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Thread: 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:

    $\displaystyle M_1$-$\displaystyle f_1$->$\displaystyle M_2$-$\displaystyle f_2$->$\displaystyle M_3$-$\displaystyle f_3$->$\displaystyle M_4$-$\displaystyle f_4$->$\displaystyle M_5$

    $\displaystyle h_1$ $\displaystyle h_2$ $\displaystyle h_3 $ $\displaystyle h_4 $ $\displaystyle h_5$

    $\displaystyle N_1$-$\displaystyle g_1$->$\displaystyle N_2$-$\displaystyle g_2$->$\displaystyle N_3$-$\displaystyle g_3$->$\displaystyle N_4$-$\displaystyle g_4$->$\displaystyle N_5$

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

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

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

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

    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 $\displaystyle M_i$, $\displaystyle N_i$, $\displaystyle Ker f_i$ and $\displaystyle Ker g_i$
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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):

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

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

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

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

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

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

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

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

    Thus $\displaystyle m_3\in f_2(\ker(h_2))$. This shows that $\displaystyle \ker(h_3)\subseteq f_2(\ker(h_2))$. The reverse inclusion would be proved in a similar way.
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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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    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 $\displaystyle h_1$ is surjective and that the square commutes, so that $\displaystyle g_1h_1 = h_2f_1.$ It does not follow that $\displaystyle h_2^{-1}(\{0\}) = f_1h_1^{-1}g_1^{-1}(\{0\}).$ For example, suppose that $\displaystyle N_1 = \{0\},$ that $\displaystyle M_1,M_2,N_2$ are all equal to $\displaystyle \mathbb{R},$ that $\displaystyle g_1$ is the inclusion map and that each of $\displaystyle f_1,h_1,h_2$ is the zero map. Then $\displaystyle h_2^{-1}(\{0\}) = \mathbb{R}$ but $\displaystyle f_1h_1^{-1}g_1^{-1}(\{0\}) = \{0\}.$ 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
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    Re: Lemma of 5 and an alternative way

    Thank you for your counter examples.
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