   Chapter 1.3, Problem 5E

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# Give an example of mapping f and g , different from those in Example 3, such that one of f or g is not one-to-one but f ∘ g is one-to-one.Example 3.Let A = { − 1 ,     0 ,     1 } . Find mappings f : A → A and g : A → A such that f ∘ g ≠ g ∘ f .

To determine

An example of mapping f and g different from those in Example 3, such that one of f or g is not one-to-one but fg is one-to-one.

Explanation

Formula used:

Definition: Let g:AB and f:BC. The composite mapping fg is the mapping from A to C defined by (fg)(x)=f(g(x)) for all xA.

Explanation:

Let f(x)={x2ifxisevenx+1ifxisodd

Consider an odd integer 1 and even integer 4.

By using the given mapping,

Since f(1)=1+1=2 and

f(4)=42=2

f(1)=f(4)=2 but 14

The element 2 has two different pre-images in .

Hence f(x)={x2ifxisevenx+1ifxisodd is not one-to-one mapping.

Now let g(x)={x1ifxiseven2xifxisodd

For even integers a and b, let g(a)=g(b)a1=b1; adding 1 on both sides a=b

For

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