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Math Help - [SOLVED] A few intro to advanced math topics

  1. #1
    welcometo1984
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    [SOLVED] A few intro to advanced math topics

    Ok, so I have a few problems that I need to work on and have some progress made by Monday. The problem is that I need to work 30+ hours this weekend. So thanks in advance to anyone who assissts in solving, offers some advice, or views this topic

    1) Let f: X--> Y and let A,B be subsets of Y. Prove that f^-1(A) - f^-1(B)= f^-1 (A-B).

    Kind of a fun problem. Just show containment both ways.

    2) Let f: X-->A and g: Y-->B. Define a function f x g by (f x g)(x,y)= (f(x),g(y)) for all (x,y) as elements in X x Y.

    note that f x g is f cross g, not f times g.

    Prove the following
    a) If f and g are one-ton-one, then f x g is one-to-one.

    b) If f maps onto A and g maps onto B then f x g maps on A x B

    3) True or False? If false give a specific counter example (So indicate exactly what the sets and functions involved are) Assume f: X --> Y and
    g: Y--> Z

    a) If F^-1(Y) = X, then f maps onto Y

    b) If g composed with f maps onto Z, then f maps onto Y

    Again, thank you very much
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  2. #2
    Super Member Rebesques's Avatar
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    f^-1(A) - f^-1(B)= f^-1 (A-B).

    For f^{-1}(A) - f^{-1}(B)\subseteq f^{-1}(A-B):

    Let a\in f^{-1}(A) - f^{-1}(B).
    Then
    a\in f^{-1}(A) and a\notin f^{-1}(B),
    that is
    f(a)\in A and f(a)\notin B
    so
    f(a)\in A-B
    or
    a\in f^{-1}(A-B).

    Try the other inclusion the same way.


    If f and g are one-ton-one, then f x g is one-to-one.
    Just let f\times g(x,y)=f\times g(x',y')
    then
    (f(x),g(y))=(f(x'),g(y'))
    or
    f(x)=f(x') and g(y)=g(y')

    Since now f,g are both one to one, we get....

    If f maps onto A and g maps onto B then f x g maps on A x B
    Let (a,b)\in A\times B. We wish to find (x,y)\in X\times Y such that f\times g(x,y)=(a,b).

    Now since a\in A and f maps X onto A, there exists some x\in X such that f(x)=a.

    Similarly, there must be some y\in Y such that g(y)=b.

    Now, check what f\times g(x,y) is.


    Ok, this is where latex officialy bores me

    If F^-1(Y) = X, then f maps onto Y
    Let X={1} and Y={2,3}. Define f(1)=2; This f is not onto. But the inverse image of Y={2,3} through f is...


    If g composed with f maps onto Z, then f maps onto Y
    With X,Y as before, define now Z={4} and g(2)=g(3)=4. Then gof (composition) maps X onto Z, but...
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