Connecting Homomorphism is Functorial

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Let $A$ be a ring with unity.


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

be a commutative diagram of $A$-modules.

Suppose that the rows are exact.


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

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 $\tuple {f_1, g_1}$.