properties of injective functions

properties of injective functions

Theorem 1

.

Suppose

A

,

B

,

C

are sets and

f

:

A

→

B

,

g

:

B

→

C

are

injective functions

. Then the

composition

g

∘

f

is an injection.

Proof.

Suppose that

(

g

∘

f

)

⁢

(

x

)

=

(

g

∘

f

)

⁢

(

y

)

for some

x

,

y

∈

A

.
By definition of composition,

g

⁢

(

f

⁢

(

x

)

)

=

g

⁢

(

f

⁢

(

y

)

)

. Since

g

, is
assumed injective,

f

⁢

(

x

)

=

f

⁢

(

y

)

. Since

f

is also assumed injective,

x

=

y

. Therefore,

(

g

∘

f

)

⁢

(

x

)

=

(

g

∘

f

)

⁢

(

y

)

implies

x

=

y

, so

g

∘

f

is injective.
∎

Theorem 2

.

Suppose

f

:

A

→

B

is an injection, and

C

⊆

A

. Then
the

restriction

f

|

C

:

C

→

B

is an injection.

Proof.

Suppose

(

f

|

C

)

⁢

(

x

)

=

(

f

|

C

)

⁢

(

y

)

for some

x

,

y

∈

C

. By definition
of restriction,

f

⁢

(

x

)

=

f

⁢

(

y

)

. Since

f

is assumed injective this,
in turn, implies that

x

=

y

. Thus,

f

|

C

is also injective.
∎

Theorem 3

.

Suppose

A

,

B

,

C

are sets and that the functions

f

:

A

→

B

and

g

:

B

→

C

are such that

g

∘

f

is injective. Then

f

is
injective.

Proof.

(

direct proof

)
Let

x

,

y

∈

A

be such that

f

⁢

(

x

)

=

f

⁢

(

y

)

. Then

g

⁢

(

f

⁢

(

x

)

)

=

g

⁢

(

f

⁢

(

y

)

)

.
But as

g

∘

f

is injective, this implies that

x

=

y

, hence

f

is also injective.
∎

Proof.

(

proof by contradiction

)
Suppose that

f

were not injective. Then there would exist

x

,

y

∈

A

such that

f

⁢

(

x

)

=

f

⁢

(

y

)

but

x

≠

y

. Composing with

g

, we would
then have

g

⁢

(

f

⁢

(

x

)

)

=

g

⁢

(

f

⁢

(

y

)

)

. However, since

g

∘

f

is assumed
injective, this would imply that

x

=

y

, which contradicts a previous
statement. Hence

f

must be injective.
∎

Theorem 4

.

Suppose

f

:

A

→

B

is an injection. Then, for all

C

⊆

A

, it is the case that

f

-

1

⁢

(

f

⁢

(

C

)

)

=

C

.

1

1

In this equation, the symbols “

f

” and
“

f

-

1

” as applied to sets denote the

direct image

and the

inverse
image

, respectively

Proof.

It follows from the definition of

f

-

1

that

C

⊆

f

-

1

⁢

(

f

⁢

(

C

)

)

, whether or not

f

happens to be injective. Hence, all that
need to be shown is that

f

-

1

⁢

(

f

⁢

(

C

)

)

⊆

C

. Assume the
contrary. Then there would exist

x

∈

f

-

1

⁢

(

f

⁢

(

C

)

)

such that

x

∉

C

. By defintion,

x

∈

f

-

1

⁢

(

f

⁢

(

C

)

)

means

f

⁢

(

x

)

∈

f

⁢

(

C

)

, so there exists

y

∈

A

such that

f

⁢

(

x

)

=

f

⁢

(

y

)

. Since

f

is injective, one would have

x

=

y

, which is impossible because

y

is supposed to belong to

C

but

x

is not supposed to belong to

C

.
∎

Theorem 5

.

Suppose

f

:

A

→

B

is an injection. Then, for all

C

,

D

⊆

A

,
it is the case that

f

⁢

(

C

∩

D

)

=

f

⁢

(

C

)

∩

f

⁢

(

D

)

.

Proof.

Whether or not

f

is injective, one has

f

⁢

(

C

∩

D

)

⊆

f

⁢

(

C

)

∩

f

⁢

(

D

)

; if

x

belongs to both

C

and

D

, then

f

⁢

(

x

)

will clearly
belong to both

f

⁢

(

C

)

and

f

⁢

(

D

)

. Hence, all that needs to be shown is
that

f

⁢

(

C

)

∩

f

⁢

(

D

)

⊆

f

⁢

(

C

∩

D

)

. Let

x

be an element of

B

which belongs to both

f

⁢

(

C

)

and

f

⁢

(

D

)

. Then, there exists

y

∈

C

such that

f

⁢

(

y

)

=

x

and

z

∈

D

such that

f

⁢

(

z

)

=

x

. Since

f

⁢

(

y

)

=

f

⁢

(

z

)

and

f

is injective,

y

=

z

, so

y

∈

C

∩

D

, hence

x

∈

f

⁢

(

C

∩

D

)

.
∎

Title

properties of injective functions

Canonical name

PropertiesOfInjectiveFunctions

Date of creation

2013-03-22 16:40:20

Last modified on

2013-03-22 16:40:20

Owner

rspuzio (6075)

Last modified by

rspuzio (6075)

Numerical id

22

Author

rspuzio (6075)

Entry type

Theorem

Classification

msc 03E20

Classification

msc 03E99

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