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Let $R$ be a ring and let $$ M_1 \stackrel{f_1}{\hookrightarrow} M_2 \\ \alpha_1\downarrow \hspace{1cm} \downarrow \alpha_2 \\ N_1 \stackrel{f_2}{\hookrightarrow} N_2 $$ be a commutative diagram of $R$-modules in which the two horizontal maps are injective. I need to show that there exists an $R$-module $E$ and an exact sequence $$ \DeclareMathOperator{\coker}{coker} 0 \to \ker(\alpha_1) \to \ker (\alpha_2) \to E \to \coker (\alpha_1) \to \coker(\alpha_2) $$ of $R$-modules.

What I have tried so far: extending the diagram on the right with the projection maps $g_1 : M_2 \to \coker(f_1)$ and $g_2 : N_2 \to \coker(f_2)$, and the map $\alpha_3 : \coker(f_1) \to \coker(f_2) : \overline{x} \mapsto \overline{\alpha_2(x)}$ (which can be shown to be well-defined). Then we can use Snake's lemma to obtain an exact sequence $$ 0 \to \ker(\alpha_1) \to \ker (\alpha_2) \to \ker(\alpha_3) \to \coker(\alpha_1) \to \coker (\alpha_2) \to \coker(\alpha_3) \to 0 $$ Hence if we take $E = \ker(\alpha_3)$ and just `cut-off' the last two elements of this exact sequence, we are done. Is this correct?

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  • $\begingroup$ That is correct $\endgroup$ – Hagen von Eitzen Feb 22 at 18:06
  • $\begingroup$ Thank you for checking! $\endgroup$ – Sigurd Feb 22 at 18:18

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