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# Inverse of a Bijective Function

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#### Inverse of a Bijective Function - Lesson Summary

Let f: A â†’ B be a function. If, for an arbitrary x âˆˆ A we have f(x) = y âˆˆ B, then the function, g: B â†’ A, given by g(y) = x, where y âˆˆ B and x âˆˆ A, is called the inverse function of f.

Ex:

Define f: A â†’ B such that

f(2) = -2, f(½) = -2, f(½) = -½, f(-1) = 1, f(-1/9) = 1/9

Here f one-one and onto.

g(-2) = 2, g(-½) = 2, g(-½) = ½, g(1) = -1, g(1/9) = -1/9

g is the inverse of f.

A function, f: A â†’ B, is said to be invertible, if there exists a function, g : B â†’ A, such that g o f = IA and f o g = IB.

The function, g, is called the inverse of f, and is denoted by f -1.

Ex:

Let 2 âˆˆ A.Then gof(2) = g{f(2)} = g(-2) = 2

Let -2 âˆˆ B.Then fog(-2) = f{g(-2)} = f(2) = -2.

Example:

Let P = {y Ïµ N: y = 3x - 2 for some x ÏµN}. Show that a function, f : N â†’ P, defined by f (x) = 3x - 2, is invertible, and find f-1.

Solution:

Let us consider an arbitrary element, y Ïµ P.

â‡’ y=3x-2

â‡’ x = (y+2)/3

Let us define g : P â†’ N by g(y) = (y+2)/3

Let x âˆˆ N.

Then g o f (x) = g (f (x)) = g (3x - 2)

= (3x-2+2)/3 = x

This shows g o f = IN â€¦(i)

Let y âˆˆ P.

Then fog (y) = f (g (y)) = f((y+2)/3)

= 3((y+2)/3) - 2 = y

This shows fog = IP â€¦(ii)

Hence, f is invertible and g is the inverse of f.

Theorem:

Let f : X â†’ Y and g : Y â†’ Z be two invertible (i.e. bijective) functions. Then g o f is also invertible with (g o f)-1 = f -1o g-1.

Proof:

Given, f and g are invertible functions.

To prove that g o f is invertible, with (g o f)-1 = f -1o g-1.

It is sufficient to prove that:

i. (f -1 o g-1) o (g o f) = IX, and

ii. (g o f)o( f -1o g-1) = IZ.

Now, ( f -1 o g-1) o (g o f) = {( f -1 o g-1) o g} o f {'.' l o (m o n) = (l o m) o n}

= { f -1 o (g-1 o g)} o f

= ( f -1oIY)of {'.' g-1og = IY}

= f-1of { '.' f-1oI = f-1}

= IX

Hence, (f -1o g-1)o(g o f) =IX â€¦â€¦ ( i)

Similarly, (g o f)o( f -1o g-1)

={(g o f) o f--1} o g-1

={g 0 (f o f--1} o g-1

=(g 0 Ix) o g-1

=g o g-1 = Iz

(g o f)o( f -1o g-1) =IZ â€¦â€¦. (ii)

From equations (i) and (ii),

(g o f)-1 = f -1 o g-1

Hence, the composition of two invertible functions is also invertible.