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I have some trouble understanding the definition of homology. As the below figure says, $H_n$ is a functor for every $n\in\Bbb Z$. For example, $H_4$ is a functor from $_R\mathbf{Comp}$ to $_R\textbf{Mod}$. Since each object in $_R\mathbf{Comp}$ is a chain complex and a funtor maps each object in $_R\mathbf{Comp}$ to a corresponding object in $_R\textbf{Mod}$, then how does $H_4$ act on the chain, say $\cdots\to C_n\to C_{n-1}\to C_{n-2}\to\cdots$?

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  • $\begingroup$ It’s ignoring almost all other terms. $\endgroup$ – Randall Mar 23 at 15:53
  • $\begingroup$ @Randall I can't quite get. Can you explain more? $\endgroup$ – Eric Mar 23 at 16:00
  • $\begingroup$ if I have time later, yes. Right now my kid is screaming. $\endgroup$ – Randall Mar 23 at 16:19
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The $n$-th homology of a chain complex $(C_\bullet,d_\bullet)$ is defined as $$H_n(C_\bullet) = \ker d_n/\operatorname{im}d_{n+1}.$$

If $f\colon (C_\bullet,d_\bullet)\to (C'_\bullet,d'_\bullet)$ is a map of chain complexes (i.e. a collection of maps $\{f_n\colon C_n\to C'_n\}_n$ such that $f_{n-1}\circ d_n = d'_n\circ f_n$, for all $n$), then $H_nf\colon H_n(C_\bullet)\to H_n(C'_\bullet)$ is the map $$z_n+\operatorname{im} d_{n+1}\mapsto f_n(z_n)+\operatorname{im} d'_{n+1}.$$

Once you show that $H_nf$ is well defined and that $H_n$ respects compositions and identities, for each $n$ you have a well defined functor $H_n$ from the category of chain complexes to the category of modules.

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  • $\begingroup$ Is the way the author stated the theorem problematic? I think the word "for each $n\in\Bbb Z$" is somehow weird. $\endgroup$ – Eric Mar 23 at 16:07
  • $\begingroup$ @Eric, there is no problem, for every $n\in \mathbb Z$ there is a corresponding functor $H_n$. There are infinitely many of these functors. $\endgroup$ – Ennar Mar 23 at 16:08
  • $\begingroup$ Oh I see! I made some mistake. Thanks. $\endgroup$ – Eric Mar 23 at 16:18

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