Page 253 - HOW TO PROVE IT: A Structured Approach, Second Edition

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One-to-one and Onto 239
5. This function is both one-to-one and onto. The verification that h is one-
to-one is very similar to the verification in solution 4 that g is one-to-
one, and it is left to the reader. To see that h is onto, we must show that
∀y ∈ R∃x ∈ R(h(x) = y). Here is a brief proof of this statements. Let y
be an arbitrary real number. Let x = (y − 3)/2. Then g(x) = 2x + 3 =
2 · ((y − 3)/2) + 3 = y − 3 + 3 = y. Thus, ∀y ∈ R∃x ∈ R(h(x) = y), so
h is onto.

Although the definition of one-to-one is easiest to understand when it is
stated as a negative statement, as in Definition 5.2.1, we know from Chapter 3
that the definition will be easier to use in proofs if we reexpress it as an
equivalent positive statement. The following theorem shows how to do this. It
also gives a useful equivalence for the definition of onto.

Theorem 5.2.3. Suppose f : A → B.

1. f is one-to-one iff ∀a 1 ∈ A∀a 2 ∈ A( f (a 1 ) = f (a 2 ) → a 1 = a 2 ).
2. f is onto iff Ran( f ) = B.

Proof
1. We use the rules from Chapters 1 and 2 for reexpressing negative statements
as positive ones.
f is one-to-one iff ¬∃a 1 ∈ A∃a 2 ∈ A( f (a 1 ) = f (a 2 ) ∧ a 1 = a 2 )
iff ∀a 1 ∈ A∀a 2 ∈ A¬( f (a 1 ) = f (a 2 ) ∧ a 1 = a 2 )
iff ∀a 1 ∈ A∀a 2 ∈ A( f (a 1 ) = f (a 2 ) ∨ a 1 = a 2 )
iff ∀a 1 ∈ A∀a 2 ∈ A( f (a 1 ) = f (a 2 ) → a 1 = a 2 ).

2. First we relate the definition of onto to the definition of range.
f is onto iff ∀b ∈ B∃a ∈ A( f (a) = b)
iff ∀b ∈ B∃a ∈ A((a, b) ∈ f )
iff ∀b ∈ B(b ∈ Ran( f ))
iff B ⊆ Ran( f ).

Now we are ready to prove part 2 of the theorem.
(→) Suppose f is onto. By the equivalence just derived we have B ⊆
Ran( f ), and by the definition of range we have Ran( f ) ⊆ B. Thus, it follows
that Ran( f ) = B.
(←) Suppose Ran( f ) = B. Then certainly B ⊆ Ran( f ), so by the equiva-
lence, f is onto.