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Section Project: More on Inverses, Composition and Relations

Problem 2.32.
  1. Is Problem 1.28 valid if we define (a,b)={{a,1},{2,b}}?

  2. How would one formally define an ordered triple? An ordered quadruple?

Problem 2.33.

Show that the composition of one-to-one functions is one-to-one by letting g:BC and h:AB be one-to-one functions and showing that gh:AC is one-to-one.

Problem 2.34.

Show that the composition of onto functions is onto.

The word identity is used in a number of different contexts in mathematics. The function f:RR defined by f(x)=x is called the identity function because it identifies every number with itself. Similarly, 0 is called an additive identity because adding 0 to a number does not change the number, so 0+x is identified with x. By the same reasoning, 1 is called the multiplicative identity.

Problem 2.35.

Let S denote the set of all bijections from A to itself. According to Problems 2.33 and Problem 2.34, is a binary operation on S, that is, fgS whenever f and g are in S. We define a special bijection i:AA called the identity function on A as

i(x)=x for all xA.

Show that i is an identity for S, that is,

  1. fi=f=if and

  2. ff1=i=f1f

for all fS. This shows that i serves as an identity for composition in the same way that 0, 1, work as identities for +, × and , respectively.

Problem 2.36.

Consider the list below of all 6 different bijections from {0,1,2} onto itself. For example p maps 01, 12 and 20.

Table 2.37. Six Bijections
i p q f g h
012 012 012 012 012 012
012 120 201 021 210 102

Make a composition table for these 6 bijections, illustrating Problems 2.33, Problem 2.34 and Problem 2.35 above.

Problem 2.38.

A function f:BC is said to be left cancelative if, for all sets A and all functions g,h:AB,

fg=fh  implies   g=h.

Show that f is left cancelative if and only if f is one-to-one.

Problem 2.39.

A function f:BC is said to be right cancelative if, for all sets D and all functions g,h:CD,

gf=hf,   implies   g=h.

Show that f is right cancelative if and only if f is onto C.