Page 18 - DMTH403_ABSTRACT_ALGEBRA

Unit 1: Generating Sets

Theorem 3: Let A be a set. For every function f : A  A, we have f o I = I o f = f. Notes
A A
Proof: Since both f and I are defined from A to A, both the compositions f o I and I o f are
A
A
A
defined. Moreover, V x  A,
f o I (x) = f(I (x)) = f (x), so f o I = f.
A
A
A
Also, V X  A, I o f(x) = I (f(x)) = f(x), so I o f = f.
A
A
A
In the case of real numbers, you know that given any real number x + 0, $ y % 0 such that xy = 1.
y is called the inverse of x. Similarly, we can define an inverse function for a given function.
Definition: Let f : A  B be a given function. If there exists a function g : B  A such that f o g =
I and g o f = I , then we say that g is the inverse of f, and we write g = f .
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A
B
For example, consider f : R  R defined by f(x) = x + 3. If we define g : R  R by g(x) = x – 3, then
f o g(x) = f(g(x)) = g(x) + 3 = (x – 3) + 3 = x V x  R. Hence, f o g = I . You can also verify that
R
g o f = I . So g = f .
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R
Note that in this example f adds 3 to x and g does the opposite – it subtracts 3 from x. Thus, the
key to finding the inverse of a given function is : try to retrieve x from f(x).
For example, let f : R  R be defined by f(x) = 3x + 5. How can we retrieve x from 3x + 5? The
x 5

answer is “first subtract 5 and then divide by 3”. So, we try g(x) = (x)  . And we find
3
f(x) 5 (3x 5) 5



g o f(x) = g(f(x)) =   x.
3 3

Also, f o g(x) = 3(g(x)) + 5 = 3    (x 5)     5  x V x R.

3
Let’s see if you’ve understood the process of extracting the inverse of a function,
Do all functions have an inverse? No, as the following example shows.

Example: Let f : R  R be the constant function given by f(x) = 1 V x " R. What is the
inverse of f ?

Solution: If f has an inverse g : R  R, we have f o g = I , i.e., V x  R, f o g(x) = x. Now take
R
x = 5. We should have f o g (5) = 5, i.e., f(g(5)) = 5. But f(g(5)) = 1, since f(x) = 1 V x. So we reach a
contradiction. Therefore, f has no inverse.
In view of this example, we naturally ask for necessary and sufficient conditions for f to have an
inverse. The answer is given by the following theorem.
Theorem 4: A function f ; A  B has an inverse if and only if f is bijective.
Proof: Firstly, suppose f is bijective. We shall define a function g : B  A and prove that g = f .
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Let b  B. Since f is onto, there is some a " A such that f(a) = b. Since f is one-one, there is only one
such a  A. We take this unique element a of A as g(b). That is, given b  B, we define g(b) = a,
where f(a) = b.

Note that, since f is onto, B = { f(a) | a  A). Then, we are simply defining g : B  A by g(f(a)) =
a. This automatically ensures that g o f = I .
A
Now, let b  B and g(b) = a. Then f(a) = b, by definition of g. Therefore, f o g(b) = f(g(b)) = f(a) =
b. Hence, f o g = I .
B

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