Page 255 - HOW TO PROVE IT: A Structured Approach, Second Edition

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One-to-one and Onto 241
Let’s return now to the proof that f is not onto. If we let x = 2, then to
complete the proof we must show that ∀a ∈ A( f (a) = 2). We’ll do this by
letting a be an arbitrary element of A, assuming f (a) = 2, and then trying to
derive a contradiction. The remaining details of the proof are not hard.

Solution

Proof To see that f is one-to-one, let a 1 and a 2 be arbitrary elements of A
and assume f (a 1 ) = f (a 2 ). Applying the definition of f, it follows that 2a 1 =
a 1 +1
2a 2 . Thus, 2a 1 (a 2 + 1) = 2a 2 (a 1 + 1). Multiplying out both sides gives us
a 2 +1
2a 1 a 2 + 2a 1 = 2a 1 a 2 + 2a 2 ,so2a 1 = 2a 2 and therefore a 1 = a 2 .
To show that f is not onto we will prove that ∀a ∈ A( f (a) = 2). Suppose
a ∈ A and f (a) = 2. Applying the definition of f, we get 2a = 2. Thus,
a+1
2a = 2a + 2, which is clearly impossible. Thus, 2 /∈ Ran( f ), so Ran( f ) = R
and therefore f is not onto.

As we saw in the preceding example, when proving that a function f is
one-to-one it is usually easiest to prove the equivalent statement ∀a 1 ∈ A∀a 2 ∈
A( f (a 1 ) = f (a 2 ) → a 1 = a 2 ) given in part 1 of Theorem 5.2.3. Of course,
this is just an example of the fact that it is generally easier to prove a positive
statement than a negative one. This equivalence is also often used in proofs in
which we are given that a function is one-to-one, as you will see in the proof
of part 1 of the following theorem.

Theorem 5.2.5. Suppose f : A → B and g : B → C. As we saw in Theo-
rem 5.1.5, it follows that g ◦ f : A → C.

1. If f and g are both one-to-one, then so is g ◦ f.
2. If f and g are both onto, then so is g ◦ f.

Proof

1. Suppose f and g are both one-to-one. Let a 1 and a 2 be arbitrary elements
of A and suppose that (g ◦ f )(a 1 ) = (g ◦ f )(a 2 ). By Theorem 5.1.5 this
means that g( f (a 1 )) = g( f (a 2 )). Since g is one-to-one it follows that
f (a 1 ) = f (a 2 ), and similarly since f is one-to-one we can then conclude
that a 1 = a 2 . Thus, g ◦ f is one-to-one.
2. Suppose f and g are both onto, and let c be an arbitrary element of C. Since
g is onto, we can find some b ∈ B such that g(b) = c. Similarly, since f is
onto, there is some a ∈ A such that f (a) = b. Then (g ◦ f )(a) = g( f (a)) =
g(b) = c. Thus, g ◦ f is onto.