14.3 Inverse Functions

14.12   Definition (Injective.) Let $A$ and $B$ be sets. A function $f: A \to B$ is called injective or one-to-one if and only if for all points $a,b$ in $A$

$$(a \neq b) \mbox{$\hspace{1ex}\Longrightarrow\hspace{1ex}$}(f(a) \neq f(b)),$$

or equivalently if and only if

$$(f(a) = f(b)) \mbox{$\hspace{1ex}\Longrightarrow\hspace{1ex}$}(a=b).$$

If $f$ is a function whose domain and codomain are subsets of R then $f$ is injective if and only if each horizontal line intersects the graph of $f$ at most once.

14.13   Examples. Let $f:[0,\infty) \to \mbox{{\bf R}}$ and $g:\mbox{{\bf R}}\to \mbox{{\bf R}}$ be defined by

$$f(x) = x^2 \mbox{ for all }x \in [0,\infty)$$

$$g(x) = x^2 \mbox{ for all }x \in (-\infty,\infty).$$

Then $f$ is injective, since for all $x,y \in [0,\infty)$ we have $x+y>0$, and hence

$$\Big(x^2 = y^2\Big) \mbox{$\hspace{1ex}\Longrightarrow\hspace... ...\mbox{$\hspace{1ex}\Longrightarrow\hspace{1ex}$}\Big(x=y\Big).$$

However $g$ is not injective, since $g(-1) = g(1).$

14.14   Remark (Strictly monotonic functions are injective.) If $h$ is strictly increasing on an interval $J$, then $h$ is injective on $J$, since for all $x,y \in J$

$$x \neq y \mbox{$\Longrightarrow$} ((x < y) \mbox{ or }(y<x)) \mbox{{}}$$

$$\mbox{$\Longrightarrow$} (( h(x) < h(y) )\mbox{ or }((h(y) < h(x)) \mbox{{}}$$

$$\mbox{$\Longrightarrow$} h(x) \neq h(y). \mbox{{}}$$

Similarly, any strictly decreasing function on $J$ is injective.

14.15   Definition (Surjective.) Let $A,B$ be sets and let $f: A \to B$. We say that $f$ is surjective if and only if $B = \mbox{image}(f)$, i.e. if and only if for every $b \in B$ there is at least one element $a$ of $A$ such that $f(a) = b$.

14.16   Examples. Let $f:\mbox{{\bf R}}\to \mbox{{\bf R}}$ and $g : \mbox{{\bf R}}\to [0,\infty)$ be defined by

$$f(x) = x^2 \mbox{ for all }x \in (-\infty,\infty)$$

$$g(x) = x^2 \mbox{ for all }x \in [0,\infty).$$

Then $g$ is surjective, since if $x \in [0,\infty),$ then $x = g(\sqrt x)$, but $f$ is not surjective, since $-1$ is not in the image of $f$.

14.17   Exercise. A Give examples of functions with the following properties, or else show that no such functions exist.

$f:\mbox{{\bf R}}\to \mbox{{\bf R}}$, $f$ is injective and surjective.

$g:\mbox{{\bf R}}\to \mbox{{\bf R}}$, $g$ is injective but not surjective.

$h: \mbox{{\bf R}}\to \mbox{{\bf R}}$, $h$ is surjective but not injective.

$k : \mbox{{\bf R}}\to \mbox{{\bf R}}$, $k$ is neither injective nor surjective.

14.18   Definition (Bijective.) Let $A,B$ be sets. A function $f: A \to B$ is called bijective if and only if $f$ is both injective and surjective.

14.19   Examples. If $f : [0,\infty) \to [0,\infty)$ is defined by

$$f(x) = x^2 \mbox{ for all }x \in [0, \infty),$$

then $f$ is bijective.

The function $\ln$ is a bijective function from $\mbox{${\mbox{{\bf R}}}^{+}$}$ to R. We know that ln is strictly increasing, and hence is injective. If $y$ is any real number we know that $\ln$ takes on values greater than $y$, and values less that $y$, so by the intermediate value property (here we use the fact that $\ln$ is continuous) it also takes on the value $y$, i.e. $\ln$ is surjective.

14.20   Remark. Let $A$ and $B$ be sets, and let $f: A \to B$ be a bijective function. Let $b$ be a generic element of $B$. Since $f$ is surjective, there is an element $a$ in $A$ such that $f(a) = b$. Since $f$ is injective this element $a$ is unique, i.e. if $a$ and $c$ are elements of $A$ then

$$\Big(f(a) = b \mbox{ and }f(c) = b\Big)\mbox{$\Longrightarrow$}\Big(f(a) = f(c)\Big) \mbox{$\Longrightarrow$}\Big(a = c\Big).$$

Hence we can define a function $g:B \to A$ by the rule

$$g(b) = \mbox{ the unique element $a \in A$ such that $f(a)=b$. }$$

Then by definition

$$f(g(b)) = b \mbox{ for all }b \in B.$$

Now let $a \in A$, so that $f(a) \in B$. It is clear that the unique element $s$ in $A$ such that $f(s) = f(a)$ is $s=a$, and hence

$$g(f(a)) = a \mbox{ for all }a \in A.$$

14.21   Definition (Inverse function.) Let $A,B$ be sets, and let $f: A \to B$. An inverse function for $f$ is a function $g:B \to A$ such that

$$\Big(f(g(b)) = b \mbox{ for all }b\in B\Big) \mbox{ and }\Big( g(f(a)) = a \mbox{ for all }a \in A\Big).$$

14.22   Remark (Bijective functions have inverses.) Notice that in the definition of inverse functions, both the domain and the codomain of $f$ enter in a crucial way. It is clear that if $g$ is an inverse function for $f$, then $f$ is an inverse function for $g$. Remark 14.20 shows that every bijective function $f: A \to B$ has an inverse.

14.23   Example. Let $f:[0,\infty)$ be defined by

$$f(x) = x^2 \mbox{ for all }x \in [0,\infty).$$

We saw above that $f$ is bijective, and hence has an inverse. If

$$g(x) = \sqrt x \mbox{ for all }x \in [0,\infty)$$

Then it is clear that $g$ is an inverse function for $f$.

We also saw that $\ln: \mbox{${\mbox{{\bf R}}}^{+}$}\to \mbox{{\bf R}}$ is bijective, and so it has an inverse. This inverse is not expressible in terms of any functions we have discussed. We will give it a name.

14.24   Definition ($E(x)$.) Let $E$ denote the inverse of the logarithm function. Thus $E$ is a function from R to $\mbox{${\mbox{{\bf R}}}^{+}$}$, and it satisfies the conditions

$$\ln(E(x)) = x \mbox{ for all }x \in \mbox{{\bf R}},$$

$$E(\ln(x)) = x \mbox{ for all }x \in \mbox{${\mbox{{\bf R}}}^{+}$}.$$

We will investigate the properties of $E$ after we have proved a few general properties of inverse functions.

In order to speak of the inverse of a function, as we did in the last definition, we should note that inverses are unique.

14.25   Theorem (Uniqueness of inverses.) Let $A,B$ be sets and let
$f: A \to B$. If $g$ and $h$ are inverse functions for $f$, then $g=h$.

Proof:     If $g$ and $h$ are inverse functions for $f$ then

$$\mbox{{\rm dom}}{(g)} = \mbox{{\rm dom}}{(h)} = \mbox{ codomain}(f) = B,$$

and

$$\mbox{codomain}(g) = \mbox{codomain}(h) = \mbox{{\rm dom}}{(f)} = A.$$

Also for all $x \in B$

$$h(x) = g(f(h(x))) = g(x).$$

(I have used the facts that $y = g(f(y))$ for all $y\in A$, and $f(h(x)) = x$ for all $x \in B$).

14.26   Theorem (Reflection theorem.) Let $f: A \to B$ be a function which has an inverse function $g:B \to A$. Then for all $(a,b) \in A \times B$

$$(a,b) \in {\rm graph}(f) \hspace{1ex}\Longleftrightarrow\hspace{1ex}(b,a) \in {\rm graph}(g).$$

Proof:     Let $f: A \to B$ be a function that has an inverse function $g:B \to A$. Then for all $(a,b) \in A \times B$

$$\Big(b = f(a)\Big) \mbox{$\hspace{1ex}\Longrightarrow\hspace{... ...x{$\hspace{1ex}\Longrightarrow\hspace{1ex}$}\Big(g(b) = a\Big)$$

and

$$\Big(g(b) = a\Big) \mbox{$\hspace{1ex}\Longrightarrow\hspace{... ...(g(b)) = f(a)\Big) \mbox{$\Longrightarrow$}\Big(b = f(a)\Big).$$

Thus

$$\Big(b = f(a)\Big) \hspace{1ex}\Longleftrightarrow\hspace{1ex}\Big(a = g(b)\Big).$$

Now

$$\Big( b = f(a)\Big) \hspace{1ex}\Longleftrightarrow\hspace{1ex}\Big( (a,b) \in \mbox{ graph}(f)\Big),$$

and

$$\Big(a = g(b) \Big) \hspace{1ex}\Longleftrightarrow\hspace{1ex}\Big( (b,a) \in \mbox{ graph}(g)\Big),$$

and the theorem now follows. $\diamondsuit$

Remark: If $f$ is a bijective function with $\mbox{{\rm dom}}(f)\subset\mbox{{\bf R}}$ and codomain($f$) $\subset \mbox{{\bf R}}$ Then the reflection theorem says that if $g$ is the inverse function for $f$, then graph($g$) = $D_+$(graph$(f)$) where $D_+$ is the reflection about the line $y=x$.

Since we know what the graph of $\ln$ looks like, we can make a reasonable sketch of graph$(E)$.

It is a standard notation to denote the inverse of a function $f$ by $f^{-1}$. However since this is also a standard notation for the function $${{1\over f}}$$ which is an entirely different object, I will not use this notation.

We have shown that if $f: A \to B$ is bijective, then $f$ has an inverse function. The converse is also true.

14.27   Theorem. Let $A,B$ be sets and let $f: A \to B$. If $f$ has an inverse function, then $f$ is both injective and surjective.

Proof:     Suppose $f$ has an inverse function $g:B \to A$. Then for all $s,t$ in $A$ we have

$$\Big(f(s) = f(t) \Big) \mbox{$\hspace{1ex}\Longrightarrow\hs... ... \mbox{$\hspace{1ex}\Longrightarrow\hspace{1ex}$}\Big(s=t\Big)$$ (14.28)

and hence $f$ is injective. Also, for each $b \in B$

$$b = f(g(b)),$$

so $b \in$ image($f$), and $f$ is surjective. $\diamondsuit$