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Math Help - kernel

  1. #1
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    kernel

    i was reading my notes and it states that the kernel of a function allows you to check if the function is injective.

    how do you check with a kernel? isint elements in the kernel, elements that through the function get mapped to the identity?
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  2. #2
    MHF Contributor Swlabr's Avatar
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    Quote Originally Posted by alexandrabel90 View Post
    i was reading my notes and it states that the kernel of a function allows you to check if the function is injective.

    how do you check with a kernel? isint elements in the kernel, elements that through the function get mapped to the identity?
    What are we "in" here? Linear maps? Group homomorphisms?

    The proof is pretty much the same in every context through.

    Let T be a linear map.

    Then T is injective
    \Leftrightarrow (Tx = Ty \Rightarrow x=y)
    \Leftrightarrow (Tx-Ty = 0 \Rightarrow x=y)
    \Leftrightarrow (T(x-y) = 0 \Rightarrow x = y)
    \Leftrightarrow (T(x-y) = 0 \Rightarrow x-y = 0)
    \Leftrightarrow (Tu = 0 \Rightarrow u =0)
    \Leftrightarrow kerT = \{0\}.
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  3. #3
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    im studying group homomorphisms now.

    in this case, if it is a function from A to B, where the identity of B =0, then the element in A has to be 0 since u=0 (from your above statement) right?

    what if the element in A that get mapped to 0 in B is not 0, wont u be not equals to 0? in such case, Tu=0 but u wont be zero?

    thanks!
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  4. #4
    MHF Contributor Swlabr's Avatar
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    Quote Originally Posted by alexandrabel90 View Post
    im studying group homomorphisms now.

    in this case, if it is a function from A to B, where the identity of B =0, then the element in A has to be 0 since u=0 (from your above statement) right?

    what if the element in A that get mapped to 0 in B is not 0, wont u be not equals to 0? in such case, Tu=0 but u wont be zero?

    thanks!
    My above proof can be quite easily changed to look at group homomorphisms (just change a+b to ab, and a-b to ab^{-1}). So I'll use group notation.

    Is your question basically,

    Is it possible that 1 \not\in ket(\phi)?

    The answer is `no'.

    Let 1 \phi = a. Then a^2 = (1\phi)^2 = (1^2)\phi = 1\phi = a. Thus, a^2=a ( a is an idempotent). Then, a^2 = a \Rightarrow a^2a^{-1} = aa^{-1}  \Rightarrow a=1. Thus, 1_G\phi = 1_H for \phi: G\rightarrow H a group homomorphism.

    The same holds for linear maps, etc.
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  5. #5
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    for the proof under group homomorphism, we are assuming that it is under the operation x right? is that what we usually assume?

    becos im not sure when it should be under + and when it should be under x.

    thanks
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  6. #6
    MHF Contributor Swlabr's Avatar
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    Quote Originally Posted by alexandrabel90 View Post
    for the proof under group homomorphism, we are assuming that it is under the operation x right? is that what we usually assume?

    becos im not sure when it should be under + and when it should be under x.

    thanks
    Just go through the proof I gave for linear maps and replace + with * and T with \phi.

    So it starts,

    \phi is injective
    \Leftrightarrow (g\phi = h\phi \Rightarrow g=h)
    \Leftrightarrow ((g\phi)(h\phi)^{-1} = 1 \Rightarrow g=h)
    \vdots
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  7. #7
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    Quote Originally Posted by alexandrabel90 View Post
    for the proof under group homomorphism, we are assuming that it is under the operation x right? is that what we usually assume?

    becos im not sure when it should be under + and when it should be under x.

    thanks
    You can use any symbol you like for the operation!

    It is a general convention that \abelian (commutative) groups are written "additively" and non-abelian groups are written "multiplicatively".
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