PlanetMath: proof of 9-lemma

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proof of 9-lemma

(Proof)

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 's (as well as between the 's and the 's) are cohomology sequences, we denote all vertical maps by . 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

$\beta_1$ is surjective;
$\alpha_1$ is injective;
$\ker \beta_1 \subset \mathrm{im}\ \alpha_1$ ;
$\beta_1\circ\alpha_1=0$ (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 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 '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 . Finally, $d\alpha_1 a'=\alpha_2 da'=\alpha_2 a=db$ . is injective and thus $b=\alpha_1 a'$ .