6.6 Properties of functions

Definition 6.41 (Injective, injection).

A function $f:D\to C$ is said to be injective (or an injection) if $f(x_{1})=f(x_{2})$ implies $x_{1}=x_{2}$ for every $x_{1},x_{2}\in D$ i.e. for every $y\in C$ , there is at most one element $x\in D$ such that $f(x)=y$ .

Example 6.42.

Consider the function $f\colon\mathbb{N}_{0}\rightarrow\mathbb{N}_{0}$ , where $\mathbb{N}_{0}=\mathbb{N}\cup\{0\}$ and $f(n)=n^{2}$ .

Claim: The function $f$ is injective.

Proof: Let $m,n\in\mathbb{N}_{0}$ be such that $f(m)=f(n)$ . Then $m^{2}=n^{2}$ , and so

0=m^{2}-n^{2}=(m-n)(m+n).

If $m+n=0$ then $m=0$ and $n=0$ , so $m=n$ . Otherwise, $m-n=0$ , so $m=n$ . Thus $f$ is injective. $\square$

Definition 6.43 (Surjective, surjection).

A function $f:D\to C$ is said to be surjective (or a surjection) if, given any $y\in C$ , there is some $x\in D$ such that $f(x)=y$ .

Example 6.44.

Consider the function $g\colon\mathbb{R}\rightarrow[-5,+\infty)$ defined by $g(x)=x^{2}+2x-4$ .

Claim: The function $g$ is surjective.

Proof: Let $y\geq-5$ . We need to show there is some $x\in\mathbb{R}$ such that $g(x)=y$ . Observe that $g(x)=y\iff x^{2}+2x-4=y$ . By the quadratic formula, that means

x=\frac{-2\pm\sqrt{4+4(4+y)}}{2}

and since $y\geq-5$ , the square root is defined and this gives our value for $x$ . Thus, since we can find such an $x\in\mathbb{R}$ for every $y\in[-5,\infty)$ , this implies that $g$ is surjective. $\square$

Definition 6.45 (Bijective, bijection).

A function is said to be bijective if it is both injective and surjective.

Notice that a function $f:D\to C$ is injective if every element of $C$ has at most one element of $D$ that $f$ maps to it. And $f$ is surjective if every element of $C$ has at least one element of $D$ that maps to it. Thus, if $f$ is bijective, every element of $C$ has exactly one element of $D$ that maps to it. In particular, we can say that, for every element of $C$ , there exists a unique element of $D$ that $f$ maps to it.

We can interpret this from a graph perspective too. If $f$ is injective, then every horizontal line of the form $y=c$ where $c\in C$ must cross the graph of $f$ at most once. If $f$ is surjective, then every such horizontal line must cross the graph of $f$ at least once. So if $f$ is bijective, every such horizontal line must cross the graph of $f$ exactly once.

Exercise 6.46.

If $f:S\rightarrow T$ is a function with $A\subseteq S$ and $B\subseteq T$ , prove that

(a)

$A=f^{-1}(f(A))$ if $f$ is injective.
(b)

$f(f^{-1}(B))=B$ if $f$ is surjective.

Solution (please try for yourself before looking)

We first note that Proposition 6.39 already does half of the work for us since it says $A\subseteq f^{-1}(f(A))$ and $f(f^{-1}(B))\subseteq B$ .

(a)

Suppose that $f$ is injective. To prove that $A=f^{-1}(f(A))$ , it suffices to show that $A\supseteq f^{-1}(f(A))$ . Indeed, suppose $x\in f^{-1}(f(A))$ . Then $f(x)\in f(A)$ but this doesn’t necessarily mean $x\in A$ . All this tells us is there exists some $s\in A$ such that $f(s)=f(x)$ . However, since $f$ is injective, this means $s=x$ . Hence, $x\in A$ and thus, $A\supseteq f^{-1}(f(A))$ . Together with the result in Proposition 6.39, we conclude that $A=f^{-1}(f(A))$ .
(b)

Suppose that $f$ is surjective and let $y\in B$ . Then, since $B\subseteq T$ and $f$ is surjective, there exists some $x\in S$ such that $f(x)=y$ . In particular, since $y\in B$ , this means $x\in f^{-1}(B)$ . But then $y$ is mapped to be an element in $f^{-1}(B)$ by $f$ . So this means $y\in f(f^{-1}(B))$ . Hence, $f(f^{-1}(B))\supseteq B$ . Together with our result in Proposition 6.39, this means $f(f^{-1}(B))=B$ .

The following results show how we can deduce the injectivity and surjectivity of compositions of functions with these properties.

Proposition 6.47.

Let $f:A\to B$ , $g:B\to C$ be functions.

1.

If $f$ , $g$ are injective then so is $g\circ f$ .
2.

If $f$ , $g$ are surjective then so is $g\circ f$ .

Hence if $f$ , $g$ are bijective then so is $g\circ f$ .

Proof.

1.

Suppose $f$ and $g$ are injective and suppose $x,x^{\prime}\in X$ satisfies $(g\circ f)(x)=(g\circ f)(x^{\prime})$ . Then $g(f(x))=g(f(x^{\prime}))$ . Since $g$ is injective, this means $f(x)=f(x^{\prime})$ . Since $f$ is injective, this implies $x=x^{\prime}$ .

This completes the proof that $g\circ f$ is injective.
2.

Now suppose $f$ and $g$ are surjective and let $z\in Z$ . Since $g$ is surjective, there is some $y\in Y$ such that $z=g(y)$ . Since $f$ is surjective, there is some $x\in X$ such that $f(x)=y$ . Then

$(g\circ f)(x)=g(f(x))=g(y)=z.$

This finishes the proof that $g\circ f$ is surjective.

Thus, if $g$ and $f$ are bijective, then they are both injective and surjective, so by the previous two parts, $g\circ f$ is also both injective and surjective, and hence $g\circ f$ is bijective.

The proof above is an example of a “definition chase” style proof, where you work from one definition to another.

Inverse Functions

Suppose that we have a function $f:D\to C$ with graph $G$ , where

G=\{(x,y):y=f(x),x\in D\}.

Under what circumstances will the set of ordered pairs

G^{\prime}:=\{(y,x):(x,y)\in G\}

also define a legitimate function $F:C\to D$ ? Note, we have reversed the role of $C$ and $D$ and also the order of the pairs in the graph so that

(x,y)\in G\Leftrightarrow(y,x)\in G^{\prime}.

$G^{\prime}$ gives a new function if, and only if, for each $y\in C$ there exists a unique point $x\in D$ with $y=f(x)$ . Hence,

•

the image of $f$ must be the whole of the codomain so that for each $y\in C$ there is at least one point $x\in D$ with $y=f(x)$ i.e. $f$ must be surjective, AND
•

further, $f$ must be injective so that the $x\in D$ with $y=f(x)$ is unique.

When such a function exists we call it the inverse of $f$ , and write this as $f^{-1}$ . If the inverse of $f$ exists, we say that $f$ is invertible. This discussion shows that an inverse exists if and only if the original function is a bijection.

Proposition 6.48.

A function is bijective if and only if it is invertible.

Proposition 6.49.

If $f:D\to C$ is invertible, then $f^{-1}:C\to D$ is its inverse function if and only if $f\circ f^{-1}=id_{C}$ and $f^{-1}\circ f=id_{D}$ .

Proof.

Let $G=\{(x,y):y=f(x),x\in D\}$ be the graph of $f$ . Then, the graph of its inverse $f^{-1}$ is defined to be $G^{\prime}:=\{(y,x):(x,y)\in G\}=\{(a,b):b=f^{-1}(a),a\in C\}$ such that

(x,y)\in G\iff(y,x)\in G^{\prime}.

Therefore,

\displaystyle x=f^{-1}(y)=f^{-1}(f(x))=f^{-1}\circ f(x)=id_{D}(x).

Since $f^{-1}\circ f:D\to D$ and $id_{D}:D\to D$ , this means that $f^{-1}\circ f=id_{D}$ .

Also,

\displaystyle y=f(x)=f(f^{-1}(y))=f\circ f^{-1}(y)=id_{C}(y).

Since, $f\circ f^{-1}:C\to C$ and $id_{C}:C\to C$ , this means that $f\circ f^{-1}=id_{C}$ .

For the other direction, suppose that $f^{-1}:C\to D$ is some function where $f\circ f^{-1}=id_{C}$ and $f^{-1}\circ f=id_{D}$ . We claim that its graph $G^{\prime}=\{(a,b):b=f^{-1}(a),a\in C\}$ satisfies

(x,y)\in G\iff(y,x)\in G^{\prime}.

Indeed,

(y,x)\in G^{\prime}\implies f(x)=f(f^{-1}(y))=f\circ f^{-1}(y)=id_{C}(y)=y% \implies(x,y)\in G,

and

(x,y)\in G\implies f^{-1}(y)=f^{-1}(f(x))=f^{-1}\circ f(x)=id_{D}(x)=x\implies% (y,x)\in G^{\prime}.

Thus, $f^{-1}$ is an inverse for $f$ .

Inverse vs Preimage

Suppose $B$ is a subset of the codomain of some function $f$ . Rather confusingly, the notation for the inverse function $f^{-1}$ of $f$ (if it exists) leads to the image of $B$ under $f^{-1}$ being denoted by $f^{-1}(B)$ which is the same notation as the preimage of $B$ under $f$ . So watch out! The preimage $f^{-1}(B)$ of the set $B$ under $f$ is always defined, regardless of whether or not $f$ has an inverse. But if $f$ is bijective, its inverse function $f^{-1}$ exists and $f^{-1}(B)$ denotes both the preimage of the set $B$ under $f$ and the image of $B$ under $f^{-1}$ , and both sets are equal in this case.