Snake Lemma

From ProofWiki
Jump to navigation Jump to search



Theorem

Let $A$ be a commutative ring with unity.

Let:

$\begin{xy}\[email protected][email protected]+1em{ & M_1 \ar[r]_*{\alpha_1} \ar[d]^*{\phi_1} & M_2 \ar[r]_*{\alpha_2} \ar[d]^*{\phi_2} & M_3 \ar[d]^*{\phi_3} \ar[r] & 0 \\ 0 \ar[r] & N_1 \ar[r]_*{\beta_1} & N_2 \ar[r]_*{\beta_2} & N_3 & }\end{xy}$

be a commutative diagram of $A$-modules.

Suppose that the rows are exact.


Then we have a commutative diagram:


$\begin{xy}\[email protected][email protected]+1em{ & \ker \phi_1 \ar[r]_*{\tilde\alpha_1} \ar[d]^*{\iota_1} & \ker \phi_2 \ar[r]_*{\tilde\alpha_2} \ar[d]^*{\iota_2} & \ker \phi_3 \ar[d]^*{\iota_3} & \\ & M_1 \ar[r]_*{\alpha_1} \ar[d]^*{\phi_1} & M_2 \ar[r]_*{\alpha_2} \ar[d]^*{\phi_2} & M_3 \ar[d]^*{\phi_3} \ar[r] & 0 \\ 0 \ar[r] & N_1 \ar[r]_*{\beta_1} \ar[d]^*{\pi_1} & N_2 \ar[r]_*{\beta_2} \ar[d]^*{\pi_2} & N_3 \ar[d]^*{\pi_3} \\ & \operatorname{coker} \phi_1 \ar[r]_*{\bar\beta_1} & \operatorname{coker} \phi_2 \ar[r]_*{\bar\beta_2} & \operatorname{coker} \phi_3 & }\end{xy}$

where, for $i = 1, 2, 3$ respectively $i = 1, 2$:

$\forall n_i + \Img {\phi_i} \in \operatorname {coker} \phi_i : \map {\bar \beta_i} {n_i + \Img {\phi_i} } = \map {\beta_i} {n_i} + \Img {\phi_{i + 1} }$


Moreover there exists a morphism $\delta : \ker \phi_3 \to \operatorname{coker} \phi_1$ such that we have an exact sequence:


$\begin{xy}\[email protected][email protected]+1em{ \ker \phi_1 \ar[r]_*{\tilde\alpha_1} & \ker \phi_2 \ar[r]_*{\tilde\alpha_2} & \ker\phi_3 \ar[r]_*{\delta} & \operatorname{coker}\phi_1 \ar[r]_*{\bar\beta_1} & \operatorname{coker}\phi_2 \ar[r]_*{\bar\beta_2} & \operatorname{coker}\phi_3 }\end{xy}$


Proof