LECTURE 23: INVERSES AND PERMUTATIONS (
§
9.69.7)
Let all things be done decently and in order. Paul the apostle
1.
More on inverses
From last time:
Theorem 1.
Let
f
:
A
→
B
be a function. The inverse relation
f

1
is a function from
B
to
A
if and
only if
f
is a bijection.
Corollary 2.
If
f

1
is a function then
f

1
is a bijection. Further,
f

1
◦
f
= id
A
and
f
◦
f

1
= id
B
.
Sometimes it is easy to figure out
f

1
because
f
has only finitely many pairs!
In other cases, it
might take more work.
Example 3.
The function
f
:
R
 {
2
} →
R
 {
3
}
given by
f
(
x
) = 3
x/
(
x

2) is a bijection. Prove this,
and find the inverse.
Proof.
First we show
f
is injective. Let
x, y
∈
R
 {
2
}
and assume
f
(
x
) =
f
(
y
). Then 3
x/
(
x

2) =
3
y/
(
y

2). Clearing denominators, we have 3
xy

6
x
= 3
xy

6
y
. So

6
x
=

6
y
. Thus
x
=
y
.
Next we show that
f
is surjective. Let
b
∈
R
 {
3
}
. We want to find
a
∈
R
 {
2
}
so that
f
(
a
) =
b
.
In other words, we want 3
a/
(
a

2) =
b
. Our scratch work suggests that
a
= 2
b/
(
b

3) should work.
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 Spring '11
 Smith
 Permutations, Inverse function, Inverses, Bijection

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