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雨雨 and Smartiecat, here is an answer of question 1 from a professor

Yes. This can be proved by contradiction

For example, Let g: A ®B and ¦: B®C

A={x1, x2}, B={y}, C={z}, and g (x1) = g (x2) = y, and ¦ (y)= z.

Thus, ¦o g (x1) = ¦o g (x2) =z. Hence ¦o g cannot be one-to-one if g is not one-to-one.

Example:

Let ¦(x) = x, and g (x) = x2, we can see that ¦(x) is a one-to-one function and g (x) is not a one-to-one function. And ¦o g (x) = x2, which is also not one-to-one. Hence ¦o g cannot be one-to-one if g is not one-to-one.
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Replies, comments and Discussions:

  • 工作学习 / 求学深造 / 雨雨 and Smartiecat, here is an answer of question 1 from a professor
    Yes. This can be proved by contradiction

    For example, Let g: A ®B and ¦: B®C

    A={x1, x2}, B={y}, C={z}, and g (x1) = g (x2) = y, and ¦ (y)= z.

    Thus, ¦o g (x1) = ¦o g (x2) =z. Hence ¦o g cannot be one-to-one if g is not one-to-one.

    Example:

    Let ¦(x) = x, and g (x) = x2, we can see that ¦(x) is a one-to-one function and g (x) is not a one-to-one function. And ¦o g (x) = x2, which is also not one-to-one. Hence ¦o g cannot be one-to-one if g is not one-to-one.
    • It's ok ar... I don't need to know this since I'm not touching any discrete math in the future.
    • 谢谢~~~
      • 问你点事儿
        • en?
          • 我记得是你吧,买过clinique all about eyes?觉得如何?
            • 比较清爽。不过所有的眼霜俺感觉都一样,当然除了油性特别大的
      • 发给我的文件有乱码,里面是修正过的。
        Yes. This can be proved by contradiction

        For example, Let g: A -> B and f: B->C

        A={x1, x2}, B={y}, C={z}, and g (x1) = g (x2) = y, and f (y)= z.

        Thus, f o g (x1) = f o g (x2) =z. Hence f o g cannot be one-to-one if g is not one-to-one.

        Example:

        Let f(x) = x, and g (x) = x2, we can see that f(x) is a one-to-one function and g (x) is not a one-to-one function. And f o g (x) = x2, which is also not one-to-one. Hence f o g cannot be one-to-one if g is not one-to-one.
    • 有点意思。离散数学中很多都只能用反证法
    • 谢谢大家的帮助,俺明天可以交差了,这个周末再好好看看书。感激涕零,谢谢谢谢谢谢~~~^_^
    • 最后一个问题。
      1: f(S) = {y| y = f(x), x <- A}

      2: f ^(-1)(S) = {x | x = f ^(-1)(y) AND y in S}

      3: f ^(-1) (S U T) Subset f ^(-1)(S) U f ^(-1)(T)
      Let x In f ^(-1) (S U T) is E.T.S x In f ^(-1)(S) U f ^(-1)(T)

      4: f ^(-1)(S) U f ^(-1)(T) Subset f ^(-1) (S U T)
      Let x In n f ^(-1)(S) U f ^(-1)(T) is E.T.S x In f ^(-1) (S U T)

      Similarly f ^(-1) (S /\ T) = f ^(-1)(S) /\ f(-1)(T).