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McCombs Math 381
Function Composition
1.
Suppose
g
:
A
→
B
and
f
:
B
→
C
where
A
=
{
1
,
2
,
3
,
4
}
,
B
=
{
a
,
b
,
c
}
,
C
=
{
2
,
7
,
10
}
,
and
f
and
g
are defined by
g
=
{
(1
,
b
)
,
(2
,
a
)
,
(3
,
a
)
,
(4
,
b
)
}
and
f
=
{
(
a
,
10)
,
(
b
,
7)
,
(
c
,
2)
}
.
Find
f
g
.
Ans:
{
(1
,
7)
,
(2
,
10)
,
(3
,
10)
,
(4
,
7)
}
.
In the questions below suppose
g
:
A
→
B
and
f
:
B
→
C
where
A
=
B
=
C
=
{
1
,
2
,
3
,
4
}
,
g
=
{
(1
,
4)
,
(2
,
1)
,
(3
,
1)
,
(4
,
2)
}
and
f
=
{
(1
,
3)
,
(2
,
2)
,
(3
,
4)
,
(4
,
2)
}
.
2.
Find
f
g
.
Ans:
{
(1
,
2)
,
(2
,
3)
,
(3
,
3)
,
(4
,
2)
}
.
3.
Find
g
f
.
Ans:
{
(1
,
1)
,
(2
,
1)
,
(3
,
2)
,
(4
,
1)
}
.
4.
g
g
.
Ans:
{
(1
,
2)
,
(2
,
4)
,
(3
,
4)
,
(4
,
1)
}
.
5.
For any function
f
:
A
→
B
, define a new function
g
:
P
(
A
)
→
P
(
B
) as follows:
for every
S
⊆
A
,
g
(
S
)
=
{
f
(
x
)

x
∈
S
}
.
Prove that
f
is onto if and only if
g
is onto.
Ans: Suppose
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Unformatted text preview: f is onto. Let T ∈ P ( B ) and let S = { x ∈ A  f ( x ) ∈ T } . Then g ( S ) = T , and g is onto. If f is not onto B , let y ∈ B − f ( A ). Then there is no subset S of A such that g ( S ) = { y } . 6. Suppose g : A → B and f : B → C , where f g is 11 and g is 11. Must f be 11? Ans: No. 7. Suppose g : A → B and f : B → C , where f g is 11 and f is 11. Must g be 11? Ans: Yes....
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This note was uploaded on 06/16/2011 for the course MATH 381 taught by Professor Na during the Summer '11 term at University of North Carolina School of the Arts.
 Summer '11
 NA
 Math

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