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Let $X$ be a topological space and $F$ be a field. I want to understand why the homology $H_n(X,F)$ is an $F$-vector space : that is how do we define the multiplication of a class in $H_n(X,F)$ by a scalar $\alpha\in F$?

My guess: if we take a class $c\in H_n(X,F)$ as a formal sum $\sum{a_ic_i}$ where $a_i\in F$ and $c_i$ are $n$-chains. then $\alpha.c=\sum{(\alpha\,a_i)c_i}$.

does this generalize if we take the coefficients to be a ring $R$ then the homology group $H_n(X,R)$ is an $R$-module?

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You are right, and it generalizes as you say. A class in $H_n(X;R)$ is represented by a formal sum of $n$-simplices in $X$ with coefficients in $R$, and scalar multiplication of coefficients gives the $R$-module structure. (It's easy to check that this is well-defined on homology.)

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is it true that ring coefficient is the most general case of coefficients? and why are group coefficients are more used than ring coefficients? –  palio Sep 2 '12 at 19:02
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Well, ring coefficients can be multiplied, which is necessary to define cup products in cohomology. I'm not sure what you mean when you say that group coefficients are more common; it all depends what you're doing. If you only need the group structure, then it doesn't matter. –  Aaron Mazel-Gee Sep 2 '12 at 19:10
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As for saying that ring coefficients are the "most general", in fact there's a vast generalization to "extraordinary (co)homology theories", among which singular (co)homology theories are characterized by the fact that their output when applied to the one-point space $\{*\}$ is concentrated in dimension 0. If such a cohomology theory has a notion of a cup product, then its value on $\{*\}$ will be a graded ring. So... yes and no. –  Aaron Mazel-Gee Sep 2 '12 at 19:11

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