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Math Help - surjection

  1. #1
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    surjection

    In my book they say that if f:A->B is a surjection then f[A]=B,
    but the definition is that for all b in B there is an element a in A sucht that f(a)=b. So we can have A={1,2,3} and B={2,3} then f(x)=x is a surjection but f(A) is not B. Am I correct?
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  2. #2
    MHF Contributor Swlabr's Avatar
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    Quote Originally Posted by sung View Post
    In my book they say that if f:A->B is a surjection then f[A]=B,
    but the definition is that for all b in B there is an element a in A sucht that f(a)=b. So we can have A={1,2,3} and B={2,3} then f(x)=x is a surjection but f(A) is not B. Am I correct?

    No, because then f:A \rightarrow A. The definition should
    read something along the lines of,


    f:A \rightarrow B is a surjection if \forall b \in B \exists a \in A \text{s.t} f(a)=b. Although your function takes care of the second part of the definition, it omits the first part.
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  3. #3
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    Quote Originally Posted by sung View Post
    In my book they say that if f:A->B is a surjection then f[A]=B,
    but the definition is that for all b in B there is an element a in A sucht that f(a)=b. So we can have A={1,2,3} and B={2,3} then f(x)=x is a surjection but f(A) is not B. Am I correct?
    No. f(x)= x is not "f:A-> B" because f(1)= 1 is not in B. If f:A->B, then f(A) must be a subset of B. That is, without any reference to "surjective", the notation f:->B implies that f(x) is in B for all x in A. "surjection" then implies the other way around: that B is a subset of f(A). Those two together, f(A) is a subset of B and B is a subset of f(A) say that f(A)= B.
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  4. #4
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    aah yes of course
    f[A] has to be a subset of B

    thanks
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