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16 June, 05:21

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

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  1. 16 June, 08:04
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    Answer with explanation: We are given two functions f (x) and g (y) such that:

    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.

    Now, example to show that gof is one-to-one but g is not 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.
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