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Math Help - Show that a function is injective

  1. #1
    Super Member Showcase_22's Avatar
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    Show that a function is injective

    Suppose that f:A \rightarrow B and g:B \rightarrowC are mappings. Show that:

    1. f and g are both injective \Rightarrow g \circf is injective
    I did it like this:

    Let a \in A, b \in B and  c \in C.

    If f and g are injective \forall x,y \inA, f(x)=f(y) \Rightarrow x=y.

    Therefore: f(a)=b and g(b)=c where a is unique to b and b is unique to c (one to one function by the injective property).

    g \circ f(x)=g(f(x))

    Let x=a:

    g \circ f(a)=g(f(a))=g(b)=c

    Since a is unique to be and b is unique to c then g\circ f(a) is also a one to one function. Hence g \circ f is injective.

    __________________________________________________ _________

    My issue here is the part where I mention that f(a)=b and g(b)=c. Is this a valid claim?
    Last edited by Showcase_22; November 17th 2008 at 10:49 AM.
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  2. #2
    MHF Contributor

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    Quote Originally Posted by Showcase_22 View Post
    Suppose that f:A B and g:BC are mappings. Show that: 1. f and g are both subjective g f is injective
    Look at the question you posted.
    Should both be injective?
    It seems that you are assuming that.
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  3. #3
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    I don't know what "subjective" means here. If it was supposed to be "surjective", the statement is not true:
    Let A= B= {x, y, z}, B= {a, b}, C= {p} f:A->B defined by f(x)= a, f(y)= a, f(z)= b, g:B->C defined by g(a)= p, g(b)= p. Then gof(x)= g(a)= p, gof(y)= f(a)= p, gof(z)= g(b)= p so both f and g are surjective but gof is not injective.

    Since you asserted in your proof that f and g are both injective, is it possible that the problem supposed to be "if f and g are both injective prove that gof is injective"?
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  4. #4
    Super Member Showcase_22's Avatar
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    I did mean "surjective" instead of "subjective". However, I did mean what you posted back. I've changed the question now.

    I was doing the second question (it's the same as this one with "surjective" instead of "injective") and I had "surjective" in my head while I was typing.

    Thinking about it using diagrams, if g and f are surjective then gof cannot be injective as well.

    So is my proof about right?
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  5. #5
    Moo
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    Hello,

    it's not very difficult for the first one

    use this property of injective functions :
    f is injective <=> [f(x)=f(y) => x=y]

    so start from g \circ f(x)=g \circ f(y) where x and y belong to C.

    so g(f(x))=g(f(y))

    but since g is injective, we have f(x)=f(y)

    and since f is injective, we have ....
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  6. #6
    Super Member Showcase_22's Avatar
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    Since f is injective we have x=y.

    Therefore gof(x)=gof(y) \Rightarrow x=y so gof is injective.

    I think I get it now.
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