# Proof-verification: Show that there exists an exact sequence

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?

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