   Chapter 1.3, Problem 6TFE

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# Label each of the following statements as either true or false.Let g : A → B and f : B → C such that f ∘ g is one-to-one. Then both f and g are one-to-one.

To determine

Whether the statement, “Let g:AB and f:BC such that fg is one-to-one. Then both f and g are one-to-one” is true or false.

Explanation

Formula used:

1) To show that f is not one-to-one, find two elements a1A and a2A such that a1a2 and f(a1)=f(a2)

2) To show that f is one-to-one by assuming that f(a1)=f(a2) and proving that this implies that a1=a2

Explanation:

Consider the statement, “Let g:AB and f:BC such that fg is one-to-one. Then both f and g are one-to-one.”

Let f: defined by f(x)=x2 and

g: defined by g(x)=ex

By using the definition of a composite function, fg: is defined by

(fg)(x)=f(g(x))=f(ex)=(ex)2=e2x

Now we check whether these mappings are one-to-one.

Let x1,x2.

Consider g(x1)=g(x2)

By using the given mapping,

ex1=ex2

By taking natural logarithm function from both sides, get

ln(ex1)=ln(ex2)

By using the rule of logarithm function,

x1ln(e)=x2ln(e)

ln(e)=1 implies that

x1=x2

This shows that mapping g: defined by g(x)=ex is one-to-one

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