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

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.

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?

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.

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.

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.

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.

Exponents and Logarithms The functions f(x) = bx and g(x) = logb x are inverses of each other. (What are their domains and ranges?) It follows that blogb x = x and logb bx = x.

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

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)).

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.

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)).

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?

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?

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.

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.

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.

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?

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?

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.