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Math Help - Please help me to prove f is surjective iff f has a right inverse.

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    Please help me to prove f is surjective iff f has a right inverse.

    Prove: f is surjective iff f has a right inverse.

    f is surjective if for all b in B there is some a in A such that f(a) = b.
    f has a right inverse if there is a function h: B ---> A such that f(h(b)) = b for every b in B.

    i. Suppose f has a right inverse h: B --> A such that f(h(b)) = b for every b in B. Then for each b in B, let a in A such that f(a) = b. Let h(b) = a, then f(h(b)) = f(a) = b. Thus f is surjective.

    ii. Suppose f is surjective. That is, for every b in B, there is some a in A such that f(a) = b. How to show that f has a right inverse? Please help. Thank you.
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    Super Member Matt Westwood's Avatar
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    Let f: S \to T be a surjection.

    then \forall y \in T: f^{-1} \{y\} \ne \varnothing

    Let f^{-1} \{y\} = X_y = \{x_{y_1}, x_{y_2}, ...\}

    Using the Axiom of Choice, for each y \in T we can choose any of the elements x_{y_1}, x_{y_2}, \ldots to be identified as x_y, and thereby define:

    g: T \to S: g (y) = x_y

    Then we see that f \circ g (y) = f (x_y) = y

    and thus f \circ  g = I_T.
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    Is f^{-1} \{y\} the same as f^{-1}(y)?
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    Super Member Matt Westwood's Avatar
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    Excellent question.

    We know that f is a surjection, but f is not necessarily bijective, thus there could be more than one element of S mapped to one element of T. Thus the inverse of f is not necessarily a mapping, as an element of T does not necessarily map to a single element of S.

    So technically speaking you can not talk about the inverse element of y, but you can talk about the set of elements of S that the inverse of f relates to.

    What I mean by f^{-1}\{y\} is actually the preimage of f, and in this context I find that considering y as a set (admittedly a singleton one) rather than an element enhances the understanding of what's going on.
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