DEFINITIONS For a function F : A  B, the inverse of F is the
following relation from B to A:
F–1 = {(x, y) : (y, x)  F}.
For functions F : A  B and G : B  C the composite of F and G is
the following relation from A to C:
G ◦ F = {(x, z)  A  C : (x, y)  F and (y, z)  G, for some y  B}.
Look at the examples on page 196.
Theorem 4.2.1 Let A, B, and C be sets, and suppose functions
F : A  B and G : B  C are defined. Then G ◦ F is a function from A
to C where Dom(G ◦ F) = A.
Proof We know that G ◦ F is a relation from A to C. We must show
that (i) Dom(G ◦ F) = A, and (ii) if (x, y)  G ◦ F and (x, z)  G ◦ F,
then y = z.
Proof We know that G ◦ F is a relation from A to C. We must show
that (i) Dom(G ◦ F) = A, and (ii) if (x, y)  G ◦ F and (x, z)  G ◦ F,
then y = z.
We first show that Dom(G ◦ F) = A
Dom(G ◦ F)  Dom(F) = A
Exercise ___________
3.1-9(a)
We now want to show that A  Dom(G ◦ F)
Let a  A = Dom(F)
(a, b)  F for some b  B
_____________________
definition of Dom(F)
(b, c)  G for some c  C
(a, c)  G ◦ F
b  B = Dom(G)
_____________________
_____________________
definition of G ◦ F
a  Dom(G ◦ F)
A  Dom(G ◦ F)
definition of Dom(G ◦ F)
_____________________
a  A  a  Dom(G ◦ F)
Dom(G ◦ F) = A
Dom(G ◦ F)  A and A  Dom(G ◦ F)
To prove (ii), let (x, y)  G ◦ F and (x, z)  G ◦ F; we must show y = z.
(x, y)  G ◦ F
_____________________
(x, u)  F /\ (u, y)  G for some u  B
(x, v)  F /\ (v, z)  G for some v  B
(x, z)  G ◦ F
_____________________
u=v
y=z
G ◦ F is a function
_____________________
F is a function
_____________________
u = v and G is a function
(x, y)  G ◦ F /\ (x, z)  G ◦ F  y = z
Note: Theorem 3.1.3(b) tells us that the composition of relations is
associative.
Theorem 4.2.2
Let A, B, C, and D be sets, and suppose the functions
f : A  B, g : B  C, and h : C  D are defined.
Then (h ◦ g) ◦ f = h ◦ (g ◦ f), that is, the composition of functions is
associative.
Proof We must show that Dom((h ◦ g) ◦ f) = Dom(h ◦ (g ◦ f)) and that
((h ◦ g) ◦ f)(x) = (h ◦ (g ◦ f))(x).
Dom((h ◦ g) ◦ f) = Dom(f) = A
Theorem_____________
4.2.1
Dom(h ◦ (g ◦ f)) = Dom(g ◦ f) = Dom(f) = A Theorem_____________
4.2.1
Now, suppose x  A
((h ◦ g) ◦ f)(x) = (h ◦ g)(f(x)) =
previously defined
h(g(f(x))) = h((g ◦ f)(x)) = (h ◦ (g ◦ f))(x) notation
(h ◦ g) ◦ f = h ◦ (g ◦ f)
Theorem_____________
4.1.1
Theorem 4.2.3
Let f : A  B. Then f ◦ IA = f and IB ◦ f = f.
Proof
Dom(f ◦ IA) = Dom(IA)
Theorem_____________
4.2.1
Dom(f ◦ IA) = A
substitution using Dom(IA) = A
Dom(f ◦ IA) = Dom(f)
supposition that ______________
f:AB
Now, suppose x  A.
(f ◦ IA)(x) = f(IA(x)) = f(x)
previously defined notation
f ◦ IA = f
the two conditions of Theorem 4.1.1
_____
are satisfied
Theorem 4.2.3
Let f : A  B. Then f ◦ IA = f and IB ◦ f = f.
Dom(IB ◦ f) = Dom(f)
Theorem_____________
4.2.1
Now, suppose x  A.
(IB ◦ f)(x) = IB(f(x)) = f(x)
previously defined notation
IB ◦ f = f
the two conditions of Theorem 4.1.1
_____
are satisfied
Theorem 4.2.4 Let f : A  B with Rng(f) = C. If f –1 is a function,
then f –1 ◦ f = IA and f ◦ f –1 = IC .
Proof:
Suppose f : A  B with Rng(f) = C, and f –1 is a function
Dom(f –1 ◦ f) = Dom(f)
Theorem_____________
4.2.1
Dom(f –1 ◦ f) = A
substitution using Dom(f) = A
Dom(f –1 ◦ f) = Dom(IA)
____________________________
Dom(IA) = A
Now, suppose x  A.
(x, f(x))  f
(f(x), x)  f –1
(f –1 ◦ f)(x) = f –1(f(x))
(f –1 ◦ f)(x) = x
(f –1 ◦ f)(x) = IA(x)
f –1 ◦ f = IA
definition of Dom(f) = A
definition of _____________
f –1
(x, f(x))  f
(f(x), x)  f –1
IA(x) = x
the two conditions of Theorem 4.1.1
_____
are satisfied
Theorem 4.2.4
Let f : A  B with Rng(f) = C. If f –1 is a function,
then f –1 ◦ f = IA and f ◦ f –1 = IC .
Proof:
Suppose f : A  B with Rng(f) = C, and f –1 is a function
Dom(f ◦ f –1) = Dom(f –1)
Theorem______________
4.2.1
Dom(f ◦ f –1) = Rng(f)
Theorem______________
3.1.2(a)
Dom(f ◦ f –1) = C
supposition that ______________
Rng(f) = C
Dom(f –1 ◦ f) = Dom(IC)
____________________________
Dom(IC) = C
Now, suppose x  C.
(x, f –1(x))  f –1
(f –1(x), x)  f
definition of Dom(f –1) = Rng(f) = C
definition of _____________
f –1
(f ◦ f –1)(x) = f(f –1(x))
(f –1 ◦ f)(x) = x
(f –1 ◦ f)(x) = IC(x)
(x, f –1(x))  f –1
(f –1(x), x)  f
IC(x) = x
f ◦ f –1 = IC
the two conditions of Theorem 4.1.1
_____
are satisfied
DEFINITIONS
D is the function
Let f : A  B, and let D  A. The restriction of f to
f |D = {(x, y) : y = f(x) and x  D}.
If g and h are functions, and g is a restriction of h, then we say h is an
extension of g.
Look at the examples on pages 199 and 200.
Theorem 4.2.5 Let h and g be functions with Dom(h) = A and
Dom(g) = B. If A  B = , then h  g is a function with domain A  B.
Furthermore,
h(x) if x  A
(h  g)(x) = g(x) if x  B
Proof We know that h  g is a relation. We must show that
(i) Dom(h  g) = A  B, and (ii) if (x, y), (x, z)  h  g, then y = z
We first show that Dom(h  g) = A  B
Let x  Dom(h  g)
y such that (x, y)  h  g
(x, y)  h or (x, y)  g
x  Dom(h) = A or x  Dom(g) = B
xAB
Dom(h  g)  A  B
__________________________
definition of Dom(h  g)
__________________________
definition of h  g
__________________________
definitions of Dom(h) & Dom(g)
__________________________
definition of A  B
x  Dom(h  g)  x  A  B
We first show that Dom(h  g) = A  B
Let x  Dom(h  g)
y such that (x, y)  h  g
(x, y)  h or (x, y)  g
x  Dom(h) = A or x  Dom(g) = B
xAB
Dom(h  g)  A  B
__________________________
definition of Dom(h  g)
__________________________
definition of h  g
__________________________
definitions of Dom(h) & Dom(g)
__________________________
definition of A  B
x  Dom(h  g)  x  A  B
Let x  A  B
x  A \/ x  B
__________________________
definition of A  B
x  A = Dom(h)  y such that (x, y)  h _____________________
definition of Dom(h)
x  B = Dom(g)  y such that (x, y)  g _____________________
definition of Dom(g)
(x, y)  h  g
_____________________
(x, y)  h or (x, y)  g
x  Dom(h  g)
definition of Dom(h  g)
_____________________
x  A  B  x  Dom(h  g)
A  B  Dom(h  g)
Dom(h  g) = A  B Dom(h  g)  A  B /\ A  B  Dom(h  g)
We now show that if (x, y), (x, z)  h  g, then y = z.
Let (x, y), (x, z)  h  g. We must show that y = z.
__________________________
definition of Dom(h  g)
x  Dom(h  g)
A  B__________________________
= Dom(h  g) was just proven
xAB
x  A \/ x  B
__________________________
definition of A  B
(x  A /\ x  B) \/ (x  B /\ x  A)
x  A /\ x  B  (x, y), (x, z)  h
(x, y), (x, z)  h  y = z
x  B /\ x  A  (x, y), (x, z)  g
__________________________
supposition that A  B = 
__________________________
definition of Dom(h)
__________________________
g is a function
__________________________
definition of Dom(g)
(x, y), (x, z)  g  y = z
__________________________
h is a function
In either case, we have y = z, which is what we wanted to show.
Exercises 4.2 (pages 202-205)
1
(b)

(d)

(f)

(g)
1 - continued

(h)

(j)
2
(b)

(d)
2 - continued

(f)

(g)

(h)

(j)
3
(b)

(c)
8
14  (b)

(c)

(d)

(e)