   Chapter 7.3, Problem 29ES ### Discrete Mathematics With Applicat...

5th Edition
EPP + 1 other
ISBN: 9781337694193

#### Solutions

Chapter
Section ### Discrete Mathematics With Applicat...

5th Edition
EPP + 1 other
ISBN: 9781337694193
Textbook Problem
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# Suppose f : X → Y and g : Y → Z are both one-to-one and onto. Prove that ( g ∘ f ) − 1 exists and that ( g ∘ f ) − 1 = f − 1 ∘ g − 1 .

To determine

To prove:

If f:XY and g:YZ are both one-to-one and onto functions, then (gf)1 exist and (gf)1=f1 g1.

Explanation

Given information:

Suppose f:XY and g:YZ are both one-to-one and onto.

Concept used:

A function is said to be one-to-one function if the distinct elements in domain must be mapped with distinct elements in co-domain.

A function is onto function if each element in co-domain is mapped with atleast one element in domain.

Proof:

Consider the one-to-one and onto functions f:XY and g:YZ.

As the functions f:XY and g:YZ are one-to-one so the composite function gf is one-to-one.

As the functions f:XY and g:YZ are onto so the composite function gf is onto.

Thus the composite function gf is one-to-one and onto.

As the function gf is one-to-one correspondence, so there must exist inverse (gf)1 for the function gf from Z to X ,

Find the inverse of the composite function gf.

Let zZ.

[(gf)( f 1 g 1)](z)=(gf)[( f 1 g 1)(z)]=(gf)[f1( g 1( z))]=g(f[ f 1( g 1 ( z ))])=g(g 1(z))=z=I(z)

This is true for all zZ, thus (gf)(f1g1)=Iz

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