MAT 300, Mathematical Structures
Spring 2010
Selected Solutions for Exam 2 Review
We answered all questions from part (II) in class, and also parts (b), (f), and (g) from
part (III). The remaining parts of (III) are answered here.
III.
(a) Suppose
f
:
A
→
B
and
g
:
B
→
C
, and assume that
g
◦
f
is onetoone and onto.
Prove that
f
is onetoone and
g
is onto.
Proof:
Assume that
f
:
A
→
B
and
g
:
B
→
C
are functions and
g
◦
f
is
onetoone and onto. First we will prove that
f
is onetoone. Let
x
1
,
x
2
∈
A
and
assume that
f
(
x
1
) =
f
(
x
2
). Then, since
g
is a function,
g
(
f
(
x
1
)) =
g
(
f
(
x
2
)). But
g
◦
f
is onetoone. This means that whenever
g
(
f
(
x
1
)) =
g
(
f
(
x
2
)), we must have
x
1
=
x
2
. Since
x
1
=
x
2
whenever
f
(
x
1
) =
f
(
x
2
), we conclude that
f
is onetoone.
Now, we wish to prove that
g
is onto. Let
z
∈
C
. Since
g
◦
f
is onto, there exists
some
x
∈
A
for which
g
(
f
(
x
)) =
z
. Let
y
=
f
(
x
)
∈
B
. Then
z
=
g
(
f
(
x
)) =
g
(
y
)
for
y
=
f
(
x
)
∈
B
. This shows that
g
is onto.
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 Spring '07
 thieme
 Math, Logic, Proof by contradiction

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