# Understanding homology as a functor from $_R\mathbf{Comp}$ to $_R\textbf{Mod}$

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 $$\cdots\to C_n\to C_{n-1}\to C_{n-2}\to\cdots$$?

• It’s ignoring almost all other terms. – Randall Mar 23 at 15:53
• @Randall I can't quite get. Can you explain more? – Eric Mar 23 at 16:00
• if I have time later, yes. Right now my kid is screaming. – Randall Mar 23 at 16:19

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.
• Is the way the author stated the theorem problematic? I think the word "for each $n\in\Bbb Z$" is somehow weird. – Eric Mar 23 at 16:07
• @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. – Ennar Mar 23 at 16:08