Connecting Homomorphism is Functorial

Theorem

Let $A$ be a ring with unity.

Let:

$\begin{xy}\xymatrix{ &&& M_1 \ar@{->}[rr] \ar@{->}[dl]^{f_1} \ar@{->}[dd]^{\phi_1}|!{[d];[d]}\hole && M_2 % \ar@{->}[rr] \ar@{->}[dl]^{f_2} \ar@{->}[dd]^{\phi_2}|!{[d];[d]}\hole && M_3 \ar@{->}[dl]^{f_3} \ar@{->}[dd]^{\phi_3}|!{[d];[d]}\hole \ar@{->}[rr] && 0 \\ && M_1' \ar@{->}[rr] \ar@{->}[dd]^{\phi_1'} && M_2' \ar@{->}[rr] \ar@{->}[dd]^{\phi_2'} && M_3' \ar@{->}[dd]^{\phi_3'} \ar@{->}[rr] && 0 \\ & 0 \ar@{->}[rr]|!{[r];[r]}\hole && N_1 \ar@{->}[rr]|!{[r];[r]}\hole \ar@{->}[dl]_{g_1} && N_2 \ar@{->}[rr]|!{[r];[r]}\hole \ar@{->}[dl]_{g_2} && N_3 \ar@{->}[dl]_{g_3} \\ % 0 \ar@{->}[rr] && N_1' \ar@{->}^(.65){e’}[rr] && N_2' \ar@{->}[rr] && N_3' }\end{xy}$

be a commutative diagram of $A$-modules.

Suppose that the rows are exact.

Let

$\delta : \map \ker {\phi_3} \to \map {\operatorname {coker} } {\phi_1}$

$\delta' : \map \ker {\phi_3'} \to \map {\operatorname {coker} } {\phi_1'}$

be the boundary homomorphisms coming from Snake Lemma applied to the front diagram and the back diagram.

Then the diagram:

$\begin{xy}\xymatrix{ \map \ker {\phi_3} \ar[r]^{\delta} \ar[d] & \map {\operatorname{coker} } {\phi_1} \ar[d] \\ \map \ker {\phi_3'} \ar[r]^{\delta'} & \map {\operatorname{coker} } {\phi_1'} \\ } \end{xy}$

is commutative.

Here the vertical arrows in the commutative square are induced by $\tuple {f_3, g_3}$ and $\tuple {f_1, g_1}$.

This needs considerable tedious hard slog to complete it. To discuss this page in more detail, feel free to use the talk page. When this work has been completed, you may remove this instance of {{Finish}} from the code. If you would welcome a second opinion as to whether your work is correct, add a call to {{Proofread}} the page.

Proof

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