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I can give an argument of "$3\implies 1$" as:

Since $H_n(0)\to H_n(C.)$ is an isomorphism, and $H_n(0)=0$, then $H_n(C.)=0$, and which means $Z_n(C.)=B_n(C.)$ by the definition of quotient module, namely $C.$ is exact at $C_n$.

However, this seems to be ugly, because my argument is through $2.$ in fact. And my question is, how can I give a "direct" proof?

闫嘉琦
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  • You can't be more direct than what you did. The thing is, property $(2)$ is the definition of property $(3)$ (or is close enough to be basically indistinguishable from the definition), so any argument will automatically have to go through $(2)$. – Captain Lama Mar 06 '20 at 15:44
  • @CaptainLama I see. thx. – 闫嘉琦 Mar 07 '20 at 08:12

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