Quotient depth: descending maps and checking representatives

1. From a set to its classes

Let $\sim$ be an equivalence relation on $X$. Its quotient $X/{\sim}$ is the set of classes $[x]=\{y:y\sim x\}$. The quotient map $\pi:X\to X/{\sim}$ sends $x$ to $[x]$. It is surjective because every class has a representative. The identity $[x]=[y]$ is equivalent to $x\sim y$, and must not be confused with $x=y$.

Suppose $h:X\to Y$ is already a function. We would like to define $\overline h([x])=h(x)$. This is a proposal, not yet a valid definition.

Theorem: criterion for descending a map

There exists a function $\overline h:X/{\sim}\to Y$ with $h=\overline h\circ\pi$ if and only if $x\sim y$ implies $h(x)=h(y)$. If it exists, it is unique.

Proof. If the factorization exists and $x\sim y$, then $\pi(x)=\pi(y)$, so applying $\overline h$ gives $h(x)=h(y)$. Conversely, assume this constancy condition. For each class choose any representative $x$ and assign output $h(x)$. The condition says every choice in that class produces exactly the same output, so the rule is unambiguous and defines a function. Its factorization identity follows by substitution. Finally any factoring function must send $[x]=\pi(x)$ to $h(x)$, so there is no second possible function. This construction does not require a simultaneous arbitrary choice of a distinguished representative in each class: its value is uniquely characterized by all representatives.

The test is not merely that equivalent inputs have “similar” outputs. Equality in $Y$ is required. If $Y$ is itself a quotient, equality there means equivalence of its representatives, giving a different but equally precise condition.

2. Congruence maps with different moduli

For positive integers $m,n\ge2$, consider the proposed map

$$T:\mathbb Z/m\mathbb Z\to\mathbb Z/n\mathbb Z, \qquad T([a]_m)=[a]_n.$$

It is well-defined exactly when $n\mid m$.

Proof of sufficiency. If $[a]_m=[b]_m$, then $a-b=mk$. If $n\mid m$, the difference is divisible by $n$, so $[a]_n=[b]_n$. The descending-map criterion applies.

Proof of necessity. The representatives $0,m$ name the same class modulo $m$. Their proposed outputs agree only if $[0]_n=[m]_n$, meaning $n\mid m$. Thus one carefully chosen representative pair already detects every forbidden modulus relation.

Worked comparison. Reduction from classes modulo six to classes modulo three is valid: adding six never changes a residue modulo three. Reduction from modulo three to modulo two is invalid: $[0]_3=[3]_3$ but $[0]_2\ne[3]_2$. The fact that both quotients are finite does not justify the rule.

For an integer $k$, the more general rule $[a]_m\mapsto[ka]_n$ is well-defined exactly when $n\mid km$. The same proof replaces the difference $m$ by $km$. It is possible for scaling to make an otherwise invalid reduction valid.

3. Two representatives can change independently

For an operation on classes, both arguments require checking. If $B:X\times X\to Y$ is a rule, the proposed output $\overline B([x],[y])=B(x,y)$ is valid precisely when

$$x\sim x',\quad y\sim y'\quad\Longrightarrow\quad B(x,y)=B(x',y').$$

One can prove this by changing the first representative while holding the second fixed, then changing the second; or directly change both. Checking only a simultaneous special replacement, such as adding the same number to both representatives, is weaker and may miss a dependence.

A warning example. On integers modulo two, propose $B([a],[b])=a-b$ as an actual integer. Replacing $(a,b)$ by $(a+2,b+2)$ preserves the output, so a simultaneous test passes. But replacing only $a$ by $a+2$ changes the integer output by two. Therefore the integer-valued rule is not well-defined. If the target is instead classes modulo two, $[a-b]_2$ is well-defined: every independent change changes the difference by a multiple of two. The declared target set changes the equality being tested.

4. Rational arithmetic: why the numerator formula matters

Operations on $Q$

The familiar fraction rules suggest formulas on pairs. To define arithmetic on $Q$, those formulas must give the same output class for every choice of input representatives.

Definition

Addition, multiplication, and inverses on $Q$

For classes in $Q$, define

$$[(a, b)] + [(c, d)] := [(ad + bc, bd)],$$

and

$$[(a, b)] \cdot [(c, d)] := [(ac, bd)].$$

The additive inverse is defined by

$$-[(a, b)] := [(-a, b)].$$

If $a \neq 0$, the multiplicative inverse is defined by

$$[(a, b)]^{-1} := [(b, a)].$$

Write $0_Q=[(0,1)]$ and $1_Q=[(1,1)]$. The cross-product relation gives $[(a,b)]=0_Q$ exactly when $a=0$. Thus being zero is a property of the class, and the condition $a\ne0$ for inversion does not depend on its representative.

Each of these formulas is written on representatives, so each of them must be checked for well-definedness. Otherwise the formula might depend on the chosen pair rather than on the rational number itself.

Before checking representatives, check the output domain. Since $b,d\ne0$ and $Z$ has no zero divisors, the denominator $bd$ is nonzero for both sums and products. For inversion, the extra condition $a\ne0$ makes $(b,a)$ a valid pair. These are two different obligations: a formula must produce a valid object, and that object must be independent of representatives.

Proof: addition on $Q$ is well-defined

Assume

$$ab' = ba' \qquad\text{and}\qquad cd' = dc'.$$

We must show

$$(ad + bc, bd) \sim_Q (a'd' + b'c', b'd').$$

By definition of $\sim_Q$, it is enough to prove

$$(ad + bc)b'd' = (a'd' + b'c')bd.$$

Expanding the left-hand side gives

$$adb'd' + bcb'd'.$$

Using $ab' = ba'$, the first term becomes

$$adb'd' = ba'dd'.$$

Using $cd' = dc'$, the second term becomes

$$bcb'd' = bdb'c'.$$

Hence

$$(ad + bc)b'd' = ba'dd' + bdb'c' = (a'd' + b'c')bd.$$

Therefore

$$(ad + bc, bd) \sim_Q (a'd' + b'c', b'd'),$$

so the sum is well-defined.

Why multiplication on $Q$ is well-defined

Assume

$$(a,b)\sim_Q(a',b') \qquad\text{and}\qquad (c,d)\sim_Q(c',d').$$

The hypotheses mean

$$ab'=ba' \qquad\text{and}\qquad cd'=dc'.$$

To prove that the product is independent of representatives, use the defining test for $\sim_Q$:

$$(ac)(b'd')=(bd)(a'c').$$

The left side factors as

$$(ac)(b'd')=(ab')(cd')=(ba')(dc')=(bd)(a'c').$$

Hence $(ac,bd)\sim_Q(a'c',b'd')$, and the product formula descends from pairs to rational classes.

Theorem

Every nonzero rational has one multiplicative inverse

If $q\in Q$ and $q\ne0$, there is a unique $q^{-1}\in Q$ with

$$q\cdot q^{-1}=[(1,1)].$$

Proof: existence and uniqueness of the rational inverse

Write $q=[(a,b)]$ with $b\ne0$. The condition $q\ne0$ means $a\ne0$, so $[(b,a)]$ is a valid rational class. Then

$$[(a,b)]\cdot[(b,a)]=[(ab,ba)]=[(1,1)],$$

because $ab=ba$ in $Z$.

For uniqueness, suppose $[(u,v)]$ is another inverse. The equation

$$[(u,v)]\cdot[(a,b)]=[(1,1)]$$

means $(ua,vb)\sim_Q(1,1)$, hence $ua=vb$. This is exactly the condition $(u,v)\sim_Q(b,a)$. Thus every inverse represents the same class $[(b,a)]$.

Worked example

Invert a negative rational without changing its class

Let $q=[(-3,5)]$. Since the numerator is nonzero,

$$q^{-1}=[(5,-3)].$$

Their product is

$$[(-3,5)]\cdot[(5,-3)]=[(-15,-15)]=[(1,1)].$$

The denominator in the inverse is nonzero, and the two coordinates may be multiplied by $-1$ without changing the class. The calculation therefore works even when the chosen denominator is negative.

Common mistake

Nonzero numerator is needed before inverting

The formula $[(a,b)]^{-1}=[(b,a)]$ is valid only when $a\ne0$. If $a=0$, the second coordinate of $[(b,a)]$ would be zero, which is excluded from $Y$, and zero has no multiplicative inverse.

Represent a rational by an integer pair $(a,b)$ with $b\ne0$, where $(a,b)\sim(a',b')$ means $ab'=a'b$. Write the class as $[a,b]$. The proposed “add one to the numerator” rule

$$N([a,b])=[a+1,b]$$

fails. For $[1,2]=[2,4]$, it gives $[2,2]$ and $[3,4]$, which are unequal since $2\cdot4\ne3\cdot2$. A familiar-looking formula has changed the represented value in a way dependent on its written denominator.

The actual add-one rule is $A([a,b])=[a+b,b]$. If $ab'=a'b$, then $(a+b)b'=ab'+bb'=a'b+bb'=(a'+b')b$, proving that the new pairs are equivalent. This is an example of deriving the correct formula from the target value rather than editing one coordinate arbitrarily.

5. Geometry: a direction is a class, not a preferred vector

On $X=\mathbb R^2\setminus\{0\}$, define $u\sim v$ if $u=\lambda v$ for a nonzero real $\lambda$. This is reflexive with $\lambda=1$, symmetric using $1/\lambda$, and transitive using a nonzero product of scalars. Its classes are unoriented lines through the origin with their zero point omitted; positive and negative multiples name the same direction.

The length $\|u\|$ is not a well-defined function of such a class: $u$ and $2u$ have different lengths. The proposed unit vector $u/\|u\|$ also fails because $u$ and $-u$ give opposite unit vectors. But for classes containing a representative $(x,y)$ with $x\ne0$, the slope $y/x$ is well-defined: nonzero scaling cancels. The vertical direction has no such slope, so this formula defines a function only on a restricted quotient domain.

If we instead identify only positive multiples, then $u/\|u\|$ is well-defined, and classes become oriented rays. Changing the equivalence relation changes which information is intentionally forgotten. A quotient is therefore a mathematical modeling decision, not only a notation shortcut.

Graded construction and diagnosis

1. Criterion. On $\mathbb Z/2\mathbb Z$, determine whether $[a]\mapsto(-1)^a$ defines a real-valued function, and prove your conclusion by the descending-map criterion.
2. Classification. Prove the general scaling criterion $[a]_m\mapsto[ka]_n$ is valid exactly when $n\mid km$. Apply it to $m=3,n=2,k=2$.
3. Operation audit. On classes modulo three, compare the proposals $[a],[b]\mapsto ab$ in $\mathbb Z$ and $[a],[b]\mapsto[ab]_3$ in the quotient. Provide both a counterexample and the complete valid representative check.
4. Fractions. Is $[a,b]\mapsto[a,b+1]$ a rational-class function on all fractions? Identify separately an output-membership failure and a representative-independence failure.
5. Geometry. For unoriented direction classes, prove $u\mapsto uu^{\mathsf T}/\|u\|^2$ is well-defined as a real $2\times2$ matrix-valued function, while the normalized vector is not. Interpret the matrix's action on a vector.
6. Uniqueness. Prove that whenever a function on classes satisfies $\overline h([x])=h(x)$ for every representative, it must be the unique factoring map in the theorem, even if no preferred representatives are chosen.

Full solutions

1. If $a\equiv b\pmod2$, then $a-b=2r$ and $(-1)^a=(-1)^b(-1)^{2r}=(-1)^b$. Thus the rule is constant on each class and descends to a real-valued function, taking values one on the even class and minus one on the odd class. Integer negative exponents of $-1$ are defined because $-1\ne0$.
2. If $n\mid km$, equivalent inputs differ by $mr$, so their scaled representatives differ by $kmr$, divisible by $n$. Conversely the equal input classes $[0]_m,[m]_m$ require outputs $[0]_n,[km]_n$ to agree, forcing the divisibility. For $m=3,n=2,k=2$, $2\mid6$, so the map is valid; indeed every output is the zero class modulo two.
3. As an integer-valued proposal it fails: $[0]_3=[3]_3$, but with the second input $[1]_3$, the products using these representatives are zero and three. For the class-valued rule, if $a-a'=3r$ and $b-b'=3s$, then $ab-a'b'=(a-a')b+a'(b-b')=3(rb+a's)$, so the outputs are equivalent modulo three. Both representatives have been allowed to change independently.
4. At $(a,b)=(1,-1)$, the proposed output denominator is zero, so it is not even a representative of a rational number. Restricting to positive denominators avoids that failure but not the second: $[1,2]=[2,4]$ would yield $[1,3]$ and $[2,5]$, unequal because $1\cdot5\ne2\cdot3$. Membership and independence are distinct requirements.
5. If $u=\lambda v$ with $\lambda\ne0$, both the numerator matrix and squared denominator gain the same factor $\lambda^2$, so the quotient matrix is unchanged. In contrast $(-u)/\|-u\|=-u/\|u\|$ differs from the normalized vector of $u$. Acting on $w$, the matrix gives $u(u\cdot w)/\|u\|^2$, exactly the orthogonal projection onto the unoriented line. Projection forgets the sign of its chosen direction, as the quotient requires.
6. Every class equals $\pi(x)$ for at least one $x$. Any factoring map must assign $\pi(x)$ the value $h(x)$, so two such maps agree on every class. The constancy condition ensures every possible representative gives that same forced value. Existence and uniqueness therefore concern outputs determined by classes, not a global choice of names for those classes.

Summary and terms

A map descends when it is constant on each equivalence class. A quotient map sends representatives to their classes. Well-definedness includes valid output membership and independence from every allowed representative change. A target quotient changes equality of outputs from literal representative equality to the target equivalence. Class constructions preserve exactly the information their equivalence relation allows; each proposed formula must respect that information loss.