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

heian(黑暗®-谢幕中)

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.

(#779121@0)
Last Updated: 2002-10-3
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工作学习 / 求学深造 / 雨雨 and Smartiecat， here is an answer of question 1 from a professor -heian(黑暗®-谢幕中); 2002-10-3 {547} (#779121@0)
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. -jeffrey815(Smartiecat); 2002-10-3 (#779123@0)
谢谢~~~ -rainrain(雨雨秋风何故乱翻书); 2002-10-3 (#779126@0)

问你点事儿 -pandora(pandora); 2002-10-3 (#779127@0)

en? -rainrain(雨雨秋风何故乱翻书); 2002-10-3 (#779144@0)

我记得是你吧，买过clinique all about eyes?觉得如何？ -pandora(pandora); 2002-10-3 (#779147@0)

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 -rainrain(雨雨秋风何故乱翻书); 2002-10-3 (#779171@0)

发给我的文件有乱码，里面是修正过的。 -heian(黑暗®-谢幕中); 2002-10-3 {484} (#779129@0)
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 -noexit(江湖▲狐仙▲省油▲灯); 2002-10-3 (#779132@0)
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~~~^_^ -rainrain(雨雨秋风何故乱翻书); 2002-10-3 (#779145@0)
最后一个问题。 -heian(黑暗®-谢幕中); 2002-10-3 {379} (#779146@0)
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).

@Calgary

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

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