4.1 Quasi-inverses

4.1 Quasi-inverses

We have said that

𝗊𝗂𝗇𝗏

⁢

(

f

)

is unsatisfactory because it is not a mere proposition, whereas we would rather that a given

function

can “be an

equivalence

” in at most one way.
However, we have given no evidence that

𝗊𝗂𝗇𝗏

⁢

(

f

)

is not a mere proposition.
In this

section

we exhibit a specific counterexample.

Lemma 4.1.1

.

If

f

:

A

→

B

is such that

qinv

⁢

(

f

)

is inhabited, then

𝗊𝗂𝗇𝗏

(

f

)

≃

(

∏

x

:

A

(

x

=

x

)

)

.

Proof.

By

assumption

,

f

is an equivalence; that is, we have

e

:

𝗂𝗌𝖾𝗊𝗎𝗂𝗏

⁢

(

f

)

and so

(

f

,

e

)

:

A

≃

B

.
By univalence,

𝗂𝖽𝗍𝗈𝖾𝗊𝗏

:

(

A

=

B

)

→

(

A

≃

B

)

is an equivalence, so we may assume that

(

f

,

e

)

is of the form

𝗂𝖽𝗍𝗈𝖾𝗊𝗏

⁢

(

p

)

for some

p

:

A

=

B

.
Then by path

induction

, we may assume

p

is

𝗋𝖾𝖿𝗅

A

, in which case

𝗂𝖽𝗍𝗈𝖾𝗊𝗏

⁢

(

p

)

is

𝗂𝖽

A

.
Thus we are

reduced

to proving

𝗊𝗂𝗇𝗏

(

𝗂𝖽

A

)

≃

(

∏

(

x

:

A

)

(

x

=

x

)

)

.
Now by definition we have

𝗊𝗂𝗇𝗏

(

𝗂𝖽

A

)

≡

∑

g

:

A

→

A

(

(

g

∼

𝗂𝖽

A

)

×

(

g

∼

𝗂𝖽

A

)

)

.

By function extensionality, this is

equivalent

to

∑

g

:

A

→

A

(

(

g

=

𝗂𝖽

A

)

×

(

g

=

𝗂𝖽

A

)

)

.

And by

http://planetmath.org/node/87641

Exercise 2.10, this is equivalent to

∑

h

:

∑

(

g

:

A

→

A

)

(

g

=

𝗂𝖽

A

)

(

𝗉𝗋

1

(

h

)

=

𝗂𝖽

A

)

However, by Lemma 3.11.8 (

http://planetmath.org/311contractibility#Thmprelem5

),

∑

(

g

:

A

→

A

)

(

g

=

𝗂𝖽

A

)

is

contractible

with center

𝗂𝖽

A

; therefore by Lemma 3.11.9 (

http://planetmath.org/311contractibility#Thmprelem6

) this type is equivalent to

𝗂𝖽

A

=

𝗂𝖽

A

.
And by function extensionality,

𝗂𝖽

A

=

𝗂𝖽

A

is equivalent to

∏

(

x

:

A

)

x

=

x

.
∎

We remark that

http://planetmath.org/node/87864

Exercise 4.3 asks for a proof of the above lemma which avoids univalence.

Thus, what we need is some

A

which admits a

nontrivial element

of

∏

(

x

:

A

)

(

x

=

x

)

.
Thinking of

A

as a higher

groupoid

, an inhabitant of

∏

(

x

:

A

)

(

x

=

x

)

is a

natural transformation

from the

identity functor

of

A

to itself.
Such

transformations

are said to form the

center of a

category

,
since the naturality axiom requires that they commute with all

morphisms

.
Classically, if

A

is simply a group regarded as a one-object groupoid, then this yields precisely its center in the usual group-theoretic sense.
This provides some motivation for the following.

Lemma 4.1.2

.

Suppose we have a type

A

with

a

:

A

and

q

:

a

=

a

such that

1.

The type

a

=

a

is a set.

2.

For all

x

:

A

we have

∥

a

=

x

∥

.

3.

For all

p

:

a

=

a

we have

p

⁢

\centerdot

⁢

q

=

q

⁢

\centerdot

⁢

p

.

Then there exists

f

:

∏

(

x

:

A

)

(

x

=

x

)

with

f

⁢

(

a

)

=

q

.

Proof.

Let

g

:

∏

(

x

:

A

)

∥

a

=

x

∥

be as given by

2

. First we
observe that each type

x

=

A

y

is a set. For since being a set is a mere
proposition, we may apply the induction principle of propositional truncation, and assume that

g

(

x

)

=

|

p

|

and

g

(

y

)

=

|

q

|

for

p

:

a

=

x

and

q

:

a

=

y

. In this case, composing with

p

and

q

-

1

yields an equivalence

(

x

=

y

)

≃

(

a

=

a

)

. But

(

a

=

a

)

is
a set by

1

, so

(

x

=

y

)

is also a set.

Now, we would like to define

f

by assigning to each

x

the path

g

⁢

(

x

)

-

1

⁢

\centerdot

⁢

q

⁢

\centerdot

⁢

g

⁢

(

x

)

, but this does not work because

g

⁢

(

x

)

does not inhabit

a

=

x

but rather

∥

a

=

x

∥

, and the type

(

x

=

x

)

may not be a mere proposition,
so we cannot use induction on propositional truncation. Instead we can apply
the technique mentioned in §3.9 (

http://planetmath.org/39theprincipleofuniquechoice

): we characterize
uniquely the object we wish to

construct

. Let us define, for each

x

:

A

, the
type

B

(

x

)

:

≡

∑

(

r

:

x

=

x

)

∏

(

s

:

a

=

x

)

(

r

=

s

-

1

\centerdot

q

\centerdot

s

)

.

We claim that

B

⁢

(

x

)

is a mere proposition for each

x

:

A

.
Since this claim is itself a mere proposition, we may again apply induction on
truncation and assume that

g

(

x

)

=

|

p

|

for some

p

:

a

=

x

.
Now suppose given

(

r

,

h

)

and

(

r

′

,

h

′

)

in

B

⁢

(

x

)

; then we have

h

⁢

(

p

)

⁢

\centerdot

⁢

h

′

⁢

(

p

)

-

1

:

r

=

r

′

.

It remains to show that

h

is identified with

h

′

when transported along this equality, which by transport in identity types and function types (§2.11 (

http://planetmath.org/211identitytype

),§2.9 (

http://planetmath.org/29pitypesandthefunctionextensionalityaxiom

)), reduces to showing

h

⁢

(

s

)

=

h

⁢

(

p

)

⁢

\centerdot

⁢

h

′

⁢

(

p

)

-

1

⁢

\centerdot

⁢

h

′

⁢

(

s

)

for any

s

:

a

=

x

.
But each side of this is an equality between elements of

(

x

=

x

)

, so it follows from our above observation that

(

x

=

x

)

is a set.

Thus, each

B

⁢

(

x

)

is a mere proposition; we claim that

∏

(

x

:

A

)

B

⁢

(

x

)

.
Given

x

:

A

, we may now invoke the induction principle of propositional truncation to assume that

g

(

x

)

=

|

p

|

for

p

:

a

=

x

.
We define

r

:

≡

p

-

1

\centerdot

q

\centerdot

p

; to inhabit

B

⁢

(

x

)

it remains to show that for any

s

:

a

=

x

we have

r

=

s

-

1

⁢

\centerdot

⁢

q

⁢

\centerdot

⁢

s

.
Manipulating paths, this reduces to showing that

q

⁢

\centerdot

⁢

(

p

⁢

\centerdot

⁢

s

-

1

)

=

(

p

⁢

\centerdot

⁢

s

-

1

)

⁢

\centerdot

⁢

q

.
But this is just an

instance

of

3

.
∎

Theorem 4.1.3

.

There exist types

A

and

B

and a function

f

:

A

→

B

such that

qinv

⁢

(

f

)

is not a mere proposition.

Proof.

It suffices to exhibit a type

X

such that

∏

(

x

:

X

)

(

x

=

x

)

is not a mere proposition.
Define

X

:

≡

∑

(

A

:

𝒰

)

∥

𝟐

=

A

∥

, as in the proof of Lemma 3.8.5 (

http://planetmath.org/38theaxiomofchoice#Thmprelem2

).
It will suffice to exhibit an

f

:

∏

(

x

:

X

)

(

x

=

x

)

which is unequal to

λ

⁢

x

.

𝗋𝖾𝖿𝗅

x

.

Let

a

:

≡

(

𝟐

,

|

𝗋𝖾𝖿𝗅

𝟐

|

)

:

X

, and let

q

:

a

=

a

be the path corresponding to the nonidentity equivalence

e

:

𝟐

≃

𝟐

defined by

e

(

0

𝟐

)

:

≡

1

𝟐

and

e

(

1

𝟐

)

:

≡

0

𝟐

.
We would like to apply Lemma 4.1.2 (

http://planetmath.org/41quasiinverses#Thmprelem2

) to build an

f

.
By definition of

X

, equalities in subset types (§3.5 (

http://planetmath.org/35subsetsandpropositionalresizing

)), and univalence, we have

(

a

=

a

)

≃

(

𝟐

≃

𝟐

)

, which is a set, so

1

holds.
Similarly, by definition of

X

and equalities in subset types we have

2

.
Finally,

http://planetmath.org/node/87644

Exercise 2.13 implies that every equivalence

𝟐

≃

𝟐

is equal to either

𝗂𝖽

𝟐

or

e

, so we can show

3

by a four-way case

analysis

.

Thus, we have

f

:

∏

(

x

:

X

)

(

x

=

x

)

such that

f

⁢

(

a

)

=

q

.
Since

e

is not equal to

𝗂𝖽

𝟐

,

q

is not equal to

𝗋𝖾𝖿𝗅

a

, and thus

f

is not equal to

λ

⁢

x

.

𝗋𝖾𝖿𝗅

x

.
Therefore,

∏

(

x

:

X

)

(

x

=

x

)

is not a mere proposition.
∎

More generally, Lemma 4.1.2 (

http://planetmath.org/41quasiinverses#Thmprelem2

) implies that any “Eilenberg–Mac Lane space”

K

⁢

(

G

,

1

)

, where

G

is a nontrivial

abelian group

, will provide a counterexample; see

http://planetmath.org/node/87582

Chapter 8.
The type

X

we used turns out to be equivalent to

K

⁢

(

ℤ

2

,

1

)

.
In

http://planetmath.org/node/87579

Chapter 6 we will see that the circle

𝕊

1

=

K

⁢

(

ℤ

,

1

)

is another easy-to-describe example.

We now move on to describing better notions of equivalence.

Title

4.1

Quasi-inverses

\metatable

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