Another proof involving composition
If
f
:
A
→
B
is onto and
g
:
B
→
C
is onto then
g
o
f
:
A
→
C
is onto
Proof
:
Let
y
∈
C
.
Since
g
is onto,
∃
z
∈
B
such that
g
(
z
) =
y
Also since
f
is onto,
∃
x
∈
A
such that
f
(
x
) =
z
Now
g
o
f
(
x
)
=
g
(
f
(
x
))
=
g
(
z
)
=
y
Hence,
∃
x
∈
A
such that
g
o
f
(
x
) =
y
and hence
g
o
f
is onto
Are the converse statements true?
— If
g
o
f
:
A
→
C
is 1to1,
is
f
:
A
→
B
1to1 and
g
:
B
→
C
1to1?
— If
g
o
f
:
A
→
C
is onto,
is
f
:
A
→
B
onto and
g
:
B
→
C
onto?
(No, for both questions!)
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View Full DocumentWithout loss of generality
Recall that we sometimes do proofs
by cases
. Consider an example where
two cases are very similar:
Function
f
:
A
→
B
is increasing if
∀
x
∈
A
,
∀
y
∈
A
,
x
<
y
→
f
(
x
)
<
f
(
y
)
Claim
:
Any increasing function is onetoone
Proof
:
We need to show that
∀
x
∈
A
,
∀
y
∈
A
,
f
(
x
) = f(
y
)
→
x
=
y
or equivalently,
∀
x
∈
A
,
∀
y
∈
A
,
x
≠
y
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 Spring '10
 fleck
 Mathematical Induction, Recursion, Inductive Reasoning, Mathematical logic, Mathematical proof

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