$L\xrightarrow{\alpha} M\xrightarrow{\beta} N\rightarrow 0$ is exact if and only if $N\cong\mathrm{coker}(\alpha)$

Question

Clearly, if a sequence of module homomorphisms $L\xrightarrow{\alpha} M\xrightarrow{\beta} N\rightarrow 0$ is exact, $N\cong\mathrm{coker}(\alpha)$ as $\mathrm{coker}(\alpha)=M/\mathrm{im}(\alpha)$, $\beta$ is surjective and $\mathrm{im}(\alpha)=\mathrm{ker}(\beta)$. But I am not sure if the other direction is also true. Is there a counter example?

Answer 1

This is true in any abelian category. Suppose $\operatorname{coker} \alpha = \beta$. This immediately implies that $\beta$ is epi. By taking kernels of both sides, we get $$ \ker(\operatorname{coker} \alpha) = \ker \beta. $$

But $\ker(\operatorname{coker} \alpha) = \operatorname{im} \alpha$. We conclude that $\operatorname{im} \alpha = \ker \beta$ and the sequence is exact.

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If you want to stick to module theory, first note that the canonical map $M \to M / \operatorname{im} \alpha$ is surjective. Now, suppose $m \in \operatorname{im} \alpha$. It is immediate that $\beta(m) = 0$ and thus $m \in \ker \beta$. Next, suppose $m \in \ker \beta$. Likewise, it follows that $m \in \operatorname{im} \alpha$ since $\operatorname{im} \alpha$ is the kernel of the canonical map $M \to M / \operatorname{im} \alpha$. We conclude that the sequence is exact.