surjection and axiom of choice

surjection and axiom of choice

In this entry, we show the statement that

(*) every

surjection

has a

right inverse

is

equivalent

to the

axiom of choice

(AC).

Proposition 1

.

AC implies (*).

Proof.

Let

f

:

A

→

B

be a surjection. Then the set

C

:=

{

f

-

1

⁢

(

y

)

∣

y

∈

B

}

partitions

A

. By the axiom of choice, there is a

function

g

:

C

→

⋃

C

such that

g

⁢

(

f

-

1

⁢

(

y

)

)

∈

f

-

1

⁢

(

y

)

for every

y

∈

B

. Since

⋃

C

=

A

,

g

is a function from

C

to

A

. Define

h

:

B

→

A

by

h

⁢

(

y

)

=

g

⁢

(

f

-

1

⁢

(

y

)

)

. Then

h

⁢

(

y

)

∈

f

-

1

⁢

(

y

)

, and therefore

(

f

∘

h

)

⁢

(

y

)

=

f

⁢

(

h

⁢

(

y

)

)

=

y

, implying that

f

has a right inverse.
∎

Remark

. The function

h

is easily seen to be an

injection

: if

h

⁢

(

y

1

)

=

h

⁢

(

y

2

)

, then

y

1

=

f

⁢

(

h

⁢

(

y

1

)

)

=

f

⁢

(

h

⁢

(

y

2

)

)

=

y

2

.

Proposition 2

.

(*) implies AC.

Before proving this, let us remark that, in the

collection

C

of non-empty sets of the axiom of choice, there is no

assumption

that the sets in

C

be

pairwise disjoint

. The statement

(**) given a set

C

of pairwise disjoint non-empty sets, there is a

choice function

f

:

C

→

⋃

C

seemingly weaker than AC, turns out to be equivalent to AC, and we will prove this fact first.

Proof.

Obviously AC implies (**).

Conversely

, assume (**). Let

C

be a collection of non-empty sets. We assume

C

≠

∅

. For each

a

∈

C

, define a set

A

a

:=

{

(

x

,

a

)

∣

x

∈

a

}

. Since

a

≠

∅

,

A

a

≠

∅

. In

addition

,

A

a

∩

A

b

=

∅

iff

a

≠

b

(true since elements of

A

a

and elements of

A

b

have distinct second coordinates). So the collection

D

:=

{

A

a

∣

a

∈

C

}

is a set consisting of pairwise disjoint non-empty sets. By (**), there is a function

f

:

D

→

⋃

D

such that

f

⁢

(

A

a

)

∈

A

a

for every

a

∈

C

. Now, define two functions

g

:

C

→

D

and

h

:

⋃

D

→

⋃

C

by

g

⁢

(

a

)

=

A

a

and

h

⁢

(

x

,

a

)

=

x

Then, for any

a

∈

C

, we have

(

h

∘

f

∘

g

)

⁢

(

a

)

=

h

⁢

(

f

⁢

(

A

a

)

)

. Since

f

⁢

(

A

a

)

∈

A

a

, its first coordinate is an element of

a

. Therefore

h

⁢

(

f

⁢

(

A

a

)

)

∈

a

, and hence

h

∘

f

∘

g

is the desired choice function.
∎

Proof of Propositon 2.

We show that (*) implies (**), and since (**) implies AC as shown above, the proof of

Proposition

2 is then

complete

.

Let

C

be a collection of pairwise disjoint non-empty sets. Each element of

⋃

C

belongs to a unique set in

C

. Then the function

g

:

⋃

C

→

C

taking each element of

⋃

C

to the set it belongs in

C

, is a well-defined function. It is clearly

surjective

. Hence, by assumption, there is a function

f

:

C

→

⋃

C

such that

g

∘

f

=

1

C

(a right inverse of

g

). For each

x

∈

C

,

g

⁢

(

f

⁢

(

x

)

)

=

x

, which is the same as saying that

f

⁢

(

x

)

is an element of

x

by the definition of

g

.
∎

Remark

. In the

category of sets

, AC is equivalent to saying that every epimorophism is a

split epimorphism

. In general, a

category

is said to have the axiom of choice if every

epimorphism

is a split epimorphism.

Title

surjection and axiom of choice

Canonical name

SurjectionAndAxiomOfChoice

Date of creation

2013-03-22 18:44:37

Last modified on

2013-03-22 18:44:37

Owner

CWoo (3771)

Last modified by

CWoo (3771)

Numerical id

9

Author

CWoo (3771)

Entry type

Derivation

Classification

msc 03E25

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