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Math Help - Injective if and only if Surjective

  1. #1
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    Injective if and only if Surjective

    Hi there, I have this problem and while I think what I've put is correct I'm not sure that it is rigorous enough. Any help much appreciated!!

    (a) Let A be a finite set. Show that a function f : A → A is injective if
    and only if it is surjective.

    (b) Show that this result is false for infinite sets by exhibiting
    An injective map f : N → N that is not surjective;
    A surjective map f : N → N that is not injective.

    For a) I've put that a function is injective if each unique element x in the domain X is mapped to an element y in some co-domain Y, and a function is surjective if for every element y in the co-domain Y there is atleast one element x in the domain X such that f(x)=y. As the cardinality of the domain and co-domain are the same if the function is injective it must also be surjective.

    For b) An injective map which is not surjective would be 2x=y
    For a surjective map which is not injective I'm thinking something like ┌ x/2┐=y (those symbols are meant to denote the ceiling function!!)
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  2. #2
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    So the interesting thing about this function is that it is a reflexive relation on the set A.

    Injective iff. surjective means:
    injective \Rightarrow surjective and surjective \Rightarrow injective.

    A function is onto if the range of f equals the codomain of f.
    A function is one-to-one if no member of the codomain is the image under f of two distinct elements of the domain.

    That is (\forall a)(\exists a')(a \mapsto a').

    if f is onto then it follows that \Vert A \Vert \geq \Vert A' \Vert but from (f: A \mapsto A \rightarrow \Vert A \Vert = \Vert A' \Vert ) \Rightarrow (\forall a')(\forall b')(a' = b' \rightarrow f'(a) = f'(b)) and f is also one-to-one.

    And then you have to show it the other way around.

    An injective map that is not surjective is f:x \mapsto 2x.
    A surjective map that is not injective is f: x \mapsto \lfloor \frac{1}{2}x \rfloor
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