1549901369-Elements_of_Real_Analysis__Denlinger_

B.3 Algebra of Real-Valued Functions 631

Definition B.3.8 Let A denote an arbitrary set. The identity function on

A is the function iA : A~ A defined by the rule

Vx EA, iA(x) = x.

Note: The identity function is a 1-1 correspondence.

Theorem B.3.9 Let A and B be arbitrary sets. The identity function on A ,

iA : A~ A, satisfies the following rules:

(a) VJ: A~ B , f o iA = f;

(b) Vg: B ~A, iA o g = g.

Proof. Exercise 15. •

Theorem B.3.10 Suppose f: A~ B.

(a) If 3 g : B ~ C 3 g o f is 1-1, then f is 1-1.

(b) If 3 h : D ~ A 3 f o h is onto B, then f is onto B.

Proof. (a) Suppose 3 g: B ~ C 3 go f is 1-1. Let a "I-a' in A. Since go f

is 1-1,

(go J)(a) "I-(go !)(a'), which means that

g(f(a)) "I-g(f(a')).

This cannot happen unless f(a) "I-f(a'). Therefore, f is 1-1.

(b) Suppose 3 h : D ~ A 3 f o h is onto B. Let b E B. Since f o h is onto
B , 3 d ED 3 (f o h)(d) = b. Let a = h(d). Then a EA and f(a) = f(h(d)) = b.
That is, Vb EB, b E R(f). Therefore, f is onto B. •

Definition B.3.11 Suppose f: A~ B. If 3 function g: B ~A 3 go f = iA
and fog= iB, then we say that f is invertible, and we say that the function
g is the inverse function of f. In symbols,

Theorem B.3.12 A function f : A ~ B is invertible iff f is 1-1 and onto
B (that is, f is a 1-1 correspondence). Moreover, if g = 1-^1 , then f = g-^1.

Proof. First, t he =? direction. Suppose f : A ~ B is invertible. Then 3
function g : B ~ A 3 go f = iA and fog = iB. Since iA is 1-1, Theorem
B.3.10 (a) says that f is 1-1. Since iB is onto B, Theorem B .3. 10 (b) says that
f is onto B.