9.4 Equivalences

9.4 Equivalences

It is usual in

category theory

to define an

equivalence of categories

to be a

functor

F

:

A

→

B

such that there exists a functor

G

:

B

→

A

and

natural isomorphisms

F

⁢

G

≅

1

B

and

G

⁢

F

≅

1

A

.
Unlike the

property

of being an

adjunction

, however, this would not be a mere proposition without truncating it, for the same reasons that the type of

quasi-inverses

is ill-behaved (see

\autoref

sec:quasi-inverses).
And as in

\autoref

sec:hae, we can avoid this by using the usual notion of

adjoint

equivalence

.

Definition 9.4.1

.

A functor

F

:

A

→

B

is an

equivalence of (pre)

categories

if it is a left adjoint for which

η

and

ϵ

are

isomorphisms

.
We write

A

≃

B

for the type of equivalences of categories from

A

to

B

.

By

\autoref

ct:adjprop,

\autoref

ct:isoprop, if

A

is a category, then the type “

F

is an equivalence of

precategories

” is a mere proposition.

Lemma 9.4.2

.

If for

F

:

A

→

B

there exists

G

:

B

→

A

and isomorphisms

G

⁢

F

≅

1

A

and

F

⁢

G

≅

1

B

, then

F

is an equivalence of precategories.

Proof.

Just like the proof of

\autoref

thm:equiv-iso-adj for equivalences of types.
∎

Definition 9.4.3

.

We say a functor

F

:

A

→

B

is

faithful

if for all

a

,

b

:

A

, the

function

F

a

,

b

:

hom

A

⁡

(

a

,

b

)

→

hom

B

⁡

(

F

⁢

a

,

F

⁢

b

)

is

injective

, and

full

if for all

a

,

b

:

A

this function is

surjective

.
If it is both (hence each

F

a

,

b

is an equivalence) we say

F

is

fully faithful

.

Definition 9.4.4

.

We say a functor

F

:

A

→

B

is

split

essentially surjective

if for all

b

:

B

there exists an

a

:

A

such that

F

⁢

a

≅

b

.

Lemma 9.4.5

.

For any precategories

A

and

B

and functor

F

:

A

→

B

, the following types are

equivalent

.

1.

F

is an equivalence of precategories.

2.

F

is fully faithful and split essentially surjective.

Proof.

Suppose

F

is an equivalence of precategories, with

G

,

η

,

ϵ

specified.
Then we have the function

hom

B

⁡

(

F

⁢

a

,

F

⁢

b

)

→

hom

A

⁡

(

a

,

b

)

,

g

↦

η

b

-

1

∘

G

⁢

(

g

)

∘

η

a

.

For

f

:

hom

A

⁡

(

a

,

b

)

, we have

η

b

-

1

∘

G

⁢

(

F

⁢

(

f

)

)

∘

η

a

=

η

b

-

1

∘

η

b

∘

f

=

f

while for

g

:

hom

B

⁡

(

F

⁢

a

,

F

⁢

b

)

we have

F

⁢

(

η

b

-

1

∘

G

⁢

(

g

)

∘

η

a

)

=

F

⁢

(

η

b

-

1

)

∘

F

⁢

(

G

⁢

(

g

)

)

∘

F

⁢

(

η

a

)

=

ϵ

F

⁢

b

∘

F

⁢

(

G

⁢

(

g

)

)

∘

F

⁢

(

η

a

)

=

g

∘

ϵ

F

⁢

a

∘

F

⁢

(

η

a

)

=

g

using naturality of

ϵ

, and the triangle identities twice.
Thus,

F

a

,

b

is an equivalence, so

F

is fully faithful.
Finally, for any

b

:

B

, we have

G

⁢

b

:

A

and

ϵ

b

:

F

⁢

G

⁢

b

≅

b

.

On the other hand, suppose

F

is fully faithful and split essentially surjective.
Define

G

0

:

B

0

→

A

0

by sending

b

:

B

to the

a

:

A

given by the specified essential splitting, and write

ϵ

b

for the likewise specified isomorphism

F

⁢

G

⁢

b

≅

b

.

Now for any

g

:

hom

B

⁡

(

b

,

b

′

)

, define

G

⁢

(

g

)

:

hom

A

⁡

(

G

⁢

b

,

G

⁢

b

′

)

to be the unique

morphism

such that

F

⁢

(

G

⁢

(

g

)

)

=

(

ϵ

b

′

)

-

1

∘

g

∘

ϵ

b

(which exists since

F

is fully faithful).
Finally, for

a

:

A

define

η

a

:

hom

A

⁡

(

a

,

G

⁢

F

⁢

a

)

to be the unique morphism such that

F

⁢

η

a

=

ϵ

F

⁢

a

-

1

.
It is easy to verify that

G

is a functor and that

(

G

,

η

,

ϵ

)

exhibit

F

as an equivalence of precategories.

Now consider the composite

1

→

2

→

1

.
We clearly recover the same function

G

0

:

B

0

→

A

0

.
For the action of

G

on hom-sets, we must show that for

g

:

hom

B

⁡

(

b

,

b

′

)

,

G

⁢

(

g

)

is the (necessarily unique) morphism such that

F

⁢

(

G

⁢

(

g

)

)

=

(

ϵ

b

′

)

-

1

∘

g

∘

ϵ

b

.
But this

equation

holds by the assumed naturality of

ϵ

.
We also clearly recover

ϵ

, while

η

is uniquely characterized by

F

⁢

η

a

=

ϵ

F

⁢

a

-

1

(which is one of the triangle identities assumed to hold in the

structure

of an equivalence of precategories).
Thus, this composite is equal to the

identity

.

Finally, consider the other composite

2

→

1

→

2

.
Since being fully faithful is a mere proposition, it suffices to observe that we recover, for each

b

:

B

, the same

a

:

A

and isomorphism

F

⁢

a

≅

b

.
But this is clear, since we used this function and isomorphism to define

G

0

and

ϵ

in

1

, which in turn are precisely what we used to recover

2

again.
Thus, the composites in both directions are equal to identities, hence we have an equivalence

1

≃

2

.
∎

However, if

B

is not a category, then neither type in

\autoref

ct:ffeso may necessarily be a mere proposition.
This suggests considering as well the following notions.

Definition 9.4.6

.

A functor

F

:

A

→

B

is

essentially surjective

if for all

b

:

B

, there

merely

exists an

a

:

A

such that

F

⁢

a

≅

b

.
We say

F

is a

weak equivalence

if it is fully faithful and essentially surjective.

Being a weak equivalence is

always

a mere proposition.
For categories, however, there is no

difference

between equivalences and weak ones.

Lemma 9.4.7

.

If

F

:

A

→

B

is fully faithful and

A

is a category, then for any

b

:

B

the type

∑

(

a

:

A

)

(

F

a

≅

b

)

is a mere proposition.
Hence a functor between categories is an equivalence if and only if it is a weak equivalence.

Proof.

Suppose given

(

a

,

f

)

and

(

a

′

,

f

′

)

in

∑

(

a

:

A

)

(

F

a

≅

b

)

.
Then

f

′

-

1

∘

f

is an isomorphism

F

⁢

a

≅

F

⁢

a

′

.
Since

F

is fully faithful, we have

g

:

a

≅

a

′

with

F

⁢

g

=

f

′

-

1

∘

f

.
And since

A

is a category, we have

p

:

a

=

a

′

with

𝗂𝖽𝗍𝗈𝗂𝗌𝗈

⁢

(

p

)

=

g

.
Now

F

⁢

g

=

f

′

-

1

∘

f

implies

(

(

F

0

)

(

p

)

)

*

(

f

)

=

f

′

, hence (by the

characterization

of equalities in dependent pair types)

(

a

,

f

)

=

(

a

′

,

f

′

)

.

Thus, for fully

faithful functors

whose domain is a category, essential surjectivity is equivalent to split essential surjectivity, and so being a weak equivalence is equivalent to being an equivalence.
∎

This is an important advantage of our category theory over set-based approaches.
With a purely set-based

definition

of category, the statement “every fully faithful and essentially surjective functor is an equivalence of categories” is equivalent to the

axiom of choice

𝖠𝖢

.
Here we have it for free, as a category-theoretic version of the principle of unique choice (

\autoref

sec:unique-choice).
(In fact, this property characterizes categories among precategories; see

\autoref

sec:rezk.)

On the other hand, the following characterization of equivalences of categories is perhaps even more useful.

Definition 9.4.8

.

A functor

F

:

A

→

B

is an

isomorphism of (pre)categories

if

F

is fully faithful and

F

0

:

A

0

→

B

0

is an equivalence of types.

This definition is an exception to our general rule (see

\autoref

sec:basics-equivalences) of only using the word “isomorphism” for sets and set-like objects.
However, it does carry an appropriate

connotation

here, because for general precategories, isomorphism is stronger than equivalence.

Note that being an isomorphism of precategories is always a mere property.
Let

A

≅

B

denote the type of isomorphisms of (pre)categories from

A

to

B

.

Lemma 9.4.9

.

For precategories

A

and

B

and

F

:

A

→

B

, the following are equivalent.

1.

F

is an isomorphism of precategories.

2.

There exist

G

:

B

→

A

and

η

:

1

A

=

G

⁢

F

and

ϵ

:

F

⁢

G

=

1

B

such that

𝖺𝗉

(

λ

H

.

F

H

)

⁢

(

η

)

=

𝖺𝗉

(

λ

K

.

K

F

)

⁢

(

ϵ

-

1

)

.

(9.4.9)

3.

There merely exist

G

:

B

→

A

and

η

:

1

A

=

G

⁢

F

and

ϵ

:

F

⁢

G

=

1

B

.

Note that if

B

0

is not a 1-type, then (

9.4.9

) may not be a mere proposition.

Proof.

First note that since hom-sets are sets, equalities between equalities of functors are uniquely determined by their object-parts.
Thus, by function extensionality, (

9.4.9

) is equivalent to

(

F

0

)

(

η

0

)

a

=

(

ϵ

0

)

-

1

F

0

⁢

a

.

(9.4.11)

for all

a

:

A

0

.
Note that this is precisely the triangle identity for

G

0

,

η

0

, and

ϵ

0

to be a proof that

F

0

is a half adjoint equivalence of types.

Now suppose

1

.
Let

G

0

:

B

0

→

A

0

be the

inverse

of

F

0

, with

η

0

:

𝗂𝖽

A

0

=

G

0

⁢

F

0

and

ϵ

0

:

F

0

⁢

G

0

=

𝗂𝖽

B

0

satisfying the triangle identity, which is precisely (

9.4.11

).
Now define

G

b

,

b

′

:

hom

B

⁡

(

b

,

b

′

)

→

hom

A

⁡

(

G

0

⁢

b

,

G

0

⁢

b

′

)

by

G

b

,

b

′

(

g

)

:

≡

(

F

G

0

⁢

b

,

G

0

⁢

b

′

)

-

1

(

𝗂𝖽𝗍𝗈𝗂𝗌𝗈

(

(

ϵ

0

)

-

1

b

′

)

∘

g

∘

𝗂𝖽𝗍𝗈𝗂𝗌𝗈

(

(

ϵ

0

)

b

)

)

(using the

assumption

that

F

is fully faithful).
Since

𝗂𝖽𝗍𝗈𝗂𝗌𝗈

takes inverses to inverses and

concatenation

to composition, and

F

is a functor, it follows that

G

is a functor.

By definition, we have

(

G

⁢

F

)

0

≡

G

0

⁢

F

0

, which is equal to

𝗂𝖽

A

0

by

η

0

.
To obtain

1

A

=

G

⁢

F

, we need to show that when transported along

η

0

, the

identity function

of

hom

A

⁡

(

a

,

a

′

)

becomes equal to the composite

G

F

⁢

a

,

F

⁢

a

′

∘

F

a

,

a

′

.
In other words, for any

f

:

hom

A

⁡

(

a

,

a

′

)

we must have

𝗂𝖽𝗍𝗈𝗂𝗌𝗈

⁢

(

(

η

0

)

a

′

)

∘

f

∘

𝗂𝖽𝗍𝗈𝗂𝗌𝗈

⁢

(

(

η

0

)

-

1

a

)

=

(

F

G

⁢

F

⁢

a

,

G

⁢

F

⁢

a

′

)

-

1

⁢

(

𝗂𝖽𝗍𝗈𝗂𝗌𝗈

⁢

(

(

ϵ

0

)

-

1

F

⁢

a

′

)

∘

F

a

,

a

′

⁢

(

f

)

∘

𝗂𝖽𝗍𝗈𝗂𝗌𝗈

⁢

(

(

ϵ

0

)

F

⁢

a

)

)

.

But this is equivalent to

(

F

G

⁢

F

⁢

a

,

G

⁢

F

⁢

a

′

)

⁢

(

𝗂𝖽𝗍𝗈𝗂𝗌𝗈

⁢

(

(

η

0

)

a

′

)

∘

f

∘

𝗂𝖽𝗍𝗈𝗂𝗌𝗈

⁢

(

(

η

0

)

-

1

a

)

)

=

𝗂𝖽𝗍𝗈𝗂𝗌𝗈

⁢

(

(

ϵ

0

)

-

1

F

⁢

a

′

)

∘

F

a

,

a

′

⁢

(

f

)

∘

𝗂𝖽𝗍𝗈𝗂𝗌𝗈

⁢

(

(

ϵ

0

)

F

⁢

a

)

.

which follows from functoriality of

F

, the fact that

F

preserves

𝗂𝖽𝗍𝗈𝗂𝗌𝗈

, and (

9.4.11

).
Thus we have

η

:

1

A

=

G

⁢

F

.

On the other side, we have

(

F

⁢

G

)

0

≡

F

0

⁢

G

0

, which is equal to

𝗂𝖽

B

0

by

ϵ

0

.
To obtain

F

⁢

G

=

1

B

, we need to show that when transported along

ϵ

0

, the identity function of

hom

B

⁡

(

b

,

b

′

)

becomes equal to the composite

F

G

⁢

b

,

G

⁢

b

′

∘

G

b

,

b

′

.
That is, for any

g

:

hom

B

⁡

(

b

,

b

′

)

we must have

F

G

⁢

b

,

G

⁢

b

′

⁢

(

(

F

G

⁢

b

,

G

⁢

b

′

)

-

1

⁢

(

𝗂𝖽𝗍𝗈𝗂𝗌𝗈

⁢

(

(

ϵ

0

)

-

1

b

′

)

∘

g

∘

𝗂𝖽𝗍𝗈𝗂𝗌𝗈

⁢

(

(

ϵ

0

)

b

)

)

)

=

𝗂𝖽𝗍𝗈𝗂𝗌𝗈

⁢

(

(

ϵ

0

-

1

)

b

′

)

∘

g

∘

𝗂𝖽𝗍𝗈𝗂𝗌𝗈

⁢

(

(

ϵ

0

)

b

)

.

But this is just the fact that

(

F

G

⁢

b

,

G

⁢

b

′

)

-

1

is the inverse of

F

G

⁢

b

,

G

⁢

b

′

.
And we have remarked that (

9.4.9

) is equivalent to (

9.4.11

), so

2

holds.

Conversely

, suppose given

2

; then the object-parts of

G

,

η

, and

ϵ

together with (

9.4.11

) show that

F

0

is an equivalence of types.
And for

a

,

a

′

:

A

0

, we define

G

¯

a

,

a

′

:

hom

B

⁡

(

F

⁢

a

,

F

⁢

a

′

)

→

hom

A

⁡

(

a

,

a

′

)

by

G

¯

a

,

a

′

(

g

)

:

≡

𝗂𝖽𝗍𝗈𝗂𝗌𝗈

(

η

-

1

)

a

′

∘

G

(

g

)

∘

𝗂𝖽𝗍𝗈𝗂𝗌𝗈

(

η

)

a

.

(9.4.12)

By naturality of

𝗂𝖽𝗍𝗈𝗂𝗌𝗈

⁢

(

η

)

, for any

f

:

hom

A

⁡

(

a

,

a

′

)

we have

G

¯

a

,

a

′

⁢

(

F

a

,

a

′

⁢

(

f

)

)

=

𝗂𝖽𝗍𝗈𝗂𝗌𝗈

⁢

(

η

-

1

)

a

′

∘

G

⁢

(

F

⁢

(

f

)

)

∘

𝗂𝖽𝗍𝗈𝗂𝗌𝗈

⁢

(

η

)

a

=

𝗂𝖽𝗍𝗈𝗂𝗌𝗈

⁢

(

η

-

1

)

a

′

∘

𝗂𝖽𝗍𝗈𝗂𝗌𝗈

⁢

(

η

)

a

′

∘

f

=

f

.

On the other hand, for

g

:

hom

B

⁡

(

F

⁢

a

,

F

⁢

a

′

)

we have

F

a

,

a

′

⁢

(

G

¯

a

,

a

′

⁢

(

g

)

)

=

F

⁢

(

𝗂𝖽𝗍𝗈𝗂𝗌𝗈

⁢

(

η

-

1

)

a

′

)

∘

F

⁢

(

G

⁢

(

g

)

)

∘

F

⁢

(

𝗂𝖽𝗍𝗈𝗂𝗌𝗈

⁢

(

η

)

a

)

=

𝗂𝖽𝗍𝗈𝗂𝗌𝗈

⁢

(

ϵ

)

F

⁢

a

′

∘

F

⁢

(

G

⁢

(

g

)

)

∘

𝗂𝖽𝗍𝗈𝗂𝗌𝗈

⁢

(

ϵ

-

1

)

F

⁢

a

=

𝗂𝖽𝗍𝗈𝗂𝗌𝗈

⁢

(

ϵ

)

F

⁢

a

′

∘

𝗂𝖽𝗍𝗈𝗂𝗌𝗈

⁢

(

ϵ

-

1

)

F

⁢

a

′

∘

g

=

g

.

(There are lemmas needed here regarding the compatibility of

𝗂𝖽𝗍𝗈𝗂𝗌𝗈

and whiskering, which we leave it to the reader to state and prove.)
Thus,

F

a

,

a

′

is an equivalence, so

F

is fully faithful; i.e.

1

holds.

Now the composite

1

→

2

→

1

is equal to the identity since

1

is a mere proposition.
On the other side, tracing through the above constructions we see that the composite

2

→

1

→

2

essentially preserves the object-parts

G

0

,

η

0

,

ϵ

0

, and the object-part of (

9.4.9

).
And in the latter three cases, the object-part is all there is, since hom-sets are sets.

Thus, it suffices to show that we recover the action of

G

on hom-sets.
In other words, we must show that if

g

:

hom

B

⁡

(

b

,

b

′

)

, then

G

b

,

b

′

⁢

(

g

)

=

G

¯

G

0

⁢

b

,

G

0

⁢

b

′

⁢

(

𝗂𝖽𝗍𝗈𝗂𝗌𝗈

⁢

(

(

ϵ

0

)

-

1

b

′

)

∘

g

∘

𝗂𝖽𝗍𝗈𝗂𝗌𝗈

⁢

(

(

ϵ

0

)

b

)

)

where

G

¯

is defined by (

9.4.12

).
However, this follows from functoriality of

G

and the

other

triangle identity, which we have seen in

\autoref

cha:equivalences is equivalent to (

9.4.11

).

Now since

1

is a mere proposition, so is

2

, so it suffices to show they are

logically equivalent

to

3

.
Of course,

2

→

3

, so let us assume

3

.
Since

1

is a mere proposition, we may assume given

G

,

η

, and

ϵ

.
Then

G

0

along with

η

and

ϵ

imply that

F

0

is an equivalence.
Moreover, we also have natural isomorphisms

𝗂𝖽𝗍𝗈𝗂𝗌𝗈

⁢

(

η

)

:

1

A

≅

G

⁢

F

and

𝗂𝖽𝗍𝗈𝗂𝗌𝗈

⁢

(

ϵ

)

:

F

⁢

G

≅

1

B

, so by

\autoref

ct:adjointification,

F

is an equivalence of precategories, and in particular fully faithful.
∎

From

\autoref

ct:isoprecat

2

and

𝗂𝖽𝗍𝗈𝗂𝗌𝗈

in

functor categories

, we conclude immediately that any isomorphism of precategories is an equivalence.
For precategories, the

converse

can fail.

Example 9.4.13

.

Let

X

be a type and

x

0

:

X

an

element

, and let

X

ch

denote the

chaotic

or

indiscrete

precategory on

X

.
By definition, we have

(

X

ch

)

0

:

≡

X

, and

hom

X

ch

(

x

,

x

′

)

:

≡

1

for all

x

,

x

′

.
Then the unique functor

X

ch

→

1

is an equivalence of precategories, but not an isomorphism unless

X

is

contractible

.

This example also shows that a precategory can be equivalent to a category without itself being a category.
Of course, if a precategory is

isomorphic

to a category, then it must itself be a category.

However, for categories, the two notions coincide.

Lemma 9.4.14

.

For categories

A

and

B

, a functor

F

:

A

→

B

is an equivalence of categories if and only if it is an isomorphism of categories.

Proof.

Since both are mere properties, it suffices to show they are logically equivalent.
So first suppose

F

is an equivalence of categories, with

(

G

,

η

,

ϵ

)

given.
We have already seen that

F

is fully faithful.
By

\autoref

ct:functor-cat, the natural isomorphisms

η

and

ϵ

yield identities

1

A

=

G

⁢

F

and

F

⁢

G

=

1

B

, hence in particular identities

𝗂𝖽

A

=

G

0

∘

F

0

and

F

0

∘

G

0

=

𝗂𝖽

B

.
Thus,

F

0

is an equivalence of types.

Conversely, suppose

F

is fully faithful and

F

0

is an equivalence of types, with inverse

G

0

, say.
Then for each

b

:

B

we have

G

0

⁢

b

:

A

and an identity

F

⁢

G

⁢

b

=

b

, hence an isomorphism

F

⁢

G

⁢

b

≅

b

.
Thus, by

\autoref

ct:ffeso,

F

is an equivalence of categories.
∎

Of course, there is yet a third notion of sameness for (pre)categories: equality.
However, the univalence axiom implies that it coincides with isomorphism.

Lemma 9.4.15

.

If

A

and

B

are precategories, then the function

(

A

=

B

)

→

(

A

≅

B

)

(defined by

induction

from the

identity functor

) is an equivalence of types.

Proof.

As usual for dependent sum types, to give an element of

A

=

B

is equivalent to giving

•

an identity

P

0

:

A

0

=

B

0

,

•

for each

a

,

b

:

A

0

, an identity

P

a

,

b

:

hom

A

(

a

,

b

)

=

hom

B

(

P

0

*

(

a

)

,

P

0

*

(

b

)

)

,

•

identities

(

P

a

,

a

)

*

(

1

a

)

=

1

P

0

*

(

a

)

and

(

P

a

,

c

)

*

(

g

f

)

=

(

P

b

,

c

)

*

(

g

)

∘

(

P

a

,

b

)

*

(

f

)

.

(Again, we use the fact that the identity types of hom-sets are mere propositions.)
However, by univalence, this is equivalent to giving

•

an equivalence of types

F

0

:

A

0

≃

B

0

,

•

for each

a

,

b

:

A

0

, an equivalence of types

F

a

,

b

:

hom

A

⁡

(

a

,

b

)

≃

hom

B

⁡

(

F

0

⁢

(

a

)

,

F

0

⁢

(

b

)

)

,

•

and identities

F

a

,

a

⁢

(

1

a

)

=

1

F

0

⁢

(

a

)

and

F

a

,

c

⁢

(

g

⁢

f

)

=

F

b

,

c

⁢

(

g

)

∘

F

a

,

b

⁢

(

f

)

.

But this consists exactly of a functor

F

:

A

→

B

that is an isomorphism of categories.
And by induction on identity, this equivalence

(

A

=

B

)

≃

(

A

≅

B

)

is equal to the one obtained by induction.
∎

Thus, for categories, equality also coincides with equivalence.
We can interpret this as saying that categories, functors, and

natural transformations

form, not just a pre-2-category, but a

2-category

.

Theorem 9.4.16

.

If

A

and

B

are categories, then the function

(

A

=

B

)

→

(

A

≃

B

)

(defined by induction from the identity functor) is an equivalence of types.

Proof.

By

\autoref

ct:cat-eq-iso,

\autoref

ct:eqv-levelwise.
∎

As a

consequence

, the type of categories is a 2-type.
For since

A

≃

B

is a subtype of the type of functors from

A

to

B

, which are the objects of a category, it is a 1-type; hence the identity types

A

=

B

are also 1-types.

Title

9.4 Equivalences

\metatable

Generated on Thu Feb 8 19:42:39 2018 by

LaTeXML