Bridge to Abstract Mathematics: Mathematical Proof and Structures

268 RELATIONS: FUNCTIONS AND MAPPINGS Chapter 8

Each of the remaining results (a), (b), and (c) can be proved in a similar

direct manner. You, in fact, may already have proved (a) and (c) in

Exercise 10, Article 8.1. You are asked to provide a similar elementary

(i.e., direct from the definitions involved) proof of (b) in Exercise 3(a).

An alternative, and far more sophisticated, approach to each of the four

results may be taken by employing our two theorems on right and left

cancellation. The former, Theorem 1, relating to the surjective property,

is useful for (b) and (d). The latter, Theorem 4 of Article 8.1, concerning

injectivity, may be employed to prove (a) and (c). Let us see how such

an argument would go by proving (c).

Given that g 0 f is injective, we use the condition in Theorem 4, Article

8.1, to prove f is injective. To set up such a proof, we let W be a set

and let h and k be mappings of W into X such that f 0 h = f 0 k. We

must prove that h = k. Now since f 0 h = f 0 k, then g 0 (f 0 h) =

g 0 (f 0 k), where both these mappings send W to 2, as shown in Figure

8.5. By associativity of composition (Theorem 3, Article 8.1), we may

deduce (g 0 f) 0 h = (g 0 f) 0 k. Since g 0 f is injective, then by Theorem

4, Article 8.1, we may "cancel" g 0 f on the left to conclude h = k, as

desired.

Exercise 5 requires you to give similar proofs of (a), (b), and (d) of Theo-

rem 2.

We next consider possible combinations of the one-to-one and onto

properties. If f(x) = x2, then f: R + R is neither one to one nor onto,

whereas f: R + [0, co) is onto, but is not one to one. On the other hand,

if g(x) = ex, then g: R + R is one to one, but is not onto. Finally, map-

pings from R to R such as y = Mx + B (M # 0), y = sinh x, and y = x3

Figure 8.5 A diagram helps to keep track of the sets between which

various mappings operate.