6.10 Quotients

6.10 Quotients

A particularly important sort of colimit of sets is the

quotient

by a

relation

.
That is, let

A

be a set and

R

:

A

×

A

→

𝖯𝗋𝗈𝗉

a family of mere propositions (a

mere relation

).
Its quotient should be the set-coequalizer of the two

projections

∑

(

a

,

b

:

A

)

R

⁢

(

a

,

b

)

⁢

\rightrightarrows

⁢

A

.

We can also describe this directly, as the

higher inductive type

A

/

R

generated by

•

A

function

q

:

A

→

A

/

R

;

•

For each

a

,

b

:

A

such that

R

⁢

(

a

,

b

)

, an equality

q

⁢

(

a

)

=

q

⁢

(

b

)

; and

•

The

0

-

truncation

constructor: for all

x

,

y

:

A

/

R

and

r

,

s

:

x

=

y

, we have

r

=

s

.

We may sometimes refer to

A

/

R

as the

set-quotient

of

A

by

R

, to emphasize that it produces a set by definition.
(There are more general notions of “quotient” in homotopy theory, but they are mostly beyond the scope of this book.
However, in §9.9 (

http://planetmath.org/99therezkcompletion

) we will consider the “quotient” of a type by a 1-groupoid, which is the next level up from set-quotients.)

Remark 6.10.1

.

It is not actually necessary for the definition of set-quotients, and most of their properties, that

A

be a set.
However, this is generally the case of most interest.

Lemma 6.10.2

.

The function

q

:

A

→

A

/

R

is

surjective

.

Proof.

We must show that for any

x

:

A

/

R

there merely exists an

a

:

A

with

q

⁢

(

a

)

=

x

.
We use the

induction

principle of

A

/

R

.
The first case is

trivial

: if

x

is

q

⁢

(

a

)

, then of course there merely exists an

a

such that

q

⁢

(

a

)

=

q

⁢

(

a

)

.
And since the goal is a mere proposition, it automatically respects all path constructors, so we are done.
∎

Lemma 6.10.3

.

For any set

B

, precomposing with

q

yields an equivalence

(

A

/

R

→

B

)

≃

(

∑

(

f

:

A

→

B

)

∏

(

a

,

b

:

A

)

R

(

a

,

b

)

→

(

f

(

a

)

=

f

(

b

)

)

)

.

Proof.

The

quasi-inverse

of

–

∘

q

, going from right to left, is just the

recursion

principle for

A

/

R

.
That is, given

f

:

A

→

B

such that

∏

(

a

,

b

:

A

)

R

(

a

,

b

)

→

(

f

(

a

)

=

f

(

b

)

)

,

we define

f

¯

:

A

/

R

→

B

by

f

¯

(

q

(

a

)

)

:

≡

f

(

a

)

.
This defining equation says precisely that

(

f

↦

f

¯

)

is a

right inverse

to

(

–

∘

q

)

.

For it to also be a

left inverse

, we must show that for any

g

:

A

/

R

→

B

and

x

:

A

/

R

we have

g

⁢

(

x

)

=

g

∘

q

¯

.
However, by Lemma 6.10.2 (

http://planetmath.org/610quotients#Thmprelem1

) there merely exists

a

such that

q

⁢

(

a

)

=

x

.
Since our desired equality is a mere proposition, we may assume there purely exists such an

a

, in which case

g

⁢

(

x

)

=

g

⁢

(

q

⁢

(

a

)

)

=

g

∘

q

¯

⁢

(

q

⁢

(

a

)

)

=

g

∘

q

¯

⁢

(

x

)

.
∎

Of course, classically the usual case to consider is when

R

is an

equivalence relation

, i.e. we have

•

reflexivity

:

∏

(

a

:

A

)

R

⁢

(

a

,

a

)

,

•

symmetry

:

∏

(

a

,

b

:

A

)

R

⁢

(

a

,

b

)

→

R

⁢

(

b

,

a

)

, and

•

transitivity

:

∏

(

a

,

b

,

c

:

C

)

R

⁢

(

a

,

b

)

×

R

⁢

(

b

,

c

)

→

R

⁢

(

a

,

c

)

.

In this case, the set-quotient

A

/

R

has additional good properties, as we will see in §10.1 (

http://planetmath.org/101thecategoryofsets

): for

instance

, we have

R

(

a

,

b

)

≃

(

q

(

a

)

=

A

/

R

q

(

b

)

)

.

We often write an equivalence relation

R

⁢

(

a

,

b

)

infix as

a

∼

b

.

The quotient by an equivalence relation can also be constructed in other ways.
The set theoretic approach is to consider the set of

equivalence classes

, as a subset of the

power set

of

A

.
We can mimic this “

impredicative

” construction in

type theory

as well.

Definition 6.10.4

.

A

predicate

P

:

A

→

Prop

is an

equivalence class

of a relation

R

:

A

×

A

→

Prop

if there merely exists an

a

:

A

such that for all

b

:

A

we have

R

⁢

(

a

,

b

)

≃

P

⁢

(

b

)

.

As

R

and

P

are mere propositions, the equivalence

R

⁢

(

a

,

b

)

≃

P

⁢

(

b

)

is the same thing as

implications

R

⁢

(

a

,

b

)

→

P

⁢

(

b

)

and

P

⁢

(

b

)

→

R

⁢

(

a

,

b

)

.
And of course, for any

a

:

A

we have the canonical equivalence class

P

a

(

b

)

:

≡

R

(

a

,

b

)

.

Definition 6.10.5

.

We define

A

⫽

R

:

≡

\setof

P

:

A

→

𝖯𝗋𝗈𝗉

|

P

is an equivalence class of

R

.

The function

q

′

:

A

→

A

⫽

R

is defined by

q

′

(

a

)

:

≡

P

a

.

Theorem 6.10.6

.

For any equivalence relation

R

on

A

, the two set-quotients

A

/

R

and

A

⫽

R

are

equivalent

.

Proof.

First, note that if

R

⁢

(

a

,

b

)

, then since

R

is an equivalence relation we have

R

⁢

(

a

,

c

)

⇔

R

⁢

(

b

,

c

)

for any

c

:

A

.
Thus,

R

⁢

(

a

,

c

)

=

R

⁢

(

b

,

c

)

by univalence, hence

P

a

=

P

b

by function extensionality, i.e.

q

′

⁢

(

a

)

=

q

′

⁢

(

b

)

.
Therefore, by Lemma 6.10.3 (

http://planetmath.org/610quotients#Thmprelem2

) we have an induced map

f

:

A

/

R

→

A

⫽

R

such that

f

∘

q

=

q

′

.

We show that

f

is

injective

and surjective, hence an equivalence.
Surjectivity follows immediately from the fact that

q

′

is surjective, which in turn is true essentially by definition of

A

⫽

R

.
For injectivity, if

f

⁢

(

x

)

=

f

⁢

(

y

)

, then to show the mere proposition

x

=

y

, by surjectivity of

q

we may assume

x

=

q

⁢

(

a

)

and

y

=

q

⁢

(

b

)

for some

a

,

b

:

A

.
Then

R

⁢

(

a

,

c

)

=

f

⁢

(

q

⁢

(

a

)

)

⁢

(

c

)

=

f

⁢

(

q

⁢

(

b

)

)

⁢

(

c

)

=

R

⁢

(

b

,

c

)

for any

c

:

A

, and in particular

R

⁢

(

a

,

b

)

=

R

⁢

(

b

,

b

)

.
But

R

⁢

(

b

,

b

)

is inhabited, since

R

is an equivalence relation, hence so is

R

⁢

(

a

,

b

)

.
Thus

q

⁢

(

a

)

=

q

⁢

(

b

)

and so

x

=

y

.
∎

In §10.1.3 (

http://planetmath.org/1013quotients

) we will give an alternative proof of this

theorem

.
Note that unlike

A

/

R

, the construction

A

⫽

R

raises

universe

level: if

A

:

𝒰

i

and

R

:

A

→

A

→

𝖯𝗋𝗈𝗉

𝒰

i

, then in the definition of

A

⫽

R

we must also use

𝖯𝗋𝗈𝗉

𝒰

i

to include all the equivalence classes, so that

A

⫽

R

:

𝒰

i

+

1

.
Of course, we can avoid this if we assume the propositional resizing axiom from §3.5 (

http://planetmath.org/35subsetsandpropositionalresizing

).

Remark 6.10.7

.

The previous two constructions provide quotients in generality, but in particular cases there may be easier constructions.
For instance, we may define the integers

Z

as a set-quotient

ℤ

:

≡

(

ℕ

×

ℕ

)

/

∼

where

∼

is the equivalence relation defined by

(

a

,

b

)

∼

(

c

,

d

)

:

≡

(

a

+

d

=

b

+

c

)

.

In other words, a pair

(

a

,

b

)

represents the integer

a

-

b

.
In this case, however, there are

canonical representatives

of the equivalence classes: those of the form

(

n

,

0

)

or

(

0

,

n

)

.

The following lemma says that when this sort of thing happens, we don’t need either general construction of quotients.
(A function

r

:

A

→

A

is called

idempotent

if

r

∘

r

=

r

.)

Lemma 6.10.8

.

Suppose

∼

is an equivalence relation on a set

A

, and there exists an idempotent

r

:

A

→

A

such that, for all

x

,

y

∈

A

,

(

r

(

x

)

=

r

(

y

)

)

≃

(

x

∼

y

)

. Then the
type

(

A

/

∼

)

:

≡

∑

x

:

A

r

(

x

)

=

x

is the set-quotient of

A

by

∼

.
In other words, there is a map

q

:

A

→

(

A

/

∼

)

such that for every set

B

, the type

(

A

/

∼

)

→

B

is equivalent to

∑

(

g

:

A

→

B

)

∏

(

x

,

y

:

A

)

(

x

∼

y

)

→

(

g

(

x

)

=

g

(

y

)

)

(6.10.8)

with the map being induced by precomposition with

q

.

Proof.

Let

i

:

∏

(

x

:

A

)

r

⁢

(

r

⁢

(

x

)

)

=

r

⁢

(

x

)

witness idempotence of

r

.
The map

q

:

A

→

A

/

∼

is defined by

q

(

x

)

:

≡

(

r

(

x

)

,

i

(

x

)

)

. An equivalence

e

from

A

/

∼

→

B

to (

6.10.8

) is defined by

e

(

f

)

:

≡

(

f

∘

q

,

¯

)

,

where the underscore

¯

denotes the following proof: if

x

,

y

:

A

and

x

∼

y

then by

assumption

r

⁢

(

x

)

=

r

⁢

(

y

)

, hence

(

r

⁢

(

x

)

,

i

⁢

(

x

)

)

=

(

r

⁢

(

y

)

,

i

⁢

(

y

)

)

as

A

is a set, therefore

f

⁢

(

q

⁢

(

x

)

)

=

f

⁢

(

q

⁢

(

y

)

)

. To see that

e

is an equivalence, consider the map

e

′

in the opposite
direction,

e

′

⁢

(

g

,

p

)

⁢

(

x

,

q

)

≡

g

⁢

(

x

)

.

Given any

f

:

A

/

∼

→

B

,

e

′

⁢

(

e

⁢

(

f

)

)

⁢

(

x

,

p

)

≡

f

⁢

(

q

⁢

(

x

)

)

≡

f

⁢

(

r

⁢

(

x

)

,

i

⁢

(

x

)

)

=

f

⁢

(

x

,

p

)

where the last equality holds because

p

:

r

⁢

(

x

)

=

x

and so

(

x

,

p

)

=

(

r

⁢

(

x

)

,

i

⁢

(

x

)

)

because

A

is a set. Similarly we compute

e

⁢

(

e

′

⁢

(

g

,

p

)

)

≡

e

⁢

(

g

∘

𝗉𝗋

1

)

≡

(

f

∘

𝗉𝗋

1

∘

q

,

¯

)

.

Because

B

is a set we need not worry about the

¯

part, while for the first

component

we have

f

(

𝗉𝗋

1

(

q

(

x

)

)

)

:

≡

f

(

r

(

x

)

)

=

f

(

x

)

,

where the last

equation

holds because

r

⁢

(

x

)

∼

x

and

f

respects

∼

by
assumption.
∎

Corollary 6.10.10

.

Suppose

p

:

A

→

B

is a

retraction

between sets.
Then

B

is the quotient of

A

by the equivalence relation

∼

defined by

(

a

1

∼

a

2

)

:

≡

(

p

(

a

1

)

=

p

(

a

2

)

)

.

Proof.

Suppose

s

:

B

→

A

is a

section

of

p

.
Then

s

∘

p

:

A

→

A

is an idempotent which

satisfies

the condition of Lemma 6.10.8 (

http://planetmath.org/610quotients#Thmprelem3

) for this

∼

, and

s

induces an

isomorphism

from

B

to its set of

fixed points

.
∎

Remark 6.10.11

.

Lemma 6.10.8 (

http://planetmath.org/610quotients#Thmprelem3

) applies to

Z

with the idempotent

r

:

N

×

N

→

N

×

N

defined by

r

⁢

(

a

,

b

)

=

{

(

a

-

b

,

0

)

if

a

≥

b

,

(

0

,

b

-

a

)

otherwise.

(This is a valid definition even constructively, since the relation

≥

on

N

is

decidable

.)
Thus a non-negative integer is canonically represented as

(

k

,

0

)

and a non-positive one by

(

0

,

m

)

, for

k

,

m

:

N

.
This division into cases implies the following induction principle for integers, which will be useful in

http://planetmath.org/node/87582

Chapter 8.
(As usual, we identify

natural numbers

with the corresponding non-negative integers.)

Lemma 6.10.12

.

Suppose

P

:

Z

→

U

is a type family and that we have

•

d

0

:

P

⁢

(

0

)

,

•

d

+

:

∏

(

n

:

ℕ

)

P

⁢

(

n

)

→

P

⁢

(

𝗌𝗎𝖼𝖼

⁢

(

n

)

)

, and

•

d

-

:

∏

(

n

:

ℕ

)

P

⁢

(

-

n

)

→

P

⁢

(

-

𝗌𝗎𝖼𝖼

⁢

(

n

)

)

.

Then we have

f

:

∏

(

z

:

Z

)

P

⁢

(

z

)

such that

f

⁢

(

0

)

≡

d

0

and

f

⁢

(

succ

⁢

(

n

)

)

≡

d

+

⁢

(

f

⁢

(

n

)

)

, and

f

⁢

(

-

succ

⁢

(

n

)

)

≡

d

-

⁢

(

f

⁢

(

-

n

)

)

for all

n

:

N

.

Proof.

We identify

ℤ

with

∑

(

x

:

ℕ

×

ℕ

)

(

r

(

x

)

=

x

)

, where

r

is the above idempotent.
Now define

Q

:

≡

P

∘

r

:

ℕ

×

ℕ

→

𝒰

.
We can construct

g

:

∏

(

x

:

ℕ

×

ℕ

)

Q

⁢

(

x

)

by double induction on

n

:

g

⁢

(

0

,

0

)

:

≡

d

0

,

g

⁢

(

𝗌𝗎𝖼𝖼

⁢

(

n

)

,

0

)

:

≡

d

+

(

g

(

n

,

0

)

)

,

g

⁢

(

0

,

𝗌𝗎𝖼𝖼

⁢

(

m

)

)

:

≡

d

-

(

g

(

0

,

m

)

)

,

g

⁢

(

𝗌𝗎𝖼𝖼

⁢

(

n

)

,

𝗌𝗎𝖼𝖼

⁢

(

m

)

)

:

≡

g

(

n

,

m

)

.

Let

f

be the

restriction

of

g

to

ℤ

.
∎

For example, we can define the

n

-fold

concatenation

of a loop for any integer

n

.

Corollary 6.10.13

.

Let

A

be a type with

a

:

A

and

p

:

a

=

a

.
There is a function

∏

(

n

:

Z

)

(

a

=

a

)

, denoted

n

↦

p

n

, defined by

p

0

:

≡

𝗋𝖾𝖿𝗅

𝖻𝖺𝗌𝖾

p

n

+

1

:

≡

p

n

\centerdot

p

for

n

≥

0

p

n

-

1

:

≡

p

n

\centerdot

p

-

1

for

n

≤

0

.

We will discuss the integers further in §6.11 (

http://planetmath.org/611algebra

),§11.1 (

http://planetmath.org/111thefieldofrationalnumbers

).

Title

6.10 Quotients

\metatable

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