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      "quote": "Proof. Both sides are groups according to lemma 3.1. Define a map  $\\varphi : G_1 / \\ker f \\to \\operatorname{im} f$  by  $\\varphi([x]) = f(x)$ . This map is well defined since for  $x' \\in [x]$ , there exists  $h \\in \\ker f$  such that  $x' = x + h$  and  $f(x') = f(x + h) = f(x) + f(h) = f(x)$ . Now we show that  $\\varphi$  is an isomorphism. First,  $\\varphi$  is a homomorphism,",
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      "quote": "From lemma 3.1, it follows that  $Z_{r}(K)$  and  $B_{r}(K)$  are subgroups of  $C_{r}(K)$ . We now prove an important relation between  $Z_{r}(K)$  and  $B_{r}(K)$ , which is crucial in the definition of homology groups.",
      "citation": "From lemma 3.1",
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      "quote": "Proof. This is obvious from lemma 3.3. Any element  $c$  of  $B_r(K)$  is written as  $c = \\partial_{r+1}d$  for some  $d \\in C_{r+1}(K)$ . Then we find  $\\partial_r c = \\partial_r(\\partial_{r+1}d) = 0$ , that is,  $c \\in Z_r(K)$ . This implies  $Z_r(K) \\supset B_r(K)$ .",
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      "quote": "From lemma 3.1, it follows that  $Z_{r}(K)$  and  $B_{r}(K)$  are subgroups of  $C_{r}(K)$ . We now prove an important relation between  $Z_{r}(K)$  and  $B_{r}(K)$ , which is crucial in the definition of homology groups.",
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