I post here cause I have a doubt on my proof about the snake lemma. Actually, I have the impression that I use nowhere the commutativity of the diagram. Actually, this is the diagram I consider :

\begin{array}{c} & & M_1 & \xrightarrow{\alpha} & M_2 & \xrightarrow{\beta} & M_3 & \to & 0 \\ & & \downarrow u & & \downarrow v & & \downarrow w \\ 0 & \to & N_1 & \xrightarrow{\alpha'} & N_2 & \xrightarrow{\beta'} & N_3 \end{array}

And I have to determine a linear application : $f : \ker(w) \rightarrow coker(u) $.

This is what I did :

Let $m_3 \in \ker(w)$. As $\ker(w) = Im(\beta)$, let $m_2 \in M_2$ such that $\beta(m_2) = m_3$. We would have the uniqueness of $m_2$, and for that, I consider $\overline{m_2} \in M_2/\ker(\beta)$. Thus, for $m_2, m'_2$ such that $\beta(m_2) = \beta(m'_2) = m_3$, we have : $m_2 - m'_2 \in \ker{\beta}$, and then $\overline{m_2} = \overline{m'_2}$.

It give us a first linear application well defined : $\lambda_1 : \ker(w) \rightarrow M_2/\ker(\beta)$ which send $m_3 \in \ker(w)$ to $\overline{m_2} = m_2 + \ker(\beta)$.

Now, we have that : $\ker(\beta) = Im(\alpha)$, so let $m_3 \in \ker(w)$ and $m_2 \in M_2$ such that $\beta(m_2) = m_3$. Then, we are considering $v(m_2) \in Im(v)$. As $Im(v) = \ker(\beta')$, we have : $v(m_2) \in \ker(\beta')$. But we have as well : $\ker(\beta') = Im(\alpha')$. So : $v(m_2) \in Im(\alpha')$. Thus, let $n_1 \in N_1$ such that : $\alpha'(n_1) = v(m_2)$.

We would like to have $n_1$ independent from $m_2$, but only dependent of $\overline{m_2}$. But, let $m_2, m_2 + \hat{v}$ with $m_2 \in M_2, \hat{v} \in \ker(\beta) = Im(\alpha) = \ker(v)$. Then, let $n_1, n'_1$ such that : $\alpha'(n_1) = v(m_2)$, $\alpha'(n_1') = v(m_2 + \hat{v}) = v(m_2) + 0 = v(m_2)$.

Then : $\alpha'(n_1) = \alpha'(n_1')$, so $n_1 - n_1' \in \ker(\alpha')$. So, we are considering $\overline{n_1} = n_1 + \ker(\alpha')$. But $ker(\alpha') = Im(u)$, so : $\overline{n_1} = n_1 + Im(u)$.

It gives us a second linear application : $\lambda_2 : M_2/\ker(\beta) \rightarrow N_1/Im(u) = coker(u)$ which send $m_2 + \ker(\beta)$ to $n_1 + Im(u)$ as defined previously.

Finally : $\lambda_2 \circ \lambda_1$ is the linear application which suit.

And my question : I have the impression that my proof is right, but I also have the impression that I have not use the commutativity of the diagram, so my proof is probably wrong. But I don't see why.

Someone could help me ? :)

Thank you !

  • $\begingroup$ You seem to bend the exact sequences. In general, there is no connection between $\ker w$ and $\mathrm{im\, }\beta$ or between $\ker\beta'$ and $\mathrm{im\, } v$.. $\endgroup$ – Berci Sep 23 '18 at 10:49
  • 1
    $\begingroup$ Yes, big big mistake, my bad... Actually, only the lines constitute exact sequences... I thought It was all the paths made by the arrows (I hope i'm clear)... I'm going to start again from the beginning. $\endgroup$ – ChocoSavour Sep 23 '18 at 10:54

Actually I believe that there is a "unique" way to proving the Snake lemma, and the commutativity is used for the well-definedness of the "boundary" map $M_3\to N_1$.

I don't think you actually write down the proof of it. You will find it necessary immediately you prove the well-definedness.

Hint: The right square is used for the pull-back of $\alpha'$, and the left square is used for the independence of choice $m_2\in M_2$.


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