As in the proof of the 5-lemma, we assume without loss of generality that we are working in modules over a ring. In keeping with the notion that the maps between the $A$ 's (as well as between the $B$ 's and the $C$ 's) are cohomology sequences, we denote all vertical maps by $d$ . The map $A_i\to B_i$ is denoted $\alpha_i$ , and the map $B_i\to C_i$ is denoted $\beta_i$ . We must show that

1. $\beta_1$ is surjective;
2. $\alpha_1$ is injective;
3. $\ker \beta_1 \subset \mathrm{im}\ \alpha_1$ ;
4. $\beta_1\circ\alpha_1=0$ (i.e. $\ker \beta_1 \supset \mathrm{im}\ \alpha_1$ )

$\beta_1$ is surjective: Choose $c\in C_1$ . Then $dc = \beta_2 b$ , and $\beta_3db=d\beta_2b=d^2c=0$ , so $db=\alpha_3a=\alpha_3 da'$ . Thus $d(b-\alpha_2 a')=0$ , so $db'=b-\alpha_2 a'$ . Finally, $d\beta_1 b'=\beta_2 db'=\beta_2(b-\alpha_2a')=\beta_2 b=dc$ . But $d$ is injective, so $c=\beta_1 b'$ .

$\alpha_1$ is injective: This is clear, since $d\alpha_1=\alpha_2 d$ , and $\alpha_2$ and both $d$ 's are injective.

$\ker \beta_1\subset \mathrm{im}\ \alpha_1$ : Suppose $\beta_1(b)=0$ . Then $\beta_2 db=d\beta_1 b=0$ , so $db=\alpha_2 a$ . But then $\alpha_3 da=d\alpha_2 a=d^2b=0$ , and $\alpha_3$ is injective, so $a\in\ker d$ and $da'=a$ . Finally, $d\alpha_1 a'=\alpha_2 da'=\alpha_2 a=db$ . $d$ is injective and thus $b=\alpha_1 a'$ .

$\beta_1\circ\alpha_1=0$ : $d\beta_1\alpha_1=\beta_2\alpha_2d=0$ . But $d$ is injective, so $\beta_1\alpha_1=0$ .

Similar diagram chasing can be used to prove that if the top two rows are exact then so is the bottom row.