napkin solutions > Ch. 68
Back to Chapter Selection
Table of Contents
Problem 68A
Problem 68B
Problem 68A
#
Problem 68A.
Show that the two definitions of natural transformation (one in terms of
𝒜
×
𝟐
→
ℬ
and
one in terms of arrows
𝐹
(
𝐴
)
⟶
𝛼
𝐴
𝐺
(
𝐴
)
) are equivalent.
Solution
by
RanolP
Common Condition
Given two categories
𝒜
,
ℬ
and two functors
𝐹
,
𝐺
:
𝒜
→
ℬ
.
Given natural transformation
𝐹
(
𝐴
)
⟶
𝛼
𝐴
𝐺
(
𝐴
)
Our goal is to construct a functor
𝛽
:
𝒜
×
𝟐
→
ℬ
for each
𝛼
.
Since product category is defined as tuple of objects and arrows, we can construct
𝛽
by
sending tuple of objects
𝛽
(
(
𝐴
,
0
)
)
=
𝐹
(
𝐴
)
𝛽
(
(
𝐴
,
1
)
)
=
𝐺
(
𝐴
)
and sending tuple of morphisms
𝛽
(
(
𝑓
,
id
0
)
)
=
𝐹
(
𝑓
)
𝛽
(
(
𝑓
,
id
1
)
)
=
𝐺
(
𝑓
)
𝛽
(
(
𝐴
1
→
𝑓
𝐴
2
,
0
→
≤
1
)
)
=
𝛼
𝐴
2
∘
𝐹
(
𝑓
)
to
ℬ
Now we need to check the
𝛽
is functor.
1.
Mapping of Objects
Trivially done by
𝐹
(
𝐴
)
and
𝐺
(
𝐴
)
2.
Mapping of Morphisms
Trivially done by
𝐹
(
𝑓
)
,
𝐺
(
𝑓
)
and
𝛼
𝐴
2
∘
𝐹
(
𝑓
)
.
𝐹
(
𝑓
)
𝛼
𝐴
1
𝛼
𝐴
2
𝐺
(
𝑓
)
𝐹
(
𝐴
1
)
𝐹
(
𝐴
2
)
𝐺
(
𝐴
1
)
𝐺
(
𝐴
2
)
3.
Preservation of Identity Morphism
In
𝒜
×
𝟐
, identity is either
id
(
𝐴
,
0
)
=
(
id
𝐴
,
id
0
)
which will be sent to
𝐹
(
id
𝐴
)
or
id
(
𝐴
,
1
)
=
(
id
𝐴
,
id
1
)
which
will be sent to
𝐺
(
id
𝐴
)
. both will preserve identity by functor
𝐹
and
𝐺
.
4.
Preservation of Composition
It’s the most unsatisfying part. We will split cases for easier understanding.
i.
Both sides are
(
𝑓
,
id
0
)
form: Preserved by functor
𝐹
.
ii.
Both sides are
(
𝑓
,
id
1
)
form: Preserved by functor
𝐺
.
iii.
The one is
(
𝐴
1
→
𝑓
𝐴
2
,
0
→
≤
1
)
: Prove case by case.
a.
𝛽
(
(
𝐴
2
→
𝑔
𝐴
3
,
id
1
)
)
∘
𝛽
(
(
𝐴
1
→
𝑓
𝐴
2
,
0
→
≤
1
)
)
=
𝐺
(
𝑔
)
∘
𝛼
𝐴
2
∘
𝐹
(
𝑓
)
Abuse of Notation
: let’s call the function above as
𝛽
(
𝑔
)
∘
𝛽
(
𝑓
)
for now.
𝑓
𝑔
∘
𝑓
𝑔
𝐹
(
𝑓
)
𝛽
(
𝑔
)
∘
𝛽
(
𝑓
)
𝛼
𝐴
2
𝐺
(
𝑔
)
𝐴
1
𝐴
2
𝐴
3
𝐹
(
𝐴
1
)
𝐹
(
𝐴
2
)
𝐺
(
𝐴
2
)
𝐺
(
𝐴
3
)
The diagram above commutes.
b.
(
𝐴
1
→
𝑓
𝐴
2
,
0
→
≤
1
)
∘
(
𝐴
0
→
𝑔
𝐴
1
,
id
0
)
=
𝛼
𝐴
2
∘
𝐹
(
𝑓
)
∘
𝐹
(
𝑔
)
Abuse of Notation
: let’s call the function above as
𝛽
(
𝑓
)
∘
𝛽
(
𝑔
)
for now.
𝑔
𝑓
∘
𝑔
𝑓
𝐹
(
𝑔
)
𝛽
(
𝑓
)
∘
𝛽
(
𝑔
)
𝐹
(
𝑓
)
𝛼
𝐴
2
𝐴
0
𝐴
1
𝐴
2
𝐹
(
𝐴
0
)
𝐹
(
𝐴
1
)
𝐹
(
𝐴
2
)
𝐺
(
𝐴
2
)
The diagram above commutes.
Given functor
𝛽
:
𝒜
×
𝟐
→
ℬ
Our goal is to construct a natural transformation
𝐹
(
𝐴
)
⟶
𝛼
𝐴
𝐺
(
𝐴
)
from
𝛽
.
Consider
𝐴
1
→
𝑓
𝐴
2
, then following diagram commutes.
(
𝑓
,
id
0
)
(
id
𝐴
1
,
0
→
1
)
(
id
𝐴
2
,
0
→
1
)
(
𝑓
,
id
1
)
𝛽
𝐹
(
𝑓
)
𝛽
(
(
id
𝐴
1
,
0
→
1
)
)
𝛽
(
(
id
𝐴
2
,
0
→
1
)
)
𝐺
(
𝑓
)
(
𝐴
1
,
0
)
(
𝐴
2
,
0
)
(
𝐴
1
,
1
)
(
𝐴
2
,
1
)
𝐹
(
𝐴
1
)
𝐹
(
𝐴
2
)
𝐺
(
𝐴
1
)
𝐺
(
𝐴
2
)
Let’s say
𝛼
𝐴
=
𝛽
(
(
id
𝐴
,
0
→
1
)
)
. Then it’s the naturally requirement.
𝐹
(
𝑓
)
𝛼
𝐴
1
𝛼
𝐴
2
𝐺
(
𝑓
)
𝐹
(
𝐴
1
)
𝐹
(
𝐴
2
)
𝐺
(
𝐴
1
)
𝐺
(
𝐴
2
)
Problem 68B
#
Problem 68B.
Let
𝒜
be the category of finite sets whose arrows are bijections between sets. For
𝐴
∈
𝒜
,
let
𝐹
(
𝐴
)
be the set of
permutations
of
𝐴
and let
𝐺
(
𝐴
)
be the set of
orderings
on
𝐴
.
¹
(a)
Extend
𝐹
and
𝐺
to functors
𝒜
→
Set
.
(b)
Show that
𝐹
(
𝐴
)
≅
𝐺
(
𝐴
)
for every
𝐴
, but this isomorphism is
not
natural.
Solution
by
finalchild
(a)
𝐹
and
𝐺
act on arrows as follows:
𝐹
(
𝑓
)
=
𝑝
↦
𝑓
∘
𝑝
∘
𝑓
−
1
𝐺
(
𝑓
)
=
𝑜
↦
𝑓
∘
𝑜
(b)
|
𝐹
(
𝐴
)
|
=
|
𝐺
(
𝐴
)
|
, hence
𝐹
(
𝐴
)
≅
𝐺
(
𝐴
)
.
Consider
𝐴
=
{
0
,
1
}
, and an arrow
𝑓
:
𝐴
→
𝐴
=
𝑥
↦
1
−
𝑥
. Let
𝛼
𝐴
be an isomorphism from
𝐹
(
𝐴
)
to
𝐺
(
𝐴
)
.
Since conjugation has no effect on permutations on two-element set,
𝛼
𝐴
∘
𝐹
(
𝑓
)
=
𝛼
𝐴
. On the other hand,
𝐺
(
𝑓
)
∘
𝛼
𝐴
=
reverse
∘
𝛼
𝐴
. The two cannot be equal, thus this isomorphism is not natural.
¹
A permutation is a bijection
𝐴
→
𝐴
, and an ordering is a bijection
{
1
,
…
,
𝑛
}
→
𝐴
, where
𝑛
is the size of
𝐴
.