Page 259 - HOW TO PROVE IT: A Structured Approach, Second Edition

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Inverses of Functions 245
19. Let F ={ f | f : R → R}, and define a relation R on F as follows:
R ={( f, g) ∈ F × F |∃h ∈ F( f = h ◦ g)}.

(a) Let f , g, and h be the functions from R to R defined by the formulas
3
4
2
f (x) = x + 1, g(x) = x + 1, and h(x) = x + 1. Prove that hR f ,
but it is not the case that gR f .
(b) Prove that R is a preorder. (See exercise 24 of Section 4.6 for the
definition of preorder.)
(c) Prove that for all f ∈ F, fRi R .
(d) Prove that for all f ∈ F, i R Rf iff f is one-to-one. (Hint for right-to-
left direction: Suppose f is one-to-one. Let A = Ran( f ), and let h =
f −1 ∪ ((R \ A) ×{0}). Now prove that h : R → R and i R = h ◦ f .)
(e) Suppose that g ∈ F is a constant function; in other words, there is
some real number c such that ∀x ∈ R(g(x) = c). Prove that for all
f ∈ F, gR f . (Hint: See exercise 14 of Section 5.1.)
(f) Suppose that g ∈ F is a constant function. Prove that for all f ∈ F,
fRg iff f is a constant function.
−1
(g) As in exercise 24 of Section 4.6, if we let S = R ∩ R , then S is an
equivalence relation on F. Also, there is a unique relation T on F/S
such that for all f and g in F,[ f ] S T [g] S iff fRg, and T is a partial
order on F/S. Prove that the set of all one-to-one functions from R
to R is the largest element of F/S in the partial order T , and the set
of all constant functions from R to R is the smallest element.

5.3. Inverses of Functions

We are now ready to return to the question of whether the inverse of a function
from A to B is always a function from B to A. Consider again the function F
from part 1 of Example 5.1.2. Recall that in that example we had A ={1, 2, 3},
B ={4, 5, 6}, and F ={(1, 5), (2, 4), (3, 5)}. As we saw in Example 5.1.2, F
is a function from A to B. According to the definition of the inverse of a relation,
F −1 ={(5, 1), (4, 2), (5, 3)}, which is clearly a relation from B to A. But F −1
fails to be a function from B to A for two reasons. First of all, 6 ∈ B, but 6 isn’t
paired with any element of A in the relation F −1 . Second, 5 is paired with two
different elements of A, 1 and 3. Thus, this example shows that the inverse of
a function from A to B is not always a function from B to A.
You may have noticed that the reasons why F −1 isn’t a function from B to
A are related to the reasons why F is neither one-to-one nor onto, which were
discussed in part 1 of Example 5.2.2. This suggests the following theorem.