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Homework Help: How do you define a mapping f:K->N

  1. May 8, 2007 #1
    1. The problem statement, all variables and given/known data
    How do you define a mapping f:K->N

    with K={0}
    N is the integers.

    that maps the element 0 to every single single element in N?

    ie. 0->-n, ... , -2, -1, 0, 1, 2, ... , n

    Is that even possible?

    The mapping Z->Z by multiplying each element in Z by 0 is a legitamate mapping isn't it? So really Z->0 but I am extending the range to Z.

    In a mapping is isomorphic than there should be the same number of elements in the domain as in the range? But what happens if there are an infinite number of elements in both the range and domain?
    Last edited: May 8, 2007
  2. jcsd
  3. May 8, 2007 #2


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    The thing you suggested isn't a mapping, unless I'm missing something. Every element of the domain must be mapped to at most one element of the codomain.
  4. May 8, 2007 #3

    matt grime

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    A map is an isomorphism if it is invertible, not if there are 'the same number of elements in the domain and the range'. In fact we define (infinite) cardinals by the existence of bijections.
  5. May 8, 2007 #4
    The question is, is {0} isomorphic to Z? Where Z are the integers.

    I can see that is would be isomorphic if we can find a mapping that maps 0 to every element in Z, once.

    However mapping Z to {0} would recquire only a single mapping like map 1 to 0 or 2 to 0 or n to 0 but only once, otherwise it wouldn't be a bijection.

    I can actually show that {0} isomorphic to Z by the first isomorphism theorem. i.e. f:Z->Z where f is to multiply every element in Z by 0. So the kernel of f is all of Z. Now the quotient ring is Z/Z which is isomorphic to {0} which is isomorphic (via the equivalence relation) to Z.

    But I am not convinced. Namely the actual mappings seems a bit funny, if not incorrect.
  6. May 8, 2007 #5


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    Perhaps I am misunderstanding something but it should be obvious that {0} is not isomorphic to Z. For one thing, they don't have the same cardinality!

    While n-> 0 is a mapping from Z to K, there is NO mapping from K to Z.

    I have no idea what you think the "first isomorphism theorem" says. Yes, Z/Z contains a single member and is isomophic to {0}. That does NOT say that they are both isomorphic to Z!. In general, if G and H are groups and f is a homomorphism from G onto H, then the kernel of f, K, is a subgroup of G and G/K is isomorphic to H. That does NOT say that G is isomorphic to H.
  7. May 8, 2007 #6


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    The confusion is probably due to the fact that elements of quotient groups are cosets, and in this special case you are looking at the one and only coset eZ, which is, as HallyofIvy mentioned, isomorphic to Im(f) = {0}.
  8. May 8, 2007 #7
    1st isomorphism theorem: Let f:R->S be a surjective homomorphism of rings with kernel K. Then the quotient ring R/K is isomorphic to S.

    Let R=S=Z the integers.

    Let f be the maping multiplying each element in Z by 0 So Z->{0}
    So K = Z

    Hence Z/Z is isomorphic to Z. Z/Z is isomorphic to {0} so {0} is isomorphic to Z. But I have made a mistake because as you said they don't have the same number of elements. And I can't think of a bijective mapping from {0} to Z. But where is the mistake in my proof? Does R cannot equal S?
  9. May 8, 2007 #8


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    f isn't surjective if S=Z, f is only surjective if S={0}!
  10. May 8, 2007 #9
    Good point. So Z/Z is isomorphic to {0} as it should and not to Z.
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