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  1. #1
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    Angry prove

    Assume that alpha: S-T, and beta: T-U, and gamma: T-U. Prove the following statement

    If beta of alpha is invertible, then beta is onto and alpha is one-to-one
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by mandy123 View Post
    Assume that alpha: S-T, and beta: T-U, and gamma: T-U. Prove the following statement

    If beta of alpha is invertible, then beta is onto and alpha is one-to-one
    i might as well give a full solution. the proof for this is all over the internet and is in most textbooks that deal with functions indepth. it might as well be on here...or ami just re-inventing the wheel? ...or maybe bored.

    Proof:
    We proceed using a direct proof. Assume \beta \circ \alpha is invertible. Then \beta \circ \alpha is one-to-one and onto (there is a theorem that tells us this, it should be in your text).

    We show \beta is onto: Let z \in U. Since \beta \circ \alpha is onto, there exists x \in S such that (\beta \circ \alpha)(x) = z. But that means \beta (\alpha (x)) = z with \alpha (x) \in T. Thus \beta is onto.

    We show \alpha is one-to-one: Assume x_1, x_2 \in S with \alpha (x_1) = \alpha (x_2). Then \beta (\alpha (x_1)) = \beta (\alpha (x_2)), or in other words, (\beta \circ \alpha)(x_1) = (\beta \circ \alpha)(x_2). But \beta \circ \alpha is one-to-one, so (\beta \circ \alpha)(x_1) = (\beta \circ \alpha)(x_2) \implies x_1 = x_2. Hence, \alpha is one-to-one

    QED
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