Let f A B and g B C be onetoone and onto Show that g o f

Let f : A --> B and g : B --> C be one-to-one and onto. Show that (g o f)^-1 exists, and that (g o f)^-1 = f^-1 o g^-1

(that is, show that the inverse of the composite function \"g of f\" exists, and that the inverse of the composite function \"g of f\" is equal to the inverse of \"f\" composed with the inverse of \"g\").

Solution

To prove main condtion we will have to proof composition function f o g is an one to one and onto

(a) For one to one

Let x1, x2 A,

then (g o f ) (x1)= (g o f ) (x2)

g (f (x1)) = g (f (x2))

f (x1) = f (x2) (since g is one to one)

x1 = x2 (since f is one to one)

Hence g o f is one to one.

(b) for onto

by the definition of composition gof : A C is a function.

To prove that gof is onto we have to prove that every element u C is an image element for some x A under g o f,

since g is onto, there exists t B such that g(t) = u.

Again since f is onto from A to B, there exists x A such that f (x) = t.

Now

(g o f) (x) = g (f (x)) = g (t) = u.

which shows g o f is onto

Now

We have g o f : A C is a function. By (b) and (c) g o f : A C is a one-to-one and onto.

So its inverse (g o f) –1 exists and (g o f) –1 : C A is also a one to one and onto function.

Again, since f and g are one to one and onto functions,

f –1 and g–1 exists and f –1 : B A and g–1 : C B are also one to one and onto.

So f –1 o g–1 : C A is also one to one and onto by (b) and (c).

Now to prove (g o f) –1 = f –1 o g–1

we have to show that for every u C, the image of u C under (g o f)–1 and f –1 o g–1 are equal.

Now,

we have g(t) = u t = g –1 (u) since g is one to one and onto

f(x) = t x = f –1 ( t) since f is one to one

and onto (g o f) (x) = g(f (x)) = g(t) = u

x = (g o f) –1 (u) .....................(1)

Since gof is one to one and onto.

Again (f –1 o g–1) (u) = f –1 (g–1 (u))

= f –1 (t) = x. ............................(2)

By (1) and (2),

we have (g o f) –1 = f –1 o g-1.

Answer