Exercise A0.III.7.03

Proposition

Suppose we have an exact complex

$$\cdots \longrightarrow \; 0 \overset{f_1}\longrightarrow L \overset{f_2}\longrightarrow M \overset{\varphi}\longrightarrow M' \overset{f_3}\longrightarrow N \overset{f_4}\longrightarrow 0 \longrightarrow \cdots.$$

Then $L =\ker \varphi$ and $N = \coker \varphi,$ up to natural identifications.

Proof

Exactness aid

• $\im f_1 = 0 = \ker f_2$
• $\im f_2 = \ker \varphi$
• $\im \varphi = \ker f_3$
• $\im f_3 = N = \ker f_4$
• $\im f_4 = 0$

Since $\ker f_2$ is trivial, it is injective. Since every morphism is surjective onto its image, we have $\ker \varphi = \im f_2 \cong L.$

Note that the kernel of $f_4$ is the whole $N$, and since $\im f_3 = N = \ker f_4,$ we know that $f_3$ is an epi. By corollary 5.16 we have

$$N \cong M'/\ker f_3 = M'/\im \varphi = \coker \varphi.$$

$\blacksquare$