COT3100-01, Spring 2000
S. Lang
Solution Key to Assignment #4 (40 pts.)
3/26/2000
1.
(10 pts.) Consider a set
A
= {1, 2}, a set
B
= {
a
,
b
,
c
}, and a set
C
= {
x
,
y
}.
Define a relation
R
⊂
A
×
B
as
R
= {(1,
a
), (1,
c
), (2,
b
)}, a relation
S
⊂
B
×
C
as
S
= {(
a
,
x
), (
b
,
y
), (
c
,
x
)},
and a relation
T
⊂
C
×
A
as
T
= {(
x
, 1), (
y
, 2)}.
Now answer each of the following questions
with a short explanation.
(a)
Is
R
a function?
If so, is it an injection?
Relation
R
is not a function because 1
∈
A
is related to two elements
a
and
c
in
B
(that is,
both (1,
a
) and (1,
c
)
∈
R
.)
(b)
Is
R
–1
a function?
If so, is it a surjection?
R
–1
= {(
a
, 1), (
b
, 2), (
c
, 1)} is a function from
B
to
A
(because for each element of
B
, there
is exactly one related element of
A
).
The function
R
–1
:
B
→
A
is a surjection because its
range equals {1, 2} =
A
.
(c)
Is
S
a function?
Is
S
–1
a function?
Relation
S
is a function from
B
to
C
(because for each element of
B
, there is exactly one
related element of
C
).
S
–1
= {(
x
,
a
), (
x
,
c
), (
y
,
b
)}
⊂
C
×
B
is not a function because
element
x
∈
C
is related to two element
a
and
c
of
B
.
(d)
Is the composed
relation
S
–1
R
–1
a function?
If so, is it an injection?
Is it a surjection?
S
–1
R
–1
= {(
x
, 1), (
y
, 2)}
⊂
C
×
A
is a function from
C
to
A
(because for each element of
C
, there is exactly one related element of
A
).
It is a surjection because its range equals
{1, 2} =
A
.
(e)
Is the composed
relation
T
R
S
a function?
If so, is it an injection?
T
R
S
= (
T
R
)
S
= {(
x
,
a
), (
x
,
c
), (
y
,
b
)}
S
= {(
x
,
x
), (
y
,
y
)} is a function from
C
to
C
.
It is also an injection because each pair of distinct elements (
x
and
y
) are mapped to
distinct images (
x
and
y
, respectively).
2.
(4 pts.) Let
f
:
A
→
B
and
g
:
B
→
C
denote two functions.
If the function
g
o
f
:
A
→
C
is a
surjection, and
g
is an injection, then prove the function
f
is a surjection.
To prove
f
:
A
→
B
is a surjection, let
b
∈
B
---- (1), we need to prove there exists (i.e., to
find)
a
∈
A
such that
f
(
a
) =
b
---- (2).
Since
b
∈
B
by (1),
g
(
b
)
∈
C
---- (3) because
g
:
B
→
C
by assumption.
Since
g
o
f
:
A
→
C
is a surjection by assumption, so (3) implies there exists
a
∈
A
such that
g
o
f
(
a
) =
g