Functions and Relations

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Last update: Thursday, 6 October 2022.

Functions

Functions are the language of higher mathematics!

Let $A$ and $B$ be sets. A function $f \colon A \to B$ is a relation $f \subseteq A \times B$ such that for all $a \in A$ there is a unique $b \in B$ such that $(a, b) \in f$ . The set $A$ is called the source or domain of $f$ , written $\dom(f) = A$ . The set $B$ is called the target or codomain of $f$ , written $\cod(f) = B$ .

Usually we will write $f(a) = b$ rather than $(a,b)\in f$ or $a\, f\, b$ .

For every set $A$ there is an identity function $\id_A \colon A \to A$ having $\id_A(a) = a$ for all $a \in A$ .

Functions $f$ and $g$ with the same domain and codomain are equal when $f(x) = g(x)$ for all $x \in \dom(f)$ .

As relations, functions are special; functions take an input and produce a unique output for that input.

Given two compatible functions, we can get another function from them!

Functions $f \colon A \to B$ and $g \colon B \to C$ have composition $g \circ f \colon A \to C,~~x \mapsto g(f(x))$ .

For all $f \colon A \to B$ , $g \colon B \to C$ , and $h \colon C \to D$ we have $h \circ (g \circ f) = (h \circ g) \circ f$ .

For all $x\in\dom(f)$ we have the following equalities, completing the proof.

$\begin{align*} (h \circ (g \circ f))(x) &= h((g \circ f)(x)) \\ &= h(g(f(x))) \\ &= (h \circ g)(f(x)) \\ &= ((h \circ g) \circ f)(x) . \end{align*}$

Let $f \colon A \to B$ be a function.

The preimage of a set $S \subseteq B$ under $f$ is the set $f^{-1}(S) = \set{x \in A}{f(x) \in S}$ .
The image of a set $T \subseteq A$ under $f$ is the set $f(T) = \set{f(x) \in B}{x \in T}$ .

The next several propositions are straightforward applications of the definitions presented here. The proofs are left to you as a method of checking your understanding.

Let $f \colon A \to B$ be a function.

If $S \subseteq T \subseteq A$ , then $f(S) \subseteq f(T)$ .
If $S \subseteq T \subseteq B$ , then $f^{-1}(S) \subseteq f^{-1}(T)$ .

Let $f \colon A \to B$ be a function.

For all $S \subseteq A$ we have $S \subseteq f^{-1}(f(S))$ .
For all $T \subseteq B$ we have $f(f^{-1}(T)) \subseteq T$ .

Let $f \colon A \to B$ be a function and $S, T \subseteq A$ . The following all hold:

$f(S \cup T) = f(S) \cup f(T)$
$f(S \cap T) \subseteq f(S) \cap f(T)$
$f(S \setminus T) \supseteq f(S) \setminus f(T)$

Find examples of functions and subsets for which the above subset relations are strict.

The following proposition makes a nice exercise.

Let $f \colon A \to B$ be a function and $S, T \subseteq B$ . The following all hold.

$f^{-1}(S \cup T) = f^{-1}(S) \cup f^{-1}(T)$
$f^{-1}(S \cap T) = f^{-1}(S) \cap f^{-1}(T)$
$f^{-1}(S \setminus T) = f^{-1}(S) \setminus f^{-1}(T)$

Let $f \colon A \to B$ be a function.

Function $f$ is injective or into when for all $a, a' \in A$ we have $f(a) = f(a')$ implies $a = a'$ .
Function $f$ is surjective or onto when for all $b \in B$ there exists an $a \in A$ such that $f(a) = b$ .
Function $f$ is bijective or a one-to-one correspondence when $f$ is both injective and surjective.

The identity function $\id_A \colon A \to A$ is bijective.

Write down examples of functions which are injective, surjective, and bijective. Can you write down a function which is injective but not surjective? How about one which is surjective but not injective?

If $f$ is injective, can you strengthen Proposition 1? What if $f$ is surjective?

In Calculus, you may have studied some inverse functions (the Inverse Function Theorem needs them!).

Let $f \colon A \to B$ be a function.

A left inverse of $f$ is a function $g \colon B \to A$ such that $g \circ f = \id_A$ .
A right inverse of $f$ is a function $g \colon B \to A$ such that $f \circ g = \id_B$ .
An inverse of $f$ is a function $g \colon B \to A$ such that $g$ is both a left inverse of $f$ and a right inverse of $f$ .

The function $\id_A$ is its own inverse.

Find functions that have a left inverse but no right inverse and vice-versa.

The following proposition gives the relationship between invertibility and the properties above.

Let $f \colon A \to B$ be a function with $A \neq \emptyset$ .

Function $f$ has a left inverse if and only if $f$ is injective.
Function $f$ has a right inverse if and only if $f$ is surjective.
Function $f$ has an inverse if and only if $f$ is bijective.

Let $f \colon A \to B$ be a function.

Part 1: Supposing $f$ has a left inverse $g$ , then $(g \circ f)(a) = a$ for all $a \in A$ . Thus $g(f(a)) = a$ for all $a \in A$ . If $f(a) = f(a')$ for some $a,a' \in A$ , then $a = g(f(a)) = g(f(a')) = a'$ ; hence $f$ is injective. Supposing $f$ is injective, fix an element $a_0\in A$ (this is why we need $A\neq\emptyset$ ) and define

$\begin{equation*} g(x)= \begin{cases} a &\text{if } x = f(a) \text{ for some } a \in A \\ a_0 &\text{otherwise} \end{cases} \end{equation*}$

for all $x \in B$ . If $f(a) = f(a')$ , then $a = a'$ by injectivity; thus $g$ is well-defined. Moreover $(g \circ f)(x) = g(f(x)) = x$ for all $x \in A$ ; hence $g \circ f = \id_A$ and $g$ is a left inverse of $f$ .

Part 2: Supposing $f$ has a right inverse $g$ , then $(f \circ g)(b) = b$ for all $b \in B$ . Thus for all $b \in B$ one has $g(b) \in A$ and $f(g(b)) = b$ ; hence $f$ is surjective. Supposing $f$ is surjective, we fix for all $b \in B$ an element $a_b \in A$ with $f(a_b) = b$ .\footnote{This is possible by an abstract axiom of set theory (called the Axiom of Choice). Mathematicians argued for a long time over whether or not this is a good axiom because it has a lot of weird consequences. To learn more, email me…} Now define $g \colon B \to A$ by $g(b) = a_b$ ; note that this is well-defined by surjectivity of $f$ . Moreover $(f \circ g)(x) = f(g(x)) = f(a_x) = x$ for all $x \in B$ ; hence $f \circ g = \id_B$ and $g$ is a right inverse of $f$ .

Part 3: Supposing $f$ has an inverse, $f$ has both a left and right inverse; hence by Part 1 and Part 2, $f$ is both injective and surjective, and thus bijective. If $f$ is bijective, then $f$ is injective and surjective by definition; thus by Part 1 and Part 2 $f$ has a left inverse $g$ and a right inverse $g'$ . Now $g = g \circ \id_B = g \circ (f \circ g') = (g \circ f) \circ g' = \id_A \circ g' = g'$ and hence $g$ is both a left and right invere for $f$ .

Let $f \colon A \to B$ be a function.

If $f$ is injective, then for all $S \subseteq \dom(f)$ we have $f^{-1}(f(S)) = S$ .
If $f$ is surjective, then for all $T \subseteq \cod(f)$ we have $f(f^{-1}(S)) = S$ .

Exercise (HINT: you can use the preceeding proposition).