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Thread: 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 $\displaystyle \in $C, there exists b $\displaystyle \in $ B such that f(b) = c. Now since g is surjective, there exists an a $\displaystyle \in$ 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 $\displaystyle f\circ g$ to be injective $\displaystyle f(g(x))=f(g(y))\implies x=y$. Assume $\displaystyle f(g(x))=f(g(y))$. Since $\displaystyle f$ is injective...

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


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

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