DEFINITION A function f : A  B is onto B (called a surjection)
onto
iff Rng(f) = B. We write f : A  B to indicate that f is a surjection.
To show that f is onto B, we can show that for any b  B, there must be
some a  A such that f(a) = b.
Look at the illustration on page 205 and the examples on pages 206
and 207.
Theorem 4.3.1 If f : A  B and g : B  C, then g ◦ f : A  C. That is,
the composite of surjective functions is a surjection.
onto
onto
onto
onto
onto
4.2.1 we
Proof: Suppose f : A  B and g : B  C. By Theorem ______
have g ◦ f : A  C. Let c  C. We need to show that  a  A such that
(g ◦ f)(a) = c.
__________________________
by supposition, g is onto C
 b  B such that g(b) = c
by supposition, f is onto B
__________________________
 a  A such that f(a) = b
(g ◦ f)(a) = g(f(a)) = g(b) = c
two previous lines
onto
g◦f:AC
 a  A such that (g ◦ f)(a) = c
onto
Theorem 4.3.2 If f : A  B, g : B  C, and g ◦ f : A  C, then g is
onto C. That is, when the composite of two functions maps onto a set
C, then the second function applied must map onto the set C.
onto
Proof: Suppose f : A  B, g : B  C, and g ◦ f : A  C. Let c  C.
To show g is onto C, we must find b  B such that g(b) = c.
 a  A such that (g ◦ f)(a) = c
f (a) = b for some b  B
__________________________
by supposition, g ◦ f is onto C
__________________________
f:AB
(g ◦ f)(a) = g(f(a)) = g(b) and
(g ◦ f)(a) = c
g(b) = c
two previous lines
onto
g:BC
substitution
 b  B such that g(b) = c
DEFINITION A function f : A  B is one-to-one (called an
1-1
injection) iff whenever f(x) = f(y), then x = y. We write f : A  B to
indicate that f is an injection.
To show that f is one-to-one, we show that for any x, y  A for which
f(x) = f(y), we must have x = y.
Look at the examples on pages 208 and 209.
1-1
1-1
1-1
Theorem 4.3.3 If f : A  B and g : B  C, then g ◦ f : A  C. That
is, the composite of injective functions is an injection.
1-1
1-1
Proof: Suppose f : A  B and g : B  C. To show g ◦ f is one-to-one,
suppose (g ◦ f)(x) = (g ◦ f)(y).
g(f(x)) = g(f(y))
change of notation
by supposition g is 1-1
__________________________
f(x) = f(y)
__________________________
by supposition f is 1-1
x=y
1-1
g◦f:AC
(g ◦ f)(x) = (g ◦ f)(y)  x = y
1-1
Theorem 4.3.4 If f : A  B, g : B  C, and g ◦ f : A  C, then
1-1
f:A
B. That is, if the composite of two functions is one-to-one, then
the first function applied must be one-to-one.
1-1
Proof: Suppose f : A  B, g : B  C, and g ◦ f : A  C. To show f is
one-to-one, suppose f(x) = f(y) for some x, y  A.
by supposition g is a function
__________________________
g(f(x)) = g(f(y))
(g ◦ f)(x) = (g ◦ f)(y)
change of notation
by supposition g ◦ f is 1-1
__________________________
x=y
1-1
f:AB
f (x) = f (y)  x = y
Theorem 4.3.5
1-1
(a) Suppose f : A  B and C  A. Then f |C is one-to-one. That is, a
restriction of a one-to-one function is one-to-one.
onto
onto
1-1
1-1
onto
(b) If h : A  C , g : B  D , and A  B = , then h  g : A  B  C  D .
(c) If h : A  C , g : B  D , A  B = , and C  D = , then
1-1
hg:ABCD.
Proof of (a):
1-1
Suppose f : A  B and C  A. Let f |C(x) = f |C(y) for x, y  C.
Then f(x) = f(y). Since f is _____________,
one-to-one we have x = y.
Proof of (b):
onto
onto
Suppose h : A  C , g : B  D , and A  B = .
From Theorem ____________
4.2.5
hg:ABCD
Let y  C  D We want to show  x  A  B such that (h  g)(x) = y.
________________________________
y  C \/ y  D
definition of C  D
Proof of (b):
onto
onto
Suppose h : A  C , g : B  D , and A  B = .
From Theorem ____________
4.2.5
hg:ABCD
Let y  C  D We want to show  x  A  B such that (h  g)(x) = y.
_______________________
y  C \/ y  D
definition of C  D
Case 1: y  C
 x  A such that h(x) = y
(h  g)(x) = h(x) = y
Case 2: y  D
by supposition h is onto C
_______________________
4.2.5
A  B =  and Theorem _______
 x  B such that g(x) = y
(h  g)(x) = g(x) = y
by supposition h is onto C
_______________________
4.2.5
A  B =  and Theorem _______
In either case, we have (h  g)(x) = y for some x  A  B.
We have shown that in each case (h  g)(x) = y for some x  A  B,
and thus proven part (b).
Theorem 4.3.5
1-1
(a) Suppose f : A  B and C  A. Then f |C is one-to-one. That is, a
restriction of a one-to-one function is one-to-one.
onto
onto
1-1
1-1
onto
(b) If h : A  C , g : B  D , and A  B = , then h  g : A  B  C  D .
(c) If h : A  C , g : B  D , A  B = , and C  D = , then
1-1
hg:ABCD.
Proof of (c):
1-1
1-1
Suppose h : A  C , g : B  D , A  B = , and C  D = .
Theorem ____________
4.2.5
hg:ABCD
Suppose (h  g)(x) = (h  g)(y) where x, y  A  B .
We want to show that
One of the following cases must be true:
(i) x, y  A , (ii) x, y  B , (iii) x  A and y  B , (iv) x  B and y  A .
Case (i): x, y  A
Theorem ____________
4.2.5
(h  g)(x) = h(x) and (h  g)(y) = h(y)
h(x) = h(y)
by supposition
(h  g)(x) = (h  g)(y)
__________________________
x=y
by supposition h is 1-1
__________________________
Case (ii): x, y  B
Theorem ____________
4.2.5
(h  g)(x) = g(x) and (h  g)(y) = g(y)
by supposition
(h  g)(x) = (h  g)(y)
g(x) = g(y)
__________________________
x=y
_________________________
by supposition g is 1-1
Case (iii): x  A and y  B
Theorem ____________
4.2.5
(h  g)(x) = h(x) and (h  g)(y) = g(y)
by supposition
(h  g)(x) = (h  g)(y)
h(x) = g(y)
__________________________
h(x)  C and g(y)  D
by supposition
h : A  C and g : B  D
__________________________
by supposition C  D = 
This is a contradiction
__________________________
This case is not possible; similarly, Case (iv) is not possible.
We have shown that in each possible case x = y, and thus proven part (c).
Exercises 4.3 (pages 210-213)
1
(b)

(d)

(e)

(f)
1 - continued

(g)

(h)

(i)

(j)
1 - continued

(l)
2
(b)

(d)
2 - continued

(e)

(f)

(g)

(h)
2 - continued

(i)

(j)

(l)
9
(a)

(b)

(d)
Theorem 4.15
1-1
(a) Suppose f : A  B and C  A. Then f |C is one-to-one.
onto
onto
1-1
1-1
onto
(b) If h : A  C , g : B  D , and A  B = , then h  g : A  B  C  D .
(c) If h : A  C , g : B  D , A  B = , and C  D = , then
1-1
hg:ABCD.