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Math Help - g(f(x) is injective, is f injective?

  1. #1
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    g(f(x) is injective, is f injective?

    If g(f(x) is injective I need to determine if f is injective. If it is false, I need to give a counterexample.
    I was told that using arrow diagrams helps.
    Can some one help me get started?
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  2. #2
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    I know g doesn't have to be injective.
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    I'm think that it does but I am unsure of how to prove this.
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    So g(f(x)) is 1-1, so g(f( a_1)=g(f( a_2). Then this implies f( a_1)=f( a_2).

    Where do I go from here? Would that last statement prove that a_1=a_2? therefore, f is one to one?
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    Quote Originally Posted by kathrynmath View Post
    So g(f(x)) is 1-1, so g(f( a_1)=g(f( a_2). Then this implies f( a_1)=f( a_2).

    Where do I go from here? Would that last statement prove that a_1=a_2? therefore, f is one to one?
    hint: g(f(x)) is another way of writing (g \circ f)(x)

    (the statement is true) now you need to show that if f(x_1) = f(x_2) then x_1 = x_2
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    Quote Originally Posted by Jhevon View Post
    hint: g(f(x)) is another way of writing (g \circ f)(x)

    (the statement is true) now you need to show that if f(x_1) = f(x_2) then x_1 = x_2
    Ok, so, I'm unsure on how to prove that f(x_1)=f(x_2) that x_1=x_2. This is where I get stuck.
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    Quote Originally Posted by kathrynmath View Post
    Ok, so, I'm unsure on how to prove that f(x_1)=f(x_2) that x_1=x_2. This is where I get stuck.
    yes, and i told you how to get around it. follow my hint

    to start you off: assume f(x_1) = f(x_2). then g(f(x_1)) = g(f(x_2)). but that means ...
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    Quote Originally Posted by Jhevon View Post
    yes, and i told you how to get around it. follow my hint

    to start you off: assume f(x_1) = f(x_2). then g(f(x_1)) = g(f(x_2)). but that means ...
    Ok, so I start off by assumming f( x_1)=f( x_2)? Then g(f(x_1)=g(f(x_2). g(f) is injective, so f( x_1)=f(x_2).
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    Quote Originally Posted by kathrynmath View Post
    Ok, so I start off by assumming f( x_1)=f( x_2)? Then g(f(x_1)=g(f(x_2). g(f) is injective, so f( x_1)=f(x_2).
    ...that brings us back where we started. we know f(x_1) = f(x_2), that's what we assumed! you are supposed to apply my hint where i left off
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    Quote Originally Posted by Jhevon View Post
    ...that brings us back where we started. we know f(x_1) = f(x_2), that's what we assumed! you are to apply my hint there...
    Ok, can I just say x_1=x_2 because g(f(x_1))=g(f(x_2)) and g(f(x)) is injective?
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    Quote Originally Posted by kathrynmath View Post
    Ok, can I just say x_1=x_2 because g(f(x_1))=g(f(x_2)) and g(f(x)) is injective?
    assume f(x_1) = f(x_2). then g(f(x_1)) = g(f(x_2)). but that means (g \circ f)(x_1) = (g \circ f)(x_2). since g \circ f is injective, this means x_1 = x_2. thus, f is injective
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    Quote Originally Posted by Jhevon View Post
    assume f(x_1) = f(x_2). then g(f(x_1)) = g(f(x_2)). but that means (g \circ f)(x_1) = (g \circ f)(x_2). since g \circ f is injective, this means x_1 = x_2. thus, f is injective
    Ok, that makes sense.
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