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Math Help - Prooving surjective and injective functions

  1. #1
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    Unhappy Prooving surjective and injective functions

    Let g : A -> B and f : B -> C be functions.
    Show that if f and g are injective, then f o g is injective.
    Show that if f and g are surjective, then f o g is surjective.

    i don't know how to proove it
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  2. #2
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    since f is surjective, rng f = C. That is, for any c  \in C, there exists b  \in B such that f(b) = c. Now since g is surjective, there exists an a \in A such that g(a) = b. but then (f o g)(a) = f(g(a)) = f(b) = c, so
    f o g is surjective.
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  3. #3
    Super Member redsoxfan325's Avatar
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    Quote Originally Posted by neelpatel89 View Post
    Let g : A -> B and f : B -> C be functions.
    Show that if f and g are injective, then f o g is injective.
    Show that if f and g are surjective, then f o g is surjective.

    i don't know how to proove it
    For f\circ g to be injective f(g(x))=f(g(y))\implies x=y. Assume f(g(x))=f(g(y)). Since f is injective...

    Spoiler:
    Since f is injective, g(x)=g(y). Since g is injective, x=y, so f(g(x))=f(g(y))\implies x=y, and f\circ g is injective. \square


    For f\circ g to be surjective, \forall~y\in C, \exists~x\in A such that f(g(x))=y. Let y\in C. Because f is surjective...

    Spoiler:
    Because f is surjective, there exists z\in B such that f(z)=y. Because g is surjective, there exists x\in A such that f(x)=z. So f(g(x))=y and f\circ g is surjective. \square
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