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Why does an inverse exist only for surjective functions?

  1. May 17, 2015 #1

    kay

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    In other words, why does a function have to be onto (or surjective, i.e. Range=Codomain) for its inverse to exist?
     
  2. jcsd
  3. May 17, 2015 #2

    PeroK

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    The key property for a function to have an inverse is that it is 1-1. Any 1-1 function will have an inverse. The inverse, however, can only be defined on the range of the function. Perhaps this is best explained by an example:

    The exponential function ##e^x## maps the real number line ##(-\infty, \infty)## onto ##(0, \infty)##. The exponential is 1-1, so there exists an inverse (the natural logarithm). But, the inverse is only defined on ##(0, \infty)##. You can't define the inverse function on all of the real number line.

    This raises perhaps a technical point that if you are considering the set of functions that map ##\mathbb{R}## to ##\mathbb{R}## then the exponential function is in this set, but there is no inverse within this set of functions. From that point of view, the inverse doesn't exist.
     
  4. May 18, 2015 #3

    kay

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    Actually my question was, that why can't an inverse exist for a function whose range is not necessarily equal to but is rather a subset of the codomain.
     
  5. May 18, 2015 #4

    pasmith

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    An inverse of a function [itex]f: A \to B[/itex] is a function [itex]g: B \to A[/itex] such that [itex]g(f(a)) = a[/itex] for all [itex]a \in A[/itex] and [itex]f(g(b)) = b[/itex] for all [itex]b \in B[/itex]. An inverse is unique if it exists.

    Let [itex]f: \{1,2\} \to \{1,2,3\} : x \mapsto x[/itex]. Now the inverse of [itex]f[/itex], if it exists, is a function [itex]g: \{1,2,3\} \to \{1,2\}[/itex]. Now we must have [itex]g(1) = 1[/itex] and [itex]g(2) = 2[/itex], but what value do you assign to [itex]g(3)[/itex] such that [itex]f(g(3)) = 3[/itex]?
     
  6. May 18, 2015 #5

    kay

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    I got it. Thanks a lot for your help. :)
     
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