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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(Â½) = -2, f(Â½) = -Â½, f(-1) = 1, f(-1/9) = 1/9

Here f one-one and onto.

g(-2) = 2, g(-Â½) = 2, g(-Â½) = Â½, 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 = I_{A} and f o g = I_{B}.

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 = I_{N} 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 = I_{P} 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 ^{-1}o g^{-1}.

Proof:

Given, f and g are invertible functions.

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

It is sufficient to prove that:

i. (f ^{-1} o g^{-1}) o (g o f) = I_{X}, and

ii. (g o f)o( f ^{-1}o g^{-1}) = I_{Z}.

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 ^{-1}oI_{Y})of {'.' g^{-1}og = I_{Y}}

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

= I_{X}

Hence, (f ^{-1}o g^{-1})o(g o f) =I_{X} â€¦â€¦ ( i)

Similarly, (g o f)o( f ^{-1}o g^{-1})

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

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

=(g 0 I_{x}) o g^{-1}

=g o g^{-1 }= I_{z}

(g o f)o( f ^{-1}o g^{-1}) =I_{Z} 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.

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(Â½) = -2, f(Â½) = -Â½, f(-1) = 1, f(-1/9) = 1/9

Here f one-one and onto.

g(-2) = 2, g(-Â½) = 2, g(-Â½) = Â½, 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 = I_{A} and f o g = I_{B}.

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 = I_{N} 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 = I_{P} 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 ^{-1}o g^{-1}.

Proof:

Given, f and g are invertible functions.

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

It is sufficient to prove that:

i. (f ^{-1} o g^{-1}) o (g o f) = I_{X}, and

ii. (g o f)o( f ^{-1}o g^{-1}) = I_{Z}.

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 ^{-1}oI_{Y})of {'.' g^{-1}og = I_{Y}}

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

= I_{X}

Hence, (f ^{-1}o g^{-1})o(g o f) =I_{X} â€¦â€¦ ( i)

Similarly, (g o f)o( f ^{-1}o g^{-1})

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

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

=(g 0 I_{x}) o g^{-1}

=g o g^{-1 }= I_{z}

(g o f)o( f ^{-1}o g^{-1}) =I_{Z} 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.