Problem 5.58. supposef xy and g : y z are functions if g of is one-to-one, prove that fmust be one-to-one 2. find an example where g o f is one-to-one, but g is not one-to-one

  f : X → Y  and  g: Y → Z

Now we have to show:

If gof is one-to-one then f must be one-to-one.

Given:

gof is one-to-one

To prove:

f is one-to-one.

Proof:

Let us assume that f(x) is not one-to-one .

This means that there exist x and y such that x≠y but f(x)=f(y)

On applying both side of the function by the function g we get:

g(f(x))=g(f(y))

i.e. gof(x)=gof(y)

This shows that gof is not one-to-one which is a contradiction to the given statement.

Hence, f(x) must be one-to-one.

Let A={1,2,3,4}  B={1,2,3,4,5} C={1,2,3,4,5,6}

Let f: A → B

be defined by f(x)=x

and g: B → C be defined by:

g(1)=1,g(2)=2,g(3)=3,g(4)=g(5)=4

is not a one-to-one function.

since 4≠5 but g(4)=g(5)

Also, gof : A → C

is a one-to-one function.

4932,9180,7625 are divisible by 9