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Math Help - Lemma of 5 and an alternative way

  1. #1
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    Lemma of 5 and an alternative way

    Hello

    I've done an exercise in which the 3rd question was the lemma of 5 itself. Now, I've done it in a non conventional way and my teacher says it's wrong but she didn't point out exactly what was wrong with my solutions. I'll just post one of them and please tell me where exactly it fails, if it indeed fails (I'm still convinced it doesn't fail). I haven't done this in latex since it seems to be a nightmare. I'll try to do it without latex:

    M_1- f_1-> M_2- f_2-> M_3- f_3-> M_4- f_4-> M_5

    h_1  h_2 h_3 h_4 h_5

    N_1- g_1-> N_2- g_2-> N_3- g_3-> N_4- g_4-> N_5

    Hhmm. Imagine the h_i from M_i to N_i. Now, the exercise is: Consider the above commutative diagram in which each horizontal sequence is exact. If h_1 is an epimorphism and h_4 is a monomorphism, then Ker h_3=f_2(Ker h_2).

    My proposal of solution doesn't need that h_4 is a monomorphism, and is done in the following way:

    f_2(Ker h_2)=f_2(h_2^{-1}(\{o\}))=f_2(f_1(h_1^{-1}(g_1^{-1}(\{0\}))))=\{0\}

    h_3^{-1}(\{0\})=f_2(f_1(h_1^{-1}(g_1^{-1}(g_2^{-1}(\{0\})))))=\{0\}

    Where is this wrong, if at all?

    Thank you in advance.

    PS: a good way to visualize this is to draw a Venn diagram with circles for M_i, N_i, Ker f_i and Ker g_i
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  2. #2
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    Re: Lemma of 5 and an alternative way

    Quote Originally Posted by ModusPonens View Post
    Hello

    I've done an exercise in which the 3rd question was the lemma of 5 itself. Now, I've done it in a non conventional way and my teacher says it's wrong but she didn't point out exactly what was wrong with my solutions. I'll just post one of them and please tell me where exactly it fails, if it indeed fails (I'm still convinced it doesn't fail):

    \begin{array}{ccccccccc} M_1&\stackrel{f_1}\longrightarrow& M_2&\stackrel{f_2}\longrightarrow& M_3&\stackrel{f_3}\longrightarrow& M_4&\stackrel{f_4}\longrightarrow& M_5\\ \hspace*{9pt}\downarrow\,\scriptstyle h_1&& \hspace*{9pt}\downarrow\,\scriptstyle h_2&& \hspace*{9pt}\downarrow\,\scriptstyle h_3&& \hspace*{9pt}\downarrow\,\scriptstyle h_4&& \hspace*{9pt}\downarrow\,\scriptstyle h_5\\ N_1&\stackrel{g_1}\longrightarrow& N_2&\stackrel{g_2}\longrightarrow& N_3&\stackrel{g_3}\longrightarrow& N_4&\stackrel{g_4}\longrightarrow& N_5 \end{array}

    Hhmm. Imagine the h_i from M_i to N_i. Now, the exercise is: Consider the above commutative diagram in which each horizontal sequence is exact. If h_1 is an epimorphism and h_4 is a monomorphism, then Ker h_3=f_2(Ker h_2).

    My proposal of solution doesn't need that h_4 is a monomorphism, and is done in the following way:

    f_2(Ker h_2)=f_2(h_2^{-1}(\{0\}))=f_2(f_1(h_1^{-1}(g_1^{-1}(\{0\}))))=\{0\}

    h_3^{-1}(\{0\})=f_2(f_1(h_1^{-1}(g_1^{-1}(g_2^{-1}(\{0\})))))=\{0\}

    Where is this wrong, if at all?
    It is wrong to assume that h_2^{-1}\{0\} = f_1(h_1^{-1}(g_1^{-1}(\{0\}))). That would imply that the kernel of h_2 is contained in the image of f_1, which need not be the case.

    I think the best approach here is just tedious but routine diagram chasing. Suppose that m_3\in\ker(h_3). Then

    (1) g_3h_3(m_3) = 0,
    (2) h_4f_3(m_3) = 0,
    (3) f_3(m_3) = 0 because h_4 is injective,
    (4) m_3 = f_2(m_2) for some m_2\in M_2, and g_2h_2(m_2) = 0,
    (5) h_2(m_2) = g_1(n_1) for some n_1\in N_1,
    (6) h_2(m_2) = g_1h_1(m_1) = h_2f_1(m_1) for some m_1\in M_1 because h_1 is surjective,
    (7) m_3 = f_2(m_2-f_1(m_1)), and h_2(m_2-f_1(m_1)) = 0.

    Thus m_3\in f_2(\ker(h_2)). This shows that \ker(h_3)\subseteq f_2(\ker(h_2)). The reverse inclusion would be proved in a similar way.
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  3. #3
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    Re: Lemma of 5 and an alternative way

    Thank you for the solution, but I already had it.

    Anyway, that's not a convincing reason for the non validity of my solution. Why shouldn't be so? I don't understand. Maybe the surjectivity of h_1 implies that the kernel of h_2 is contained in the image of f_1.
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  4. #4
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    Re: Lemma of 5 and an alternative way

    Quote Originally Posted by ModusPonens View Post
    that's not a convincing reason for the non validity of my solution. Why shouldn't be so?
    What you know about the leftmost square in the diagram is that h_1 is surjective and that the square commutes, so that g_1h_1 = h_2f_1. It does not follow that h_2^{-1}(\{0\}) = f_1h_1^{-1}g_1^{-1}(\{0\}). For example, suppose that N_1  = \{0\}, that M_1,M_2,N_2 are all equal to  \mathbb{R}, that g_1 is the inclusion map and that each of f_1,h_1,h_2 is the zero map. Then h_2^{-1}(\{0\}) = \mathbb{R} but f_1h_1^{-1}g_1^{-1}(\{0\}) = \{0\}. If you think that there is some additional structure in this problem that makes such an example impossible, then it's up to you to demonstrate that.
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  5. #5
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    Re: Lemma of 5 and an alternative way

    i thought of another counter-example, with h2 not being trivial:

    let G be a 3x3 unit grid in the plane (say from (0,0) to (3,3)), and let M1 = , M2 = the group of all polygonal 2-chains on G (with addition mod 2), M3 = the group of all polygonal 1-cycles (under addition mod 2) on G, M4 = M5 = , with all the f maps being the boundary operator, defined in the usual way (the only non-null images are the k-1-chains that occur in an odd number of k-chains).

    let N1 = {0}, N2 = Z2, N3 = Z2, N4,N5 = {0}. define h1 the only way possible --->0, and define h2,h3, to be the number of simplexes in each chain (mod 2), h4, and h5 the 0-map.

    (i hope i've done this right...it's been a loooong time).

    this should define two exact rows, h1 is epi, h4 is mono. and indeed, ker(h3) = {∂C : C in M2}.

    now h2^-1(0) is the set of all 2-chains with an even number of simplexes (squares) in them. but the image of the null set (M1) is ∂ = ,

    so the kernel of h2 is not contained within the image of f1 = ∂.
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  6. #6
    Member ModusPonens's Avatar
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    Re: Lemma of 5 and an alternative way

    Thank you for your counter examples.
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