Composition of Functions

1. Composition of Functions Lecture 39 Section 7.4 Mon, Apr 9, 2007

2. Composition of Functions • Given two functions, f : AB and g : BC, where BB, the composition of f and g is the function g f : A  C which is defined by (g  f)(x) = g(f(x)). • Note that the codomain of f must be a subset of the domain of g, or else we cannot form the composition.

3. Examples of Composition • Let f : RR by f(x) = 2x + 3. • Let g : RR by g(x) = 4x + 5. • Find g  f and f  g. • Are they the same?

4. Composition with the Identity • Let f : AB be any function. • Let iA : AA be the identity function on A and let iB : BB be the identity function on B. • Then iB f = f and f  iA = f.

5. Inverses and the Identity • Let f : AB be a one-to-one correspondence and let f–1 : BA be its inverse. • Theorem: f–1f = iA and ff–1 = iB.

6. Inverses and the Identity • Furthermore, we can prove that if g : BA, then g = f–1 if and only if gf = iA and fg = iB. • We can use this as a way to verify that a function g is the inverse of a function f.

7. Inverses and the Identity • Let f : RR by f(x) = 2x + 3. • Let g : RR by g(x) = (1/2)x – 3/2. • Show that g = f–1.

8. Exponents and Logarithms • The functions f(x) = bx and g(x) = logbx are inverses of each other. • (What are their domains and ranges?) • It follows that blogbx = x and logbbx = x.

9. Exponents and Logarithms • In the equation blogbx = x, take natural logarithms of both sides: ln(blogbx) = ln x, (logbx)(ln b) = ln x, logbx = (ln x)/(ln b). • For example,

10. Composition and One-to-one-ness • Theorem: Let f : AB and g : BC be one-to-one. Then gf : AC is one-to-one. • Proof: • Assume that f and g are one-to-one. • Suppose that (gf )(a1) = (gf )(a2) for some a1, a2A. • Then g(f(a1)) = g(f(a2)).

11. Composition and One-to-one-ness • So f(a1) = f(a2) because g is one-to-one. • And then a1 = a2 because f is one-to-one. • Therefore, gf is one-to-one.

12. Composition and One-to-one-ness • Theorem: Let f : AB and g : BC. If gf : AC is one-to-one, then f is one-to-one, but g is not necessarily one-to-one. • Proof: • Assume that gf is one-to-one. • Suppose that f(a1) = f(a2) for some a1, a2A. We must show that a1 = a2. • Then g(f(a1)) = g(f(a2)).

13. Composition and One-to-one-ness • That is, (gf )(a1) = (gf )(a2). • So a1 = a2 because gf is one-to-one. • Therefore, f is one-to-one. • Now how do we show that g is not necessarily one-to-one?

14. Composition and One-to-one-ness • In the previous theorem, what if we assume also that f is onto? • Does it then follow that g is one-to-one?

15. Composition and Onto-ness. • Theorem: Let f : AB and g : BC be onto. Then gf : AC is onto. • Proof: • Assume that f and g are onto. • Let cC. • Then there is bB such that g(b) = c because g is onto.

16. Composition and Onto-ness. • Then there is aA such that f(a) = b because f is onto. • So (gf )(a) = g(f(a)) = g(b) = c. • Therefore, gf is onto.

17. Composition and Onto-ness. • Theorem: Let f : AB and g : BC. If gf : AC is onto, then g is onto, but f is not necessarily onto. • Proof: • Assume that gf is onto. • Let cC. We must show that there is bB such that g(b) = c.

18. Composition and Onto-ness. • Because gf is onto, we know that there is aA such that (gf )(a) = c. That is, g(f(a)) = c. • So let b = f(a). • Then bB and g(b) = g(f(a)) = c. • Therefore, g is onto. • Now how do we show that f is not necessarily onto?

19. Composition and Onto-ness. • In the previous theorem, what if we assume also that g is one-to-one? • Does it then follow that f is onto?

20. Composition, One-to-one-ness, and Onto-ness • Let f : R+R by f(x) = x. • Let g : R R+ by g(x) = x2. • Then g  f = iR+, so f is one-to-one and g is onto. • On the other hand, f  g = |x|, which is neither one-to-one nor onto.