Introduction
Quotients are fundamental in higher mathematics. They appear in almost every field of modern mathematics, and a firm understanding of them is crucial for further study. Unfortunately, they are often poorly explained, with their true meaning obscured behind a wall of mathematical jargon.
Motivation and Definition
At its most basic, a quotient set is no more than a partitioning of a set into subsets. We partition the set according to a rule called an equivalence relation. Loosely speaking, this tells us when two elements of our set should belong in the same subset.
A first example
Consider the set of integers, $\mathbb{Z}$. We say that two elements $a, b \in \mathbb{Z}$ are equivalent if their difference $a - b$ is contained in $p\mathbb{Z}$. We write this as $a \sim b$. In other words, $a \sim b$ if and only if $a \equiv b \pmod{p}$. We will see later that $\sim$ is an example of an equivalence relation.
Suppose $p=3$. Then, we have that $\cdots \sim -3 \sim 0 \sim 3 \sim \cdots$. In fact, if we group the elements of $\mathbb{Z}$ under the rule that two elements $a, b \in \mathbb{Z}$ are in the same subset iff $a \sim b$, we see that we can create exactly 3 subsets
\[ \begin{aligned} 3\mathbb{Z} &= [0] = \{\cdots-3, 0, 3, \cdots\}\\ 1 + 3\mathbb{Z} &= [1] = \{\cdots-2, 1, 4, \cdots\}\\ 2 + 3\mathbb{Z} &= [2] = \{\cdots-1, 2, 5, \cdots\} \end{aligned} \]We call these subsets the equivalence classes under the relation $\sim$. In particular, noting that $\mathbb{Z} = [0] \cup [1] \cup [2]$, we see that we have partioned the set $\mathbb{Z}$ into three subsets. We often write this as $\mathbb{Z}/{\sim}$, which means we have partioned $\mathbb{Z}$ under the rule $\sim$. This is a quotient set. It is important to note here that the elements of $\mathbb{Z}/{\sim}$ are not the elements of $\mathbb{Z}$ but rather $[0], [1]$ and $[2]$.
Finally, we note that this is a special quotient set, known as the integers modulo $p$. We write this as $\mathbb{Z}/p\mathbb{Z}$. Our example with $p=3$ is $\mathbb{Z}/3\mathbb{Z}$.
Rigour
We now formalise the ideas seen in the example.
The cartesian product of two sets \( A \) and \( B \) is the set of all ordered pairs,
\[ A \times B = \{(a, b) \mid a \in A, b \in B\}. \]A binary relation \( R \) on a set \( X \) is a subset of \( X \times X \). If \( (a, b) \in R \), we write \( aRb \) and say that \( a \) is related to \( b \) under \( R \).
A binary relation \( \sim \) on \( X \) is called an equivalence relation if it satisfies the following properties:
- $x \sim x$ for all $x \in X$ (reflexivity)
- \( x \sim y \Rightarrow y \sim x \) for all $x, y \in X$ (symmetry)
- \((x \sim y \text{ and } y \sim z) \Rightarrow x \sim z \) for all $x, y, z \in X$ (transitivity)
In other words, an equivalence relationship tells you when two elements can be considered to be the ‘same’.
Given an equivalence relation \( \sim \) on \( X \) and an element \( x \in X \), the equivalence class of \( x \) is the set of all elements in \( X \) that are related to \( x \):
\[ [x] = \{ y \in X \mid x \sim y \}. \]Intuitively, an equivalence class is the set of things can be considered the same under $\sim$.
The set of all equivalence classes under \( \sim \) is called the quotient set of \( X \) by \( \sim \), and is denoted by:
\[ X /{\sim} = \{ [x] \mid x \in X \}. \]It is crucially important to understand that the elements of a quotient set are not the elements of our original set. Rather, each element of our quotient set is an equivalence class.
More intuition
As seen in our first example, and made rigorous in the previous section, a quotient set is a way of creating a new set from an old set, by a special rule known as an equivalence relation.
The fundamental thing to understand about quotient sets is that they two things. Firstly, they identify elements with each other under an equivalence relation. This allows us to discard certain information that is redundant, and streamline mathematical thinking. Secondly, they offer a way to work on new spaces (which will see more of our examples).
In our first example, we begin with being able to tell apart the integers $4$ and $7$. However, in $\mathbb{Z}/3\mathbb{Z}$, both of these belong to the equivalence class $[1]$. Thus, we have lost the exact integer values. The benefit to this is we can now work in a space where only remainders modulo $3$ matters.
More Examples
We now look at a number of examples of quotients in action.
Construction of the Rationals
Suppose we have constructed the integers, $\mathbb{Z}$. Clearly, this set is closed under addition, subtraction and multiplication. However, once we get to division, it becomes necessary to construct the rationals, $\mathbb{Q}$. The rigorous way of doing so requires the notion of a quotient.
The key observation is that every rational number, $m/n$ with $m, n \in \mathbb{Z}$, can be considered as an ordered pair $(m, n)$, which lives in \(\mathbb{Z} \times \mathbb{Z} \setminus \{0\}\). However, as we know, some of these ordered pairs correspond to the same rational number. For example, $(1, 2)$ and $(2, 4)$ both correspond to the rational $1/2$. Therefore, to construct the rationals, we must have some way of identifying such pairs, and consider them to be the same.
\[ (a, b) \sim (c, d) \iff ad = bc \]This of course, is motivated by considering the pairs equal if \(a/b=c/d\), but note we have to write it as we have done above as \(a/b\) is not yet defined.
Clearly, the equivalence classes under this equivalence relation correspond bijectively with the rationals. For example:
\[ \frac{1}{2} = [(1, 2)] = \{\cdots (-1, -2), (1, 2), (2, 4), \cdots \} \]To conclude, we have seen that:
\[ \mathbb{Q} = (\mathbb{Z} \times \mathbb{Z} \setminus \{0\}) /{\sim}. \]Quotient Groups
We now turn our attention to Group Theory, where quotient groups play a large role. This section assumes a familiarity with the basics of Group Theory. We will gloss over some mathematical details in favour of intuition.
Suppose we have a group $G$ and a normal subgroup $N$. We can “collapse” $N$ down to a single point by grouping together elements that differ from $N$ by the same amount. To make this mathematically precise, we define an equivalence relation $\sim$ on $G$ by
\[ g \sim h \iff gh^{-1} \in N \]This equivalence relation underlies the construction of the quotient group.
Formally, the quotient group is the set of all cosets of \(N\) in \(G\)
\[ G/N = \{gN \mid g \in G\} \]We note that \(G/{\sim} = G/N\) is an equality on sets. Furthermore, any two cosets are either identical or disjoint. Thus, we can see that \(G\) is partitioned into cosets. We define the group operation on \(G/N\) as
\[ (gN)(hN)\overset{\text{def}}{=}ghN \]The normality of $N$ ensures that this operation is well defined.
To understand this structure more, we define the canonical map $\pi: G \to G/N$ given by $\pi(g) = gN$. We can see that this map “forgets” the internal structure of $N$ by sending all of its elements to the identity coset $\ker(\pi) = eN =N$. This illustrates the idea that $G/N$ measures elements of $G$ up to the ambiguity given by $N$.
Finally, we note that our original example is also a quotient group. We can write it now as $\mathbb{Z}/p\mathbb{Z}$. We see that here, we are identifying elements of $\mathbb{Z}$ up to how much they differ from $p\mathbb{Z}$, which is exactly in agreement with our intuition of a quotient group.
Quotient Rings
In a similar fashion to quotient groups, we can take a quotient of a ring $R$ by an ideal $I$. We write this as $R/I$.
Consider the polynomial ring $\mathbb{R}[x]$ and its ideal $I = (x^2+1)$. In the quotient ring $\mathbb{R}[x]/I$, we treat $x^2+1$ to be effectively zero, by considering elements only up to multiples of $x^2+1$. This relation, $x^2+1=0$ forces $x^2=-1$, which enables us to simplify every polynomial in $\mathbb{R}[x]$. For example
- $x^3 = x \cdot x^2 \ \rightarrow \ x \cdot (-1) = -x$
- $x^4 = (x^2)^2 \ \rightarrow \ (-1)^2 = 1$
Therefore, every equivalence class in $\mathbb{R}[x]$ can be represented uniquely by a linear polynomial $a + bx$. Furthermore, if we interpret $x$ as $i$, noticing both satisfy the same relation $x^2=i^2=-1$, we see that every equivalence class is of the form $a + bi$.
Thus, we have seen that
\[ \mathbb{R}[x]/(x^2+1) \cong \mathbb{C}. \]It can be seen in a similar fashion that the dual numbers, \(\mathbb{D}=\{a + b\epsilon \mid a, b \in \mathbb{R}, \epsilon^2 = 0\}\), which are commonly used in automatic differentiation, are isomorphic to the ring $\mathbb{R}[x]/x^2$.
The Quotient Topology
Our final example is the most visual, and highlights how we can consider a quotient space to be an identification of points under some relation. We start with the definition.
Let $X$ be a topological space, and $\sim$ be an equivalence relation on $X$. Let the canonical surjection $q: X \to Y = X/\sim$ be given by $q(x) = [x]$. The quotient space is the set $Y$ with the quotient topology. That is, the open sets are those subsets $U \subseteq Y$ such that $q^{-1}(U)$ is open.
This concept allows us to build new spaces from old in a visual way. We give an example of this by constructing the torus, $\mathbb{S}^1\times\mathbb{S}^1$, from $I \times I$, where $I$ is the closed unit interval. To see this, note that $I \times I$ looks like a square, and thus, to construct a torus all we have to do is glue the two horizontal edges first, and then glue together the two vertical edges. This ‘gluing’ is an identification of points, and is precisely the equivalence relation that we need to form our quotient topology.
Outlook
Quotients can be generalised further by short exact sequences. These can be found in many areas of math, such as Algebraic Geometry, Homological Algebra and Algebraic Topology.
A short exact sequence is a sequence of morphisms between objects in an abelian category. For the sake of simplicity, we consider homomorphisms between abelian groups:
\[ 0 \to A \xrightarrow{f} B \xrightarrow{g} C \to 0 \]such that
- The map $f:A \to B$ is injective, implying $\ker(f) = 0$.
- The map $g: B \to C$ is surjective, implying \(\operatorname{im}(g) = C\)
- \(\operatorname{im}(f)=\ker(g)\)
The first isomorphism theorem implies that $C \cong B / A$. However, this exact sequence also tell us the way in which $A$ is embedded into $B$ (via $f$) and how $B$ maps onto $C$ (via $g$). For example, in the case when $B = A \oplus C$, we say the sequence splits. Loosely speaking, this means that $f$ and $g$ act like the inclusion and projection maps into the and out of the direct sum.