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Inverse functions and null set.

  1. Apr 18, 2012 #1
    Ok,

    I understand an inverse function sends a variable in the range to the corresponding value in the domain, but am not sure if what I'm thinking is correct... : For example:

    Let A be the set


    [itex] A = \{1,2,3,7,8\} ; B = \{4,5,6\} [/itex] and the function [itex] f [/itex] map A to B s.t

    f(1) = 4
    f(2) = 5
    f(3) = 6

    so 7,8 do not have a value that is mapped one to one.

    I understand f is an surjection, but not a injection . But, does an inverse function exist?

    I would say yes, despite there not being a value in B that maps to 7 or 8.
    Is my thinking correct?


    Also, am I correct to say that a function does not have to use every element in the domain in order to have an inverse; I am confused because wouldn't one just say it maps to the null set and the inverse of the null set would contain values(7,8) that it maps to and hence not a function...?
     
  2. jcsd
  3. Apr 18, 2012 #2

    If we talk of a function [itex]f: A\to B[/itex] , we're explicitly assuming f is defined in the whole of A. Whether f is 1-1 and/or onto is another matter.

    DonAntonio
     
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