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) = logb x are inverses of each other.
(What are their domains and ranges?)
It follows that
blogb x = x
and
logb bx = x.

9. 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,

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.