7.6 Orthogonal factorization

7.6 Orthogonal factorization

In

set theory

, the

surjections

and the

injections

form a unique factorization system: every

function

factors essentially uniquely as a surjection followed by an injection.
We have seen that surjections generalize naturally to

n

-connected maps, so it is natural to inquire whether these also participate in a

factorization system

.
Here is the corresponding

generalization

of injections.

Definition 7.6.1

.

A function

f

:

A

→

B

is

n

-truncated

if the fiber

fib

f

⁢

(

b

)

is an

n

-type for all

b

:

B

.

In particular,

f

is

(

-

2

)

-truncated if and only if it is an

equivalence

.
And of course,

A

is an

n

-type if and only if

A

→

𝟏

is

n

-truncated.
Moreover,

n

-truncated maps could equivalently be defined recursively, like

n

-types.

Lemma 7.6.2

.

For any

n

≥

-

2

, a function

f

:

A

→

B

is

(

n

+

1

)

-truncated if and only if for all

x

,

y

:

A

, the map

ap

f

:

(

x

=

y

)

→

(

f

(

x

)

=

f

(

y

)

)

is

n

-truncated.
In particular,

f

is

(

-

1

)

-truncated if and only if it is an

embedding

in the sense of §4.6 (

http://planetmath.org/46surjectionsandembeddings

).

Proof.

Note that for any

(

x

,

p

)

,

(

y

,

q

)

:

𝖿𝗂𝖻

f

⁢

(

b

)

, we have

(

(

x

,

p

)

=

(

y

,

q

)

)

=

∑

r

:

x

=

y

(

p

=

𝖺𝗉

f

(

r

)

\centerdot

q

)

=

∑

r

:

x

=

y

(

𝖺𝗉

f

(

r

)

=

p

\centerdot

q

-

1

)

=

𝖿𝗂𝖻

𝖺𝗉

f

⁢

(

p

⁢

\centerdot

⁢

q

-

1

)

.

Thus, any path space in any fiber of

f

is a fiber of

𝖺𝗉

f

.
On the other hand, choosing

b

:

≡

f

(

y

)

and

q

:

≡

𝗋𝖾𝖿𝗅

f

⁢

(

y

)

we see that any fiber of

𝖺𝗉

f

is a path space in a fiber of

f

.
The result follows, since

f

is

(

n

+

1

)

-truncated if all path spaces of its fibers are

n

-types.
∎

We can now

construct

the factorization, in a fairly

obvious

way.

Definition 7.6.3

.

Let

f

:

A

→

B

be a function. The

n

-image

of

f

is defined as

𝗂𝗆

n

(

f

)

:

≡

∑

b

:

B

∥

𝖿𝗂𝖻

f

(

b

)

∥

n

.

When

n

=

-

1

, we write simply

im

⁢

(

f

)

and call it the

image

of

f

.

Lemma 7.6.4

.

For any function

f

:

A

→

B

, the

canonical

function

f

~

:

A

→

im

n

⁢

(

f

)

is

n

-connected.
Consequently, any function factors as an

n

-connected function followed by an

n

-truncated function.

Proof.

Note that

A

≃

∑

(

b

:

B

)

𝖿𝗂𝖻

f

⁢

(

b

)

. The function

f

~

is the function on total spaces induced by the canonical fiberwise

transformation

∏

b

:

B

(

𝖿𝗂𝖻

f

(

b

)

→

∥

𝖿𝗂𝖻

f

(

b

)

∥

n

)

.

Since each map

𝖿𝗂𝖻

f

(

b

)

→

∥

𝖿𝗂𝖻

f

(

b

)

∥

n

is

n

-connected by

Corollary

7.5.8 (

http://planetmath.org/75connectedness#Thmprecor2

),

f

~

is

n

-connected by Lemma 7.5.13 (

http://planetmath.org/75connectedness#Thmprelem8

).
Finally, the projection

𝗉𝗋

1

:

𝗂𝗆

n

⁢

(

f

)

→

B

is

n

-truncated, since its fibers are

equivalent

to the

n

-

truncations

of the fibers of

f

.
∎

In the following lemma we set up some machinery to prove the

unique factorization theorem

.

Lemma 7.6.5

.

Suppose we have a

commutative diagram

of functions

\includegraphics

HoTT_fig_7.6.1.png

with

H

:

h

1

∘

g

1

∼

h

2

∘

g

2

, where

g

1

and

g

2

are

n

-connected and where

h

1

and

h

2

are

n

-truncated.
Then there is an equivalence

E

⁢

(

H

,

b

)

:

𝖿𝗂𝖻

h

1

⁢

(

b

)

≃

𝖿𝗂𝖻

h

2

⁢

(

b

)

for any

b

:

B

, such that for any

a

:

A

we have an identification

E

¯

(

H

,

a

)

:

E

(

H

,

h

1

(

g

1

(

a

)

)

)

(

g

1

(

a

)

,

𝗋𝖾𝖿𝗅

h

1

⁢

(

g

1

⁢

(

a

)

)

)

=

(

g

2

(

a

)

,

H

⁢

(

a

)

-

1

)

.

Proof.

Let

b

:

B

. Then we have the following equivalences:

𝖿𝗂𝖻

h

1

⁢

(

b

)

≃

∑

w

:

𝖿𝗂𝖻

h

1

⁢

(

b

)

∥

𝖿𝗂𝖻

g

1

(

𝗉𝗋

1

w

)

∥

n

(since $g_{1}$ is $n$-connected)

≃

∥

∑

w

:

𝖿𝗂𝖻

h

1

⁢

(

b

)

𝖿𝗂𝖻

g

1

⁢

(

𝗉𝗋

1

⁢

w

)

∥

n

(by Corollary 7.3.10 (http://planetmath.org/73truncationshmprecor2), since $h_{1}$ is $n$-truncated)

≃

∥

𝖿𝗂𝖻

h

1

∘

g

1

(

b

)

∥

n

(by http://planetmath.org/node/87774Exercise 4.4)

and likewise for

h

2

and

g

2

.
Also, since we have a

homotopy

H

:

h

1

∘

g

1

∼

h

2

∘

g

2

, there is an obvious equivalence

𝖿𝗂𝖻

h

1

∘

g

1

⁢

(

b

)

≃

𝖿𝗂𝖻

h

2

∘

g

2

⁢

(

b

)

.
Hence we obtain

𝖿𝗂𝖻

h

1

⁢

(

b

)

≃

𝖿𝗂𝖻

h

2

⁢

(

b

)

for any

b

:

B

. By analyzing the underlying functions, we get the following representation of what happens to the

element

(

g

1

⁢

(

a

)

,

𝗋𝖾𝖿𝗅

h

1

⁢

(

g

1

⁢

(

a

)

)

)

after applying each of the equivalences of which

E

is composed.
Some of the identifications are definitional, but others (marked with a

=

below) are only propositional; putting them together we obtain

E

¯

⁢

(

H

,

a

)

.

(

g

1

⁢

(

a

)

,

𝗋𝖾𝖿𝗅

h

1

⁢

(

g

1

⁢

(

a

)

)

)

↦

=

(

(

g

1

(

a

)

,

𝗋𝖾𝖿𝗅

h

1

⁢

(

g

1

⁢

(

a

)

)

)

,

|

(

a

,

𝗋𝖾𝖿𝗅

g

1

⁢

(

a

)

)

|

n

)

↦

|

(

(

g

1

(

a

)

,

𝗋𝖾𝖿𝗅

h

1

⁢

(

g

1

⁢

(

a

)

)

)

,

(

a

,

𝗋𝖾𝖿𝗅

g

1

⁢

(

a

)

)

)

|

n

↦

|

(

a

,

𝗋𝖾𝖿𝗅

h

1

⁢

(

g

1

⁢

(

a

)

)

)

|

n

↦

=

|

(

a

,

H

⁢

(

a

)

-

1

)

|

n

↦

|

(

(

g

2

(

a

)

,

H

⁢

(

a

)

-

1

)

,

(

a

,

𝗋𝖾𝖿𝗅

g

2

⁢

(

a

)

)

)

|

n

↦

(

(

g

2

(

a

)

,

H

⁢

(

a

)

-

1

)

,

|

(

a

,

𝗋𝖾𝖿𝗅

g

2

⁢

(

a

)

)

|

n

)

↦

(

g

2

(

a

)

,

H

⁢

(

a

)

-

1

)

The first equality is because for general

b

, the map

𝖿𝗂𝖻

h

1

(

b

)

→

∑

(

w

:

𝖿𝗂𝖻

h

1

(

b

)

)

∥

𝖿𝗂𝖻

g

1

(

𝗉𝗋

1

w

)

∥

n

inserts the center of

contraction

for

∥

𝖿𝗂𝖻

g

1

⁢

(

𝗉𝗋

1

⁢

w

)

∥

n

supplied by the

assumption

that

g

1

is

n

-truncated; whereas in the case in question this type has the obvious inhabitant

|

(

a

,

𝗋𝖾𝖿𝗅

g

1

⁢

(

a

)

)

|

n

, which by

contractibility

must be equal to the center.
The second propositional equality is because the equivalence

𝖿𝗂𝖻

h

1

∘

g

1

⁢

(

b

)

≃

𝖿𝗂𝖻

h

2

∘

g

2

⁢

(

b

)

concatenates the second components with

H

⁢

(

a

)

-

1

, and we have

H

⁢

(

a

)

-

1

⁢

\centerdot

⁢

𝗋𝖾𝖿𝗅

=

H

⁢

(

a

)

-

1

.
The reader may check that the other equalities are definitional (assuming a reasonable solution to

http://planetmath.org/node/87774

Exercise 4.4).
∎

Combining Lemma 7.6.4 (

http://planetmath.org/76orthogonalfactorization#Thmprelem2

),Lemma 7.6.5 (

http://planetmath.org/76orthogonalfactorization#Thmprelem3

), we have the following unique factorization result:

Theorem 7.6.6

.

For each

f

:

A

→

B

, the space

fact

n

⁢

(

f

)

defined by

∑

(

X

:

𝒰

)

∑

(

g

:

A

→

X

)

∑

(

h

:

X

→

B

)

(

h

∘

g

∼

f

)

×

𝖼𝗈𝗇𝗇

n

(

g

)

×

𝗍𝗋𝗎𝗇𝖼

n

(

h

)

.

is

contractible

.
Its center of contraction is the element

(

𝗂𝗆

n

(

f

)

,

f

~

,

𝗉𝗋

1

,

θ

,

φ

,

ψ

)

:

𝖿𝖺𝖼𝗍

n

(

f

)

arising from Lemma 7.6.4 (

http://planetmath.org/76orthogonalfactorization#Thmprelem2

),
where

θ

:

pr

1

∘

f

~

∼

f

is the canonical homotopy, where

φ

is the proof of
Lemma 7.6.4 (

http://planetmath.org/76orthogonalfactorization#Thmprelem2

), and where

ψ

is the obvious proof that

pr

1

:

im

n

⁢

(

f

)

→

B

has

n

-truncated fibers.

Proof.

By Lemma 7.6.4 (

http://planetmath.org/76orthogonalfactorization#Thmprelem2

) we know that there is an element of

𝖿𝖺𝖼𝗍

n

⁢

(

f

)

, hence it is enough to
show that

𝖿𝖺𝖼𝗍

n

⁢

(

f

)

is a mere proposition. Suppose we have two

n

-factorizations

(

X

1

,

g

1

,

h

1

,

H

1

,

φ

1

,

ψ

1

)

and

(

X

2

,

g

2

,

h

2

,

H

2

,

φ

2

,

ψ

2

)

of

f

. Then we have the pointwise-concatenated homotopy

H

:

≡

(

λ

a

.

H

1

(

a

)

\centerdot

H

2

-

1

(

a

)

)

:

(

h

1

∘

g

1

∼

h

2

∘

g

2

)

.

By univalence and the

characterization

of paths and transport in

Σ

-types, function types, and path types, it suffices to show that

1.

there is an equivalence

e

:

X

1

≃

X

2

,

2.

there is a homotopy

ζ

:

e

∘

g

1

∼

g

2

,

3.

there is a homotopy

η

:

h

2

∘

e

∼

h

1

,

4.

for any

a

:

A

we have

𝖺𝗉

h

2

⁢

(

ζ

⁢

(

a

)

)

-

1

⁢

\centerdot

⁢

η

⁢

(

g

1

⁢

(

a

)

)

⁢

\centerdot

⁢

H

1

⁢

(

a

)

=

H

2

⁢

(

a

)

.

We prove these four assertions in that order.

1.

By Lemma 7.6.5 (

http://planetmath.org/76orthogonalfactorization#Thmprelem3

), we have a fiberwise equivalence

E

⁢

(

H

)

:

∏

b

:

B

𝖿𝗂𝖻

h

1

⁢

(

b

)

≃

𝖿𝗂𝖻

h

2

⁢

(

b

)

.

This induces an equivalence of total spaces, i.e. we have

(

∑

b

:

B

𝖿𝗂𝖻

h

1

⁢

(

b

)

)

≃

(

∑

b

:

B

𝖿𝗂𝖻

h

2

⁢

(

b

)

)

.

Of course, we also have the equivalences

X

1

≃

∑

(

b

:

B

)

𝖿𝗂𝖻

h

1

⁢

(

b

)

and

X

2

≃

∑

(

b

:

B

)

𝖿𝗂𝖻

h

2

⁢

(

b

)

from Lemma 4.8.2 (

http://planetmath.org/48theobjectclassifier#Thmprelem2

).
This gives us our equivalence

e

:

X

1

≃

X

2

; the reader may verify that the underlying function of

e

is given by

e

⁢

(

x

)

≡

𝗉𝗋

1

⁢

(

E

⁢

(

H

,

h

1

⁢

(

x

)

)

⁢

(

x

,

𝗋𝖾𝖿𝗅

h

1

⁢

(

x

)

)

)

.

2.

By Lemma 7.6.5 (

http://planetmath.org/76orthogonalfactorization#Thmprelem3

), we may choose

ζ

(

a

)

:

≡

𝖺𝗉

𝗉𝗋

1

(

E

¯

(

H

,

a

)

)

:

e

(

g

1

(

a

)

)

=

g

2

(

a

)

.

3.

For every

x

:

X

1

, we have

𝗉𝗋

2

⁢

(

E

⁢

(

H

,

h

1

⁢

(

x

)

)

⁢

(

x

,

𝗋𝖾𝖿𝗅

h

1

⁢

(

x

)

)

)

:

h

2

⁢

(

e

⁢

(

x

)

)

=

h

1

⁢

(

x

)

,

giving us a homotopy

η

:

h

2

∘

e

∼

h

1

.

4.

By the characterization of paths in fibers (Lemma 4.2.5 (

http://planetmath.org/42halfadjointequivalences#Thmprelem2

)), the path

E

¯

⁢

(

H

,

a

)

from Lemma 7.6.5 (

http://planetmath.org/76orthogonalfactorization#Thmprelem3

) gives us

η

⁢

(

g

1

⁢

(

a

)

)

=

𝖺𝗉

h

2

⁢

(

ζ

⁢

(

a

)

)

⁢

\centerdot

⁢

H

⁢

(

a

)

-

1

.
The desired equality follows by substituting the

definition

of

H

and rearranging paths.∎

By standard

arguments

, this yields the following orthogonality principle.

Theorem 7.6.7

.

Let

e

:

A

→

B

be

n

-connected and

m

:

C

→

D

be

n

-truncated.
Then the map

φ

:

(

B

→

C

)

→

∑

(

h

:

A

→

C

)

∑

(

k

:

B

→

D

)

(

m

∘

h

∼

k

∘

e

)

is an equivalence.

Sketch of proof.

For any

(

h

,

k

,

H

)

in the codomain, let

h

=

h

2

∘

h

1

and

k

=

k

2

∘

k

1

, where

h

1

and

k

1

are

n

-connected and

h

2

and

k

2

are

n

-truncated.
Then

f

=

(

m

∘

h

2

)

∘

h

1

and

f

=

k

2

∘

(

k

1

∘

e

)

are both

n

-factorizations of

m

∘

h

=

k

∘

e

.
Thus, there is a unique equivalence between them.
It is straightforward (if a bit tedious) to extract from this that

𝖿𝗂𝖻

φ

⁢

(

(

h

,

k

,

H

)

)

is contractible.
∎

We end by showing that images are stable under

pullback

.

Lemma 7.6.8

.

Suppose that the square

\includegraphics

HoTT_fig_7.6.2.png

is a pullback square and let

b

:

B

. Then

𝖿𝗂𝖻

f

⁢

(

b

)

≃

𝖿𝗂𝖻

g

⁢

(

h

⁢

(

b

)

)

.

Proof.

This follows from pasting of pullbacks (

http://planetmath.org/node/87643

Exercise 2.12), since the type

X

in the diagram

\includegraphics

HoTT_fig_7.6.3.png

is the pullback of the left square if and only if it is the pullback of the outer

rectangle

, while

𝖿𝗂𝖻

f

⁢

(

b

)

is the pullback of the square on the left and

𝖿𝗂𝖻

g

⁢

(

h

⁢

(

b

)

)

is the pullback of the outer rectangle.
∎

Theorem 7.6.9

.

Consider functions

f

:

A

→

B

,

g

:

C

→

D

and the diagram

\includegraphics

HoTT_fig_7.6.4.png

If the outer rectangle is a pullback, then so is the bottom square (and hence so is the top square, by

http://planetmath.org/node/87643

Exercise 2.12). Consequently, images are stable under pullbacks.

Proof.

Assuming the outer square is a pullback, we have equivalences

B

×

D

𝗂𝗆

n

⁢

(

g

)

≡

∑

(

b

:

B

)

∑

(

w

:

𝗂𝗆

n

(

g

)

)

h

⁢

(

b

)

=

𝗉𝗋

1

⁢

w

≃

∑

(

b

:

B

)

∑

(

d

:

D

)

∑

(

w

:

∥

𝖿𝗂𝖻

g

(

d

)

∥

n

)

h

⁢

(

b

)

=

d

≃

∑

b

:

B

∥

𝖿𝗂𝖻

g

(

h

(

b

)

)

∥

n

≃

∑

b

:

B

∥

𝖿𝗂𝖻

f

(

b

)

∥

n

(by

Theorem

7.6.9 (

http://planetmath.org/76orthogonalfactorization#Thmprethm3

))

≡

𝗂𝗆

n

⁢

(

f

)

.

∎

Title

7.6

Orthogonal

factorization

\metatable

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