I attempted to prove the $\it 3\times3$ Lemma (or Nine Lemma). First off, I am well aware how to proceed by means of diagram chase and I know how to prove the $\it 3\times3$ Lemma doing so. However, I am interested in an alternative way using the Snake Lemma as suggested in P. Aluffi's Algebra: Chapter $\it 0$ (exercise III.$7.15$ for reference, all appearing objects are taken to be $R$-modules over a fixed commutative, unital ring $R$ and $0$ is the trivial module).
To keep this post self-contained, let me state the Snake Lemma
Lemma (Snake Lemma). Given a commutative diagram
where the rows are exact, there is an exact sequence $$ 0\longrightarrow\ker\lambda\longrightarrow\ker\mu\longrightarrow\ker\nu\overset{\delta}\longrightarrow{\rm coker}~\lambda\longrightarrow{\rm coker}~\mu\longrightarrow{\rm coker}~\nu\longrightarrow0 $$
The $\it 3\times3$ Lemma reads as
Lemma ($\it 3\times3$ Lemma). Given a commutative diagram
where the rows and the two rightmost columns are exact, then the leftmost column is exact.
Proof. We need to prove four things:
- $\lambda_2\circ\lambda_1=0$
- $\lambda_1$ is monic
- $\lambda_2$ is epi
- $\ker\lambda_2={\rm im}~\lambda_1$
$1.$ By the commutativity of the diagramm $\gamma_1\circ(\lambda_2\circ\lambda_1)=(\mu_2\circ\mu_1)\circ\alpha_1$, but as the second column is exact $\mu_2\circ\mu_1=0$ and therefore $\gamma_1\circ(\lambda_2\circ\lambda_1)=0$. Since $\gamma_1$ is monic (as the last row is exact) it follows that $\lambda_2\circ\lambda_1=0$.
$2.$ By the commutativity of the diagram $\mu_1\circ\alpha_1=\beta_1\circ\lambda_1$. But as $\mu_1$ and $\alpha_1$ are monic by exactness of the rows and the second column it follows that $\lambda_1$ is monic aswell.
$3.$ Applying the Snake Lemma to the last two rows gives an exact sequence $$0\longrightarrow\ker\lambda_2\longrightarrow\ker\mu_2\longrightarrow\ker\nu_2\overset{\delta}\longrightarrow{\rm coker}~\lambda_2\longrightarrow{\rm coker}~\mu_2\longrightarrow{\rm coker}~\nu_2\longrightarrow0$$ As $\mu_2$ and $\nu_2$ are epi their respective cokernels are trivial. Also, by exactness of the rightmost columns $\ker\mu_2={\rm im}~\mu_1$ and $\ker\nu_2={\rm im}~\nu_1$. While leaving out unneeded parts we obtain the exact sequence $${\rm im}~\mu_1\overset{f}\longrightarrow{\rm im}~\nu_1\overset{\delta}\longrightarrow{\rm coker}~\lambda_2\overset{t}\longrightarrow0$$ On the other hand, given $x\in{\rm im}~\nu_1$ there exist an $y\in M_0$ such that $(\nu_1\circ\alpha_2)(y)=x$ as $\alpha_2$ is epi. But then $\beta_2(\mu_1(x))=y$ by commutativity of the diagram and thus the connecting arrow ${\rm im}~\mu_1\longrightarrow{\rm im}~\nu_1$ has to be epi aswell (!). As the sequence is exact, it follows that $\ker\delta={\rm im}~f={\rm im}~\nu_1$ and $\ker t={\rm im}~\delta=0$. But then $t$ is an isomorphism as it is clearly epi and monic by the aforegoing argumentation. Therefore ${\rm coker}~\lambda_2=0$, that is $\lambda_2$ is epi.
$4.$ I have a vague idea. We have the following exact sequence by the Snake Lemma
From here it should be somehow possible to identifty $\ker\lambda_2$ with ${\rm im}~\lambda_1$ but I do not know how without using a diagram chase.
$\textbf{Q1}:$ Is the proof of $3.$ fine as it stands? I am not sure whether I can conclude that $f$ like this, as the construction of a pre-image for every $x\in{\rm im}~\nu_1$ in ${\rm im}~\mu_1$ does not refer to $f$ directly. On the other hand, I do know that $f$ is uniquely defined by the universal property of $\ker\mu_2$ and I think that should to the trick.
$\textbf{Q2}:$ How can the proof be finished using the Snake Lemma? I am not sure how continue in step $4.$
Thanks in advance!