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Need Help Urgently Bijection, Injection, Subjection Functions

  1. Nov 11, 2009 #1
    Hi, I would like to say this is a great forum I found. This is my very first post yay =)

    I need help on these certain questions.

    5. (10%) Given A = {1, 2, 3} and B = {a, b, c}
    (a) list in two-line notation all one-to-one functions from A to B;
    (b) list in two-line notation all onto functions from A to B;
    (c) list in two-line notation all bijections from A to B.

    6. (10%) Given two finite sets A and B such that |A| = |B|, prove or disprove that every one-to-one function from A to B is bijective.

    7. (10%) Let f : A -> B and g : B -> C be bijections. Prove that
    (a) g f is a bijection;
    (b) (g o f)^-1 = f^-1 o g^-1

    I will scan my page for 5. Numbers 6 and 7 I have no idea how to start it^-^^^^^?-
    Number 5 i am having some troubles as well
    DSC03430.jpg
     
  2. jcsd
  3. Nov 12, 2009 #2
    if f:A-->B, where |A|=n, and |B|=m, then the total nr of functions from A to B is m^n. Now, if f is to be one to one with m=n, there will be n! functions(why?). But in this case every 1-1 function will be also an onto function (why?).
     
  4. Nov 12, 2009 #3

    lanedance

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    seems like you swapped question letters aorund?

    if the function A-> B is onto, i think it should the image of A should be all of B, simlarly if the functions is 1:1 then f(a) = f(b) iff a=b, so eitehr way you look at it, i don;t agree with what you have labelled for a)
     
  5. Nov 12, 2009 #4
    I looked at the page for wiki where injection is just each argument is mapped to at most one value.

    So It should be possible for some other values not mapped to any?
     
  6. Nov 12, 2009 #5
    For 6. since f is one-to-one you want to show tha f is onto. i.e that for any b in B there is a in A such that f(a)=b.

    let b=F^0(b),F(b),F^2(b)...., and these are all elements in a finite set,A, so cannot all be distinct. Let r,s, with say r>s, such that f^r(b)=f^s(b). Then we have:
    =>F^(r-1)(b)=F^(s-1)(b)=.....=F^(r-s)(b)=b=>F(a)=b, with a=F^(r-s-1)(b), where with F, i have denoted the inverse of f, it exists (why?)
     
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