DEFINITION A function f : A  B is a one-to-one correspondence
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(called a bijection) iff f is one-to-one and onto B. We write f : Aonto
 B to
indicate that f is a bijection.
Look at the example and illustrations on pages 214 and 215.
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Theorem 4.4.1 If f : A  B and g : B  C, then g ◦ f : A  C. That
onto
onto
onto
is, the composite of one-to-one correspondences is a one-to-one
correspondence.
Proof: This is a combination of Theorems 4.3.1 and 4.3.3.
Theorem 4.4.2 Let F : A  B (i.e., F is a function from set A to set B.)
(a) F–1 is a function from Rng(F) to A iff F is one-to-one.
(b) If F–1 is a function, then F–1 is one-to-one.
Proof of (a): Suppose F : A  B.
Suppose F–1 is a function from Rng(F) to A.
To show F is one-to-one, suppose F(x) = F(y) = z.
(x, z)  F and (y, z)  F
change of notation
definition of F–1
_______________________
(z, x)  F–1 and (z, y)  F–1
x=y
supposition
_______________________
that F–1 is a function
F is one-to-one
F(x) = F(y)  x = y
Suppose F is one-to-one.
Proof of (a): Suppose F : A  B.
Suppose F–1 is a function from Rng(F) to A.
To show F is one-to-one, suppose F(x) = F(y) = z.
(x, z)  F and (y, z)  F
change of notation
definition of F–1
_______________________
(z, x)  F–1 and (z, y)  F–1
x=y
supposition
_______________________
that F–1 is a function
F is one-to-one
F(x) = F(y)  x = y
Suppose F is one-to-one.
Dom(F–1) = Rng(F)
Theorem 3.1.2(a)
_______________________
To show F–1 is a function, suppose (x, y)  F–1 and (x, z)  F–1 .
definition of F–1
_______________________
(y, x)  F and (z, x)  F
F(y) = F(z)
follows from previous line
y=z
supposition
_______________________
that F is one-to-one
F–1 is a function
(x, y)  F–1 and (x, z)  F–1  y = z
We have now shown that F–1 is a function from Rng(F) to A
Theorem 4.4.2 Let F : A  B (i.e., F is a function from set A to set B.)
(a) F–1 is a function from Rng(F) to A iff F is one-to-one.
(b) If F–1 is a function, then F–1 is one-to-one.
Proof of (b): Suppose F : A  B.
Suppose F–1 is a function from Rng(F) to A.
Theorem 3.1.3(a)
_______________________
F = (F–1)–1
applying part (a) to _______________
F–1
F–1 is one-to-one
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Corollary 4.4.3 If F : A onto
 B, then
: B onto
 A. That is, the inverse
of a one-to-one correspondence is a one-to-one correspondence.
F–1
Theorem 4.4.4 Let F : A  B and G : B  A. Then
(a) G = F–1 iff G ◦ F = IA and F ◦ G = IB .
(b) If F is one-to-one and onto B, then G = F–1 iff G ◦ F = IA or
F ◦ G = IB .
Theorem 4.4.4 Let F : A  B and G : B  A. Then
(a) G = F–1 iff G ◦ F = IA and F ◦ G = IB .
Proof of (a): Suppose F : A  B and G : B  A.
Note: The textbook proof erroneously
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Suppose G = F
references Theorem 4.2.3.
B = Dom(G) = Dom(F–1) = Rng(F)
Theorem 3.1.2(a)
_______________
G ◦ F = IA and F ◦ G = IB
Theorem 4.2.4
_______________
Suppose G ◦ F = IA and F ◦ G = IB .
G ◦ F = IA is one-to-one
every identity function is one-to-one
F is one-to-one
Theorem 4.3.4
_______________
F ◦ G = IB is onto B
every identity function is onto
F is onto B
F–1 is a function on B
Theorem 4.3.2
_______________
F is one-to-one & Theorem ______________
4.4.2(a)
F–1 = F–1 ◦ IB = F–1 ◦ (F ◦ G) =
(F–1 ◦ F ) ◦ G = IA ◦ G = G
We have now proven part (a).
properties of the identity function
Theorem 4.4.4 Let F : A  B and G : B  A. Then
(a) G = F–1 iff G ◦ F = IA and F ◦ G = IB .
(b) If F is one-to-one and onto B, then G = F–1 iff G ◦ F = IA or
F ◦ G = IB .
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Proof of (b): Suppose F : A  B.
onto
Suppose G = F–1.
We can say G ◦ F = IA or F ◦ G = IB
from part (a)
Now suppose that G ◦ F = IA or F ◦ G = IB . We first show G  F–1.
Case 1: G ◦ F = IA
Let (b, a)  G (i.e, b  B and a  A)
(a, a)  IA
(a, a)  G ◦ F
properties of the identity function
supposition that _____________
G ◦ F = IA
definition of G ◦ F
(a, c)  F /\ (c, a)  G for some c  B
_____________________
F is one-to-one and onto B
(a, b)  F /\ (b, a)  G, that is, c = b
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Proof of (b): Suppose F : A  B.
onto
Suppose G = F–1.
We can say G ◦ F = IA or F ◦ G = IB
from part (a)
Now suppose that G ◦ F = IA or F ◦ G = IB . We first show G  F–1.
Case 1: G ◦ F = IA
Let (b, a)  G (i.e, b  B and a  A)
(a, a)  IA
(a, a)  G ◦ F
properties of the identity function
supposition that _____________
G ◦ F = IA
definition of G ◦ F
(a, c)  F /\ (c, a)  G for some c  B
_____________________
F is one-to-one and onto B
(a, b)  F /\ (b, a)  G, that is, c = b
definition of F–1
_____________________
(b, a)  F–1
G  F–1
(b, a)  G  (b, a)  F–1
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Proof of (b): Suppose F : A  B.
onto
Suppose G = F–1.
We can say G ◦ F = IA or F ◦ G = IB
from part (a)
Now suppose that G ◦ F = IA or F ◦ G = IB . We first show G  F–1.
Case 2: F ◦ G = IB
Let (b, a)  G (i.e, b  B and a  A)
(b, b)  IB
(b, b)  F ◦ G
properties of the identity function
supposition that _____________
F ◦ G = I\B
definition of F ◦ G
(b, d)  G /\ (d, b)  F for some d  A
_____________________
G is a function
(b, a)  G /\ (b, d)  G, that is, d = a
(b, a)  F–1
(a, _____________________
b)  F & definition of F–1
G  F–1
(b, a)  G  (b, a)  F–1
In either case, (b, a)  F–1 which shows that G  F–1.
We now show F–1  G.
Case 1: G ◦ F = IA
Let (b, a)  F–1 (i.e, b  B and a  A)
(a, a)  IA
(a, a)  G ◦ F
properties of the identity function
supposition that _____________
G ◦ F = IA
definition of G ◦ F
(a, c)  F /\ (c, a)  G for some c  B
_____________________
(a, c)  F /\ (a, b)  F, that is, c = b
(b, a)  G
F–1  G
a function
(b, a)  F–1 and F is ________
_____________________
(c, a)  G and c = b
(b, a)  F–1  (b, a)  G
We now show F–1  G.
Case 2: F ◦ G = IB
Let (b, a)  F–1 (i.e, b  B and a  A)
(b, b)  IB
(b, b)  F ◦ G
properties of the identity function
supposition that _____________
F ◦ G = I\B
definition of F ◦ G
(b, d)  G /\ (d, b)  F for some d  A
_____________________
(a, b)  F /\ (d, b)  F, that is, d = a (b, a)  F–1 and F is ________
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(b, a)  F–1
(a, _____________________
b)  F & definition of F–1
F–1  G
(b, a)  G  (b, a)  F–1
In either case, (b, a)  G which shows that F–1  G. Look at the
examples on
We conclude G = F–1, since G  F–1 and F–1  G.
page 216.
Exercises 4.4 (pages 218-219)
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(a)
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(b)
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(b)
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(c)
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(d)
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(d)
Section 5.1
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