2.14.1 Lifting equivalences

2.14.1 Lifting equivalences

When working loosely, one might say that a

bijection

between sets

A

and

B

“obviously” induces an

isomorphism

between

semigroup

structures

on

A

and semigroup structures on

B

. With univalence,
this is indeed

obvious

, because given an equivalence between types

A

and

B

, we can automatically derive a semigroup structure on

B

from
one on

A

, and moreover show that this

derivation

is an equivalence of
semigroup structures. The reason is that

𝖲𝖾𝗆𝗂𝗀𝗋𝗈𝗎𝗉𝖲𝗍𝗋

is a family
of types, and therefore has an action on paths between types given by

𝗍𝗋𝖺𝗇𝗌𝗉𝗈𝗋𝗍

:

𝗍𝗋𝖺𝗇𝗌𝗉𝗈𝗋𝗍

𝖲𝖾𝗆𝗂𝗀𝗋𝗈𝗎𝗉𝖲𝗍𝗋

⁢

(

𝗎𝖺

⁢

(

e

)

)

:

𝖲𝖾𝗆𝗂𝗀𝗋𝗈𝗎𝗉𝖲𝗍𝗋

⁢

(

A

)

→

𝖲𝖾𝗆𝗂𝗀𝗋𝗈𝗎𝗉𝖲𝗍𝗋

⁢

(

B

)

.

Moreover, this map is an equivalence, because

𝗍𝗋𝖺𝗇𝗌𝗉𝗈𝗋𝗍

C

⁢

(

α

)

is always an equivalence with

inverse

𝗍𝗋𝖺𝗇𝗌𝗉𝗈𝗋𝗍

C

⁢

(

α

-

1

)

, see Lemma 2.3.9 (

http://planetmath.org/23typefamiliesarefibrations#Thmprelem6

),Lemma 2.1.4 (

http://planetmath.org/21typesarehighergroupoids#Thmprelem3

).

While the univalence axiom ensures that this map exists, we need to use
facts about

𝗍𝗋𝖺𝗇𝗌𝗉𝗈𝗋𝗍

proven in the preceding

sections

to
calculate what it actually does. Let

(

m

,

a

)

be a semigroup structure on

A

, and we investigate the induced semigroup structure on

B

given by

𝗍𝗋𝖺𝗇𝗌𝗉𝗈𝗋𝗍

𝖲𝖾𝗆𝗂𝗀𝗋𝗈𝗎𝗉𝖲𝗍𝗋

⁢

(

𝗎𝖺

⁢

(

e

)

,

(

m

,

a

)

)

.

First, because

𝖲𝖾𝗆𝗂𝗀𝗋𝗈𝗎𝗉𝖲𝗍𝗋

(X) is defined to be a

Σ

-type, by

Theorem

2.7.4 (

http://planetmath.org/27sigmatypes#Thmprethm2

),

𝗍𝗋𝖺𝗇𝗌𝗉𝗈𝗋𝗍

𝖲𝖾𝗆𝗂𝗀𝗋𝗈𝗎𝗉𝖲𝗍𝗋

⁢

(

𝗎𝖺

⁢

(

e

)

,

(

m

,

a

)

)

=

(

𝗍𝗋𝖺𝗇𝗌𝗉𝗈𝗋𝗍

X

↦

(

X

→

X

→

X

)

⁢

(

𝗎𝖺

⁢

(

e

)

,

m

)

,

𝗍𝗋𝖺𝗇𝗌𝗉𝗈𝗋𝗍

(

X

,

m

)

↦

𝖠𝗌𝗌𝗈𝖼

⁢

(

X

,

m

)

(

(

𝗉𝖺𝗂𝗋

=

(

𝗎𝖺

(

e

)

,

𝗋𝖾𝖿𝗅

)

)

,

a

)

)

(2.14.2)

where

𝖠𝗌𝗌𝗈𝖼

⁢

(

X

,

m

)

is the type

∏

(

x

,

y

,

z

:

X

)

m

⁢

(

x

,

m

⁢

(

y

,

z

)

)

=

m

⁢

(

m

⁢

(

x

,

y

)

,

z

)

.
That is, the induced semigroup structure consists of an induced

multiplication

operation

on

B

m

′

:

B

→

B

→

B

m

′

(

b

1

,

b

2

)

:

≡

𝗍𝗋𝖺𝗇𝗌𝗉𝗈𝗋𝗍

X

↦

(

X

→

X

→

X

)

(

𝗎𝖺

(

e

)

,

m

)

(

b

1

,

b

2

)

together with an induced proof that

m

′

is

associative

. By function
extensionality, it suffices to investigate the

behavior

of

m

′

when
applied to

arguments

b

1

,

b

2

:

B

. By applying
(

LABEL:eq:transport-arrow

) twice, we have that

m

′

⁢

(

b

1

,

b

2

)

is equal to

𝗍𝗋𝖺𝗇𝗌𝗉𝗈𝗋𝗍

X

↦

X

⁢

(

𝗎𝖺

⁢

(

e

)

,

m

⁢

(

𝗍𝗋𝖺𝗇𝗌𝗉𝗈𝗋𝗍

X

↦

X

⁢

(

𝗎𝖺

⁢

(

e

)

-

1

,

b

1

)

,

𝗍𝗋𝖺𝗇𝗌𝗉𝗈𝗋𝗍

X

↦

X

⁢

(

𝗎𝖺

⁢

(

e

)

-

1

,

b

2

)

)

)

.

Then, because

𝗎𝖺

is

quasi-inverse

to

𝗍𝗋𝖺𝗇𝗌𝗉𝗈𝗋𝗍

X

↦

X

, this is equal to

e

⁢

(

m

⁢

(

e

-

1

⁢

(

b

1

)

,

e

-

1

⁢

(

b

2

)

)

)

.

Thus, given two elements of

B

, the induced multiplication

m

′

sends them to

A

using the equivalence

e

, multiplies them in

A

, and
then brings the result back to

B

by

e

, just as one would expect.

Moreover, though we do not show the proof, one can calculate that the
induced proof that

m

′

is associative (the second

component

of the pair
in (

2.14.2

)) is equal to a

function

sending

b

1

,

b

2

,

b

3

:

B

to a path given by the following steps:

m

′

⁢

(

m

′

⁢

(

b

1

,

b

2

)

,

b

3

)

=

e

⁢

(

m

⁢

(

e

-

1

⁢

(

m

′

⁢

(

b

1

,

b

2

)

)

,

e

-

1

⁢

(

b

3

)

)

)

(2.14.3)

=

e

⁢

(

m

⁢

(

e

-

1

⁢

(

e

⁢

(

m

⁢

(

e

-

1

⁢

(

b

1

)

,

e

-

1

⁢

(

b

2

)

)

)

)

,

e

-

1

⁢

(

b

3

)

)

)

=

e

⁢

(

m

⁢

(

m

⁢

(

e

-

1

⁢

(

b

1

)

,

e

-

1

⁢

(

b

2

)

)

,

e

-

1

⁢

(

b

3

)

)

)

=

e

⁢

(

m

⁢

(

e

-

1

⁢

(

b

1

)

,

m

⁢

(

e

-

1

⁢

(

b

2

)

,

e

-

1

⁢

(

b

3

)

)

)

)

=

e

⁢

(

m

⁢

(

e

-

1

⁢

(

b

1

)

,

e

-

1

⁢

(

e

⁢

(

m

⁢

(

e

-

1

⁢

(

b

2

)

,

e

-

1

⁢

(

b

3

)

)

)

)

)

)

=

e

⁢

(

m

⁢

(

e

-

1

⁢

(

b

1

)

,

e

-

1

⁢

(

m

′

⁢

(

b

2

,

b

3

)

)

)

)

=

m

′

⁢

(

b

1

,

m

′

⁢

(

b

2

,

b

3

)

)

.

These steps use the proof

a

that

m

is associative and the inverse
laws for

e

. From an

algebra

perspective, it may seem strange to
investigate the

identity

of a proof that an operation is associative,
but this makes sense if we think of

A

and

B

as general spaces, with
non-trivial

homotopies

between paths. In

http://planetmath.org/node/87576

Chapter 3, we will
introduce the notion of a

set

, which is a type with only

trivial

homotopies, and if we consider semigroup structures on sets, then any
two such associativity proofs are automatically equal.

Title

2.14.1 Lifting equivalences

\metatable

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