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Math Help - If gof is injective, then f is injective

  1. #1
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    If gof is injective, then f is injective

    Let f:A \rightarrow B and g: B \rightarrow C be functions.

    Show that if g \circ f is injective, then f is injective.

    Here is what I did.

    Proof. Spse. g \circ f is injective and f is not injective. Then  \exists x_1,x_2 \in A \ni f(x_1)=f(x_2) but x_1 \neq x_2.
    But then g \circ f(x_1)=g \circ f(x_2) would imply x_1 \neq x_2 thus g \circ f is not injective. A contradiction.
    Therefore if g \circ f is injective, then f is injective.

    Is this OK? for some reason I feel like I am missing something or that my logic is incorrect.

    Thanks

    dubito
    Last edited by dubito; October 17th 2009 at 10:49 PM.
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  2. #2
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    Quote Originally Posted by dubito View Post
    Let f:A \rightarrow B and g: B \rightarrow C be functions.

    Show that if g \circ f is injective, then f is injective.

    Here is what I did.

    Proof. Spse. g \circ f and f is not injective. Then  \exists x_1,x_2 \in f \ni f(x_1)=f(x_2) but x_1 \neq x_2.
    But then x_1,x_2 \in g \circ f would imply x_1 \neq x_2 thus g \circ f is not injective. A contradiction.
    Therefore if g \circ f is injective, then f is injective.

    Is this OK? for some reason I feel like I am missing something or that my logic is incorrect.

    Thanks

    dubito
    Seems you have confused yourself.
    <br />
\exists x_1,x_2 \in f \ni f(x_1)=f(x_2)<br />



    etc make no sense to me.

    gof is not a 'set', so how can you say x1,x2 belong to gof? What does it mean??
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  3. #3
    Super Member redsoxfan325's Avatar
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    Quote Originally Posted by dubito View Post
    Let f:A \rightarrow B and g: B \rightarrow C be functions.

    Show that if g \circ f is injective, then f is injective.

    Here is what I did.

    Proof. Spse. g \circ f and f is not injective. Then  \exists x_1,x_2 \in f \ni f(x_1)=f(x_2) but x_1 \neq x_2.
    But then x_1,x_2 \in g \circ f would imply x_1 \neq x_2 thus g \circ f is not injective. A contradiction.
    Therefore if g \circ f is injective, then f is injective.

    Is this OK? for some reason I feel like I am missing something or that my logic is incorrect.

    Thanks

    dubito
    Assume that f is not injective; that is, \exists~x\neq y such that f(x)=f(y). Then g(f(x))=g(f(y)), but since x\neq y, this implies that g\circ f is not injective either. \Rightarrow\Leftarrow
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  4. #4
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    Thanks guys! I reread my proof and I am like . @ aman_cc, I meant x_1,x_2 \in A sry about that (if I remember correctly though I think my teacher used to do the same thing). @ redsoxfan325 .

    dubito
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  5. #5
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    Quote Originally Posted by dubito View Post
    Thanks guys! I reread my proof and I am like . @ aman_cc, I meant x_1,x_2 \in A sry about that (if I remember correctly though I think my teacher used to do the same thing). @ redsoxfan325 .

    dubito
    Another quick way

    If f(x1)= f(x2) =>
    gof(x1) = g(f(x1))=g(f(x2)) = gof(x2) =>
    x1 = x2 (as gof is injective)

    Hence done
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