Page 268 - HOW TO PROVE IT: A Structured Approach, Second Edition

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254 Functions
11. Suppose f : A → B and g : B → A.
∗
(a) Prove that if f is one-to-one and f ◦ g = i B , then g = f −1 .
(b) Prove that if f is onto and g ◦ f = i A , then g = f −1 .
(c) Prove that if f ◦ g = i B but g ◦ f = i A , then f is onto but not
one-to-one, and g is one-to-one but not onto.
12. Suppose f : A → B and f is one-to-one. Prove that there is some set

B ⊆ B such that f −1 : B → A.
13. Suppose f : A → B and f is onto. Let R ={(x, y) ∈ A × A | f (x) =
f (y)}. By exercise 17(a) of Section 5.1, R is an equivalence relation
on A.
(a) Prove that there is a function h : A/R → B such that for all x ∈ A,
h([x] R ) = f (x). (Hint: See exercise 18 of Section 5.1.)
(b) Prove that h is one-to-one and onto. (Hint: See exercise 16 of Sec-
tion 5.2.)
(c) It follows from part (b) that h −1 : B → A/R. Prove that for all b ∈ B,
−1
h (b) ={x ∈ A | f (x) = b}.
(d) Suppose g : B → A. Prove that f ◦ g = i B iff ∀b ∈ B(g(b) ∈ h(b)).
14. Suppose f : A → B, g : B → A, and f ◦ g = i B . Let A = Ran(g) ⊆ A.
∗

(a) Prove that for all x ∈ A ,(g ◦ f )(x) = x.

(b) Prove that f A is a one-to-one, onto function from A to B and

−1
g = ( f A ) . (See exercise 7 of Section 5.1 for the meaning of the
notation used here.)
15. Let B ={x ∈ R | x ≥ 0}. Let f : R → B and g : B → R be defined by
√
2
the formulas f (x) = x and g(x) = x. As we saw in part 2 of Exam-
−1
ple 5.3.6, g = f −1 . Show that g = ( f B) . (Hint: See exercise 14.)
∗ 2
16. Let f : R → R be defined by the formula f (x) = 4x − x . Let B =
Ran( f ).
(a) Find B.
(b) Find a set A ⊆ R such that f A is a one-to-one, onto function from
−1
A to B, and find a formula for ( f A) . (Hint: See exercise 14.)
17. Let A be any set. Let P be the set of all partial orders on A, and let S be the
set of all strict partial orders on A. In exercises 4 and 5 of Section 4.5 you
showed that if R ∈ P then R \ i A ∈ S, and if R ∈ S then R ∪ i A ∈ P.
(Recall that we showed in the proof of Theorem 4.5.2 that R ∪ i A is the
reflexive closure of R.) Let f : P → S and g : S → P be defined by the
formulas

f (R) = R \ i A , g(R) = R ∪ i A .

Show that f is one-to-one and onto, and g = f −1 .