Group Theory and Linear Algebra

Arun Ram
Department of Mathematics and Statistics
University of Melbourne
Parkville, VIC 3010 Australia
aram@unimelb.edu.au

Last updated: 17 September 2014

Lecture 4: Functions

Functions are for comparing sets.

Let $S$ and $T$ be sets. A function from $S$ to $T$ is a subset $f$ of $S\times T,$ $$f=\left\{(s,f\left(s\right))\hspace{0.17em}\right|\hspace{0.17em}s\in S\},$$ such that

(a)	If $s\in S$ then there exists $t\in T$ such that $(s,t)\in f\text{.}$
(b)	If $s\in S,$ and $t_1,t_2\in T$ and $(s,t_1),(s,t_2)\in f$ then $t_1=t_2\text{.}$

The function $f$ is an assignment assigning a mark $f\left(s\right)$ from $T$ to each $s\in S\text{.}$ Write $$\begin{array}{cccc}f:& S& ⟶& T\\ & s& ⟼& f\left(s\right)\end{array}\text{or}S\stackrel{f}{⟶}T\text{.}$$

A function $f:S\to T$ is injective if it satisfies: if $s_1,s_2\in S$ and $f\left(s_1\right)=f\left(s_2\right)$ then $s_1=s_2\text{.}$

A function $f:S\to T$ is surjective if it satisfies: if $t\in T$ then there exists $s\in S$ such that $f\left(s\right)=t\text{.}$

A function $f:S\to T$ is bijective if it is injective and surjective.

Let $f:S\to T$ and $g:S\to T$ be funtions. The functions $f$ and $g$ are equal if they satisfy if $s\in S$ then $f\left(s\right)=g\left(s\right)\text{.}$

Let $f:S\to T$ and $g:T\to U$ be functions. The composition of $g$ and $f$ is the function $$g\circ f:S⟶T\text{given by}(g\circ f)\left(s\right)=g\left(f\left(s\right)\right)\text{.}$$

Let $S$ be a set. The identity function on $S$ is the function $${\text{id}}_S:S⟶S\text{given by}{\text{id}}_S\left(s\right)=s\text{.}$$

Let $f:S\to T$ be a function. An inverse function to $f:S\to T$ is a function $g:T\to S$ such that $$g\circ f={\text{id}}_S\text{and}f\circ g={\text{id}}_T\text{.}$$

Let $f:S\to T$ be a function.

(a)	An inverse function to $f$ exists if and only if $f$ is bijective.
(b)	If an inverse function to $f$ exists then it is unique.

		Proof.
	Assume $f:S\to T$ is a function. (a) To show: An inverse function to $f$ exists if and only if $f$ is bijective. $\Rightarrow$ Assume that an inverse function to $f$ exists: $g:T\to S$ such that $g\circ f={\text{id}}_S$ and $f\circ g={\text{id}}_T\text{.}$ To show: $f$ is bijective. To show: (1) $f$ is injective. (2) $f$ is surjective. (1) To show: If $s_1,s_2\in S$ and $f\left(s_1\right)=f\left(s_2\right)$ then $s_1=s_2\text{.}$ Assume $s_1,s_2\in S$ and $f\left(s_1\right)=f\left(s_2\right)\text{.}$ To show: $s_1=s_2\text{.}$ $$\begin{array}{ccc}{\displaystyle s_1}& =& {\displaystyle {\text{id}}_S\left(s_1\right)=(g\circ f)\left(s_1\right)=g\left(f\left(s_1\right)\right)}\\ & =& {\displaystyle g\left(f\left(s_2\right)\right)=(g\circ f)\left(s_2\right)={\text{id}}_S\left(s_2\right)=s_2\text{.}}\end{array}$$ (2) To show: If $t\in T$ then there exists $s\in S$ such that $f\left(s\right)=t\text{.}$ Assume $t\in T\text{.}$ To show: There exists $s\in S$ such that $f\left(s\right)=t\text{.}$ Let $s=g\left(t\right)\text{.}$ To show: $f\left(s\right)=t\text{.}$ $$f\left(s\right)=f\left(g\left(t\right)\right)=(f\circ g)\left(t\right)={\text{id}}_T\left(t\right)=t\text{.}$$ So $f$ is bijective. $\Leftarrow$ To show: If $f:S\to T$ is bijective then an inverse function $g:T\to S$ exists. Assume $f:S\to T$ is bijective. To show: There exists $g:T\to S$ such that $$g\circ f={\text{id}}_S\text{and}f\circ g={\text{id}}_T\text{.}$$ Let $g:T\to S$ be given by $$g=\{(t,s)\in T\times S\hspace{0.17em}|\hspace{0.17em}f\left(s\right)=t\}\text{.}$$ To show: (a) $g$ is a function. (b) $g\circ f={\text{id}}_S\text{.}$ (c) $f\circ g={\text{id}}_T\text{.}$ (a) To show: (aa) If $t\in T$ then there exists $s\in S$ such that $(t,s)\in g\text{.}$ (ab) If $t\in T$ and $s_1,s_2\in S$ and $(t,s_1)\in g$ and $(t,s_2)\in g$ then $s_1=s_2\text{.}$ (aa) Assume $t\in T\text{.}$ To show: There exists $s\in S$ such that $(t,s)\in g\text{.}$ To show: There exists $s\in S$ such that $f\left(s\right)=t\text{.}$ This holds since $f$ is surjective. (ab) Assume $t\in T,$ $s_1,s_2\in S$ and $(t,s_1),(t,s_2)\in g\text{.}$ To show: $s_1=s_2\text{.}$ Since $(t,s_1),(t,s_2)\in g,$ then $f\left(s_1\right)=t$ and $f\left(s_2\right)=t\text{.}$ Since $f$ is injective, $s_1=s_2\text{.}$ (b) To show: $g\circ f={\text{id}}_S\text{.}$ To show: If $s\in S$ then $(g\circ f)\left(s\right)={\text{id}}_S\left(s\right)\text{.}$ Assume $s\in S\text{.}$ To show: $(g\circ f)\left(s\right)={\text{id}}_S\left(s\right)\text{.}$ $$(g\circ f)\left(s\right)=g\left(f\left(s\right)\right)=s_1,$$ where $s_1\in S$ such that $f\left(s_1\right)=f\left(s\right)\text{.}$ Since $f$ is injective, $s_1=s\text{.}$ So $(g\circ f)\left(s\right)=s={\text{id}}_S\left(s\right)\text{.}$ (c) To show: $(f\circ g)={\text{id}}_T\text{.}$ To show: If $t\in T$ then $(f\circ g)\left(t\right)={\text{id}}_T\left(t\right)\text{.}$ Assume $t\in T\text{.}$ To show: $\left(f\circ g\right)\left(t\right)={\text{id}}_T\left(t\right)\text{.}$ $$\left(f\circ g\right)\left(t\right)=f\left(g\left(t\right)\right)=f\left(s\right),$$ where $s\in S$ such that $f\left(s\right)=t\text{.}$ So $\left(f\circ g\right)\left(t\right)=f\left(s\right)=t={\text{id}}_T\left(t\right)\text{.}$ So $g$ is an inverse function to $f\text{.}$ (b) To show: The inverse function to $f:S\to T$ is unique. Assume $g_1:T\to S$ and $g_2:T\to S$ are inverse functions to $f\text{.}$ To show: $g_1=g_2\text{.}$ To show: If $t\in T$ then $g_1\left(t\right)=g_2\left(t\right)\text{.}$ We know that $$(f\circ g_1)={\text{id}}_T,(g_1\circ f)={\text{id}}_S,(f\circ g_2)={\text{id}}_T,(g_2\circ f)={\text{id}}_S\text{.}$$ Assume $t\in T\text{.}$ To show: $g_1\left(t\right)=g_2\left(t\right)\text{.}$ $$\begin{array}{ccc}{\displaystyle g_1\left(t\right)}& =& {\displaystyle g_1\left({\text{id}}_T\left(t\right)\right)=g_1\left((f\circ g_2)\left(t\right)\right)=g_1\left(f\left(g_2\left(t\right)\right)\right)}\\ & =& {\displaystyle (g_1\circ f)\left(g_2\left(t\right)\right)={\text{id}}_S\left(g_2\left(t\right)\right)=g_2\left(t\right)\text{.}}\end{array}$$ So $g_1=g_2\text{.}$ So the inverse function to $f:S\to T$ is unique. $\square$

Notes and References

These are a typed copy of Lecture 4 from a series of handwritten lecture notes for the class Group Theory and Linear Algebra given on August 1, 2011.

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