $g$ is an inverse for $f$ (we are proving the implication $\Rightarrow$). Since $g\circ f=i_A$ is injective, so is $f$ (by 4.4.1(a)). Since $f\circ g=i_B$ is surjective, so is $f$ (by 4.4.1(b)). Therefore $f$ is injective and surjective, that is, bijective.

Conversely, suppose $f$ is bijective. Let $g\colon B\to A$ be a pseudo-inverse to $f$. From the proof of theorem 4.5.2, we know that since $f$ is surjective, $f\circ g=i_B$, and since $f$ is injective, $g\circ f= i_A$. $\qed$

We have talked about "an'' inverse of $f$, but really there is only one.

Theorem 4.6.10 If $f\colon A\to B$ has an inverse function then the inverse is unique.

Proof. Suppose $g_1$ and $g_2$ are both inverses to $f$. Then $$ g_1=g_1\circ i_B=g_1\circ (f\circ g_2)=(g_1\circ f)\circ g_2=i_A\circ g_2= g_2, $$ proving the theorem. (See exercise 7 in section 4.1.) $\qed$

Because of theorem 4.6.10, we can talk about "the'' inverse of $f$, assuming it has one; we write $f^{-1}$ for the inverse of $f$. Note well that this extends the meaning of "$f^{-1}$'', in a potentially confusing way. No matter what function $f$ we are given, the induced set function $f^{-1}$ is defined, but the inverse function $f^{-1}$ is defined only if $f$ is bijective. In other words, $f^{-1}$ is always defined for subsets of the codomain, but it is defined for elements of the codomain only if $f$ is a bijection.

We close with a pair of easy observations:

Theorem 4.6.11

    a) The composition of two bijections is a bijection.

    b) The inverse of a bijection is a bijection.

Proof. Part (a) follows from theorems 4.3.5 and 4.3.11. For part (b), if $f\colon A\to B$ is a bijection, then since $f^{-1}$ has an inverse function (namely $f$), $f^{-1}$ is a bijection.$\qed$

Exercises 4.6

Ex 4.6.1 Find an example of functions $f\colon A\to B$ and $g\colon B\to A$ such that $f\circ g=i_B$, but $f$ and $g$ are not inverse functions.

Ex 4.6.2 Suppose $[a]$ is a fixed element of $\Z_n$. Define $A_{{[ a]}}\colon \Z_n\to \Z_n$ by $A_{{[a]}}([x])=[a]+[x]$. Show this is a bijection by finding an inverse to $A_{{[a]}}$.

Ex 4.6.3 Suppose $[u]$ is a fixed element of $\U_n$. Define $M_{{[ u]}}\colon \Z_n\to \Z_n$ by $M_{{[ u]}}([x])=[u]\cdot[x]$. Show this is a bijection by finding an inverse to $M_{{[u]}}$.

Ex 4.6.4 Show that for any $m, b$ in $\R$ with $m\ne 0$, the function $L(x)=mx+b$ is a bijection, by finding an inverse.

Ex 4.6.5 Suppose $f\colon A\to A$ is a function and $f\circ f$ is bijective. Is $f$ necessarily bijective?

Ex 4.6.6 Show there is a bijection $f\colon \N\to \Z$. (Hint: define $f$ separately on the odd and even positive integers.)

Ex 4.6.7 If $f\colon A\to B$ and $g\colon B\to C$ are bijections, prove $(g\circ f)^{-1} = f^{-1}\circ g^{-1}$.

Ex 4.6.8 Suppose $f\colon A\to B$ is an injection and $X\subseteq A$. Prove $f^{-1}(f(X))=X$.