Transcript Document
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, z) G ◦ F
_____________________
(x, v) F /\ (v, z) G for some v B
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:AB
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))
(x, f –1(x)) f –1
(f –1 ◦ f)(x) = x
(f –1(x), x) f
IC(x) = x
the two conditions of Theorem 4.1.1
_____
are satisfied
(f –1 ◦ f)(x) = IC(x)
f ◦ f –1 = IC
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
__________________________
definition of Dom(h g)
__________________________
(x, y) h or (x, y) g
definition of h g
x Dom(h) = A or x Dom(g) = B __________________________
definitions of Dom(h) & Dom(g)
__________________________
xAB
definition of A B
x Dom(h g) x A B
Dom(h g) 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
xAB
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
xAB
x A \/ x 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
(x, y), (x, z) g y = z
__________________________
definition of A B
__________________________
supposition that A B =
__________________________
definition of Dom(h)
__________________________
g is a function
__________________________
definition of Dom(g)
__________________________
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)