# Snake lemma proof

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 !

• 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 $\mathrm{im\, } v$.. – Berci Sep 23 '18 at 10:49
• 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. – ChocoSavour Sep 23 '18 at 10:54

## 1 Answer

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 $$m_2\in M_2$$.