Composition and Invertible Function

An invertible function for ƒ is a function from B to A, with the property that a round trip (a composition) from A to B to A returns each element of the first set to itself. A function ƒ that has an inverse is called invertible; the inverse function is then uniquely determined by ƒ and is denoted by ƒ−1. Function composition is the applications of one function to the results of another. For instance, the functions f: X → Y and g: YZcomprised by computing the output of g when it has an argument of f(x) instead of x.  (Source. Wikipedia) .

Examples for invertible function:

To define the invertible of a function f i.e. f−1 (read as ‘f inverse’), the function f must be one-to-one and onto.

Example 1:

Let A = {1, 2, 3}, B = {a, b, c, d}. Consider a function f = {(1, a), (2, b),(3, c)}. Here the image set or the range is {a, b, c} which is not equal to the co domain {a, b, c, d}. Therefore, it is not onto.


For the inverse function f−1 the co-domain of f becomes domain of f −1.

i.e. If f : A → B then f−1 : B → A . According to the definition of domain, each element of the domain must have image in the co-domain. In f−1, the element‘d’ has no image in A. Therefore f −1 is not a function.

Example 2:

f = {(1, a), (2, a), (3, b)} where A = {1, 2, 3}, B = {a, b}

Here the two different elements ‘1’ and ‘2’ have the same image ‘a’.

Therefore the function is not one-to-one.

The range = {a, b} = B. ∴ The function is onto.

f(1) = a

f (2) = a

f(3) = b

Here all the elements in A has unique image

f −1 (a) = 1

f −1 (a) = 2

f −1 (b) = 3


The element ‘a’ has the two images 1 and 2. It violates the principle of the function that each element has a unique image. This is because the function is not one-to-one.

Thus, ‘f −1 exists if and only if f is one-to-one and onto’.

Examples for composition function:

Example :

Let A = {1, 2}, B = {3, 4} and C = {5, 6} and f : A → B and g : B → C such that f(1) = 3, f(2) = 4, g(3) = 5, g(4) = 6. Find gof.

Solution:

gof is a composition function from A → C.

Identify the images of elements of an under the composition function gof.

(gof) (1) = g(f(1)) = g(3) = 5

(gof) (2) = g(f(2)) = g(4) = 6

i.e. image of 1 is 5 and image of 2 is 6 under gof

∴ gof = {(1, 5), (2, 6)}