# Thread: Lemma of 5 and an alternative way

1. ## 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?

PS: a good way to visualize this is to draw a Venn diagram with circles for $M_i$, $N_i$, $Ker f_i$ and $Ker g_i$

2. ## Re: Lemma of 5 and an alternative way

Originally Posted by ModusPonens
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.

3. ## Re: Lemma of 5 and an alternative way

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.

4. ## Re: Lemma of 5 and an alternative way

Originally Posted by ModusPonens
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.

5. ## 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 = Ø, 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 = Ø, 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 Ø--->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 ∂Ø = Ø,

so the kernel of h2 is not contained within the image of f1 = ∂.

6. ## Re: Lemma of 5 and an alternative way

Thank you for your counter examples.