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Math Help - Surjection

  1. #1
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    Surjection

    Let  f: X \to Y be a function. Prove that there exists a function  g: Y \to X such that  f \circ g = I_Y if and only if  f is a surjection.

    So we first construct  g (using axiom of choice) by picking a point in the preimage of  y for all  y (  \overleftarrow{f}(Y) ). If  f(g(y)) = f(x) = y, \ \forall x \in X then  f is a surjection. If  f is a surjection, then  \forall y \in Y, \exists x \in X, y = f(x) . Then  f(g(y)) = f(x) = y as required.

    Is this correct?
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  2. #2
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    Quote Originally Posted by shilz222 View Post
    Let  f: X \to Y be a function. Prove that there exists a function  g: Y \to X such that  f \circ g = I_Y if and only if  f is a surjection.
    So we first construct  g (using axiom of choice) by picking a point in the preimage of  y for all  y (  \overleftarrow{f}(Y) ). If  f(g(y)) = f(x) = y, \ \forall x \in X then  f is a surjection. If  f is a surjection, then  \forall y \in Y, \exists x \in X, y = f(x) . Then  f(g(y)) = f(x) = y as required.
    Is this correct?
    No it is most definitely not correct. You have confused the two-way proof.

    First assume that  g: Y \to X such that  f \circ g = I_Y .
    In which case there is no need to construct g because you already have it assumed.
    Now just go about proving f is surjective.

    That part done, then assume that f is surjective and go about constructing g such that  f \circ g = I_Y .
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  3. #3
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    didnt I do that? Plus it says prove that there exists a function  g: Y \to X . THat is separate from the other part of the proof.

    So the two parts are: Prove existence of function  g: Y \to X , and the two-way proof.

    So I proved: (1)   f \circ g = I_y \Leftrightarrow f is surjective and (2) that such a function  g: Y \to X exists.

    So how come the question didn't given both functions  f: X \to Y and  g: Y \to X ?
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  4. #4
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    Do you have difficulty understanding the problem? Here is what this problem says.
    GIVEN  f: X \to Y be a function.
    Now we are working in a universe of discourse where  f: X \to Y is a fixed function.

    Now for the problem:
    PROBLEM Prove that there exists a function  g: Y \to X such that  f \circ g = I_Y if and only if  f is a surjection.

    1) Show that  g: Y \to X such that  f \circ g = I_Y implies that  f is a surjection.
    Here the function g implicitly given from which you are proving something about the function f.

    2) Show that  f is a surjection implies that there exists a function  g: Y \to X such that  f \circ g = I_Y .
    In this part, from an assumed property of f you are constructing a function g having a required property.

    Now if you think you have done that, then you mistaken.
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  5. #5
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    Quote Originally Posted by shilz222 View Post
    Actually the  \Leftarrow direction is as follows: Given  f is a surjection prove that there exists a function  g such that  f \circ g = I_Y . It is here where we construct the function  g .
    Well, you have finally seen the light.
    But of course, I said that in my very first reply.

    The point of f being a surjection means that the collection \left\{ {\overleftarrow f (y)} \right\}_{y \in Y} forms a partition of X.
    Now one does have to use the axiom of choice on that collection in order to construct g.
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