DEFINITION A function f : A B is onto B (called a surjection) iff Rng(f) = B. We write f : A B to indicate that f is a surjection.

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.

Proof: Suppose f : A B and g : B C. By Theorem ______ we have g ◦ f : A C. Let c C. We need to show that a A such that (g ◦ f)(a) = c.

onto

onto onto onto

onto onto

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.

b B such that g(b) = c

4.2.1

by supposition, g is onto C__________________________ a A such that f(a) = b by supposition, f is onto B__________________________ (g ◦ f)(a) = g(f(a)) = g(b) = c two previous lines

g ◦ f : A C a A such that (g ◦ f)(a) = conto

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.

onto

a A such that (g ◦ f)(a) = c by supposition, g ◦ f is onto C__________________________

f (a) = b for some b B f : A B__________________________

(g ◦ f)(a) = g(f(a)) = g(b) and (g ◦ f)(a) = c two previous lines

g(b) = c substitution

g : B C b B such that g(b) = conto

DEFINITION A function f : A B is one-to-one (called an injection) iff whenever f(x) = f(y), then x = y. We write f : A B to indicate that f is an injection.

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Look at the examples on pages 208 and 209.

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.

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.

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Proof: Suppose f : A B and g : B C. To show g ◦ f is one-to-one, suppose (g ◦ f)(x) = (g ◦ f)(y).

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g(f(x)) = g(f(y)) change of notation

f(x) = f(y) by supposition g is 1-1__________________________

x = y by supposition f is 1-1__________________________

g ◦ f : A C1-1

(g ◦ f)(x) = (g ◦ f)(y) x = y

Theorem 4.3.4 If f : A B, g : B C, and g ◦ f : A C, then 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.

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.

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g(f(x)) = g(f(y)) by supposition g is a function__________________________

(g ◦ f)(x) = (g ◦ f)(y) change of notation

x = y by supposition g ◦ f is 1-1__________________________

f : A B1-1

f (x) = f (y) x = y

Theorem 4.3.5 (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.

(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

h g : A B C D .

Suppose f : A B and C A. Let f |C(x) = f |C(y) for x, y C.

onto onto onto

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Then f(x) = f(y). Since f is _____________, we have x = y.

Proof of (a):

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one-to-one

Suppose h : A C , g : B D , and A B = .onto onto

Proof of (b):

h g : A B C D From Theorem ____________4.2.5

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

Suppose h : A C , g : B D , and A B = .onto onto

Proof of (b):

h g : A B C D From Theorem ____________4.2.5

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

x A such that h(x) = y _______________________by supposition h is onto C

Case 1: y C

(h g)(x) = h(x) = y _______4.2.5A B = and Theorem

x B such that g(x) = y _______________________by supposition h is onto C

Case 2: y D

(h g)(x) = g(x) = y _______4.2.5A 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 (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.

(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

h g : A B C D .

onto onto onto

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Proof of (c):Suppose h : A C , g : B D , A B = , and C D = .

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h g : A B C D Theorem ____________4.2.5

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

(h g)(x) = h(x) and (h g)(y) = h(y) Theorem ____________4.2.5

h(x) = h(y)

by supposition h is 1-1__________________________ x = y

Case (ii): x, y B

(h g)(x) = g(x) and (h g)(y) = g(y) Theorem ____________4.2.5

g(x) = g(y)

by supposition g is 1-1_________________________ x = y

Case (iii): x A and y B(h g)(x) = h(x) and (h g)(y) = g(y) Theorem ____________4.2.5

by supposition (h g)(x) = (h g)(y)__________________________

by supposition (h g)(x) = (h g)(y)__________________________

h(x) = g(y) by supposition (h g)(x) = (h g)(y)__________________________

h(x) C and g(y) D by supposition h : A C and g : B D__________________________

This is a contradiction __________________________by supposition C D = 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).

1 (b)

(d)

Exercises 4.3 (pages 210-213)

(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

(a) Suppose f : A B and C A. Then f |C is one-to-one.

(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

h g : A B C D .

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onto onto onto

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