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What happens if one defines homology groups $h_n(X,G)$ of the chain complex

$\cdots \rightarrow Hom(G,C_n(X)) \rightarrow Hom(G,C_{n-1}(X))\rightarrow \cdots $

?

More specifically, what are the groups $h_n(X,G)$ when $G=\Bbb Z, \Bbb Z_m, \Bbb Q$?

Now putting $G=\Bbb Z$

I will get the chain

$\cdots \rightarrow \Bbb Z^k \rightarrow \Bbb Z^r \rightarrow \cdots $.

Now what will be maps by which I can compute the homology groups? Again is there in general formulation for any $G$?

Ri-Li
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1 Answers1

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You do not get a chain $\dots \to \Bbb Z \to \Bbb Z \to \dots$ unless $X$ is a point. The chain groups $C_i$ are really, really big.

The $C_i$ are free $\Bbb Z$-modules. For $G = \Bbb Z_m, \Bbb Q$, $\text{Hom}(G, C_i) = 0$ for all $i$ (why?). So $h_n(X;G)$ is rather uninteresting here.

For $G = \Bbb Z$, nothing changes. You get back the same sequence $$\dots C_{n+1} \xrightarrow{\partial} C_n \xrightarrow{\partial} C_{n-1} \to \dots$$ so this is just singular homology with $\Bbb Z$ coefficients.

Indeed, generally, $h_n(X;G)$ is singular homology with $\text{Hom}(G,\Bbb Z)$ coefficients.

  • For $G=\Bbb Z$, why nothing changes? – Ri-Li Mar 21 '15 at 23:05
  • @user152715 There is a canonical isomorphism $\text{Hom}(\Bbb Z, A) \cong A$ for any abelian group $A$ (exercise). Now what are the induced maps $\text{Hom}(\Bbb Z, C_i) \to \text{Hom}(\Bbb Z, C_{i-1})$? –  Mar 21 '15 at 23:08
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    And for $G=\Bbb Q$ it will be $\Bbb Q^r$ then what will happen? – Ri-Li Mar 21 '15 at 23:10
  • Yeah the isomorphism I know but does this do not have any effect on the chain maps? – Ri-Li Mar 21 '15 at 23:12
  • @MikeMiller Hi, I've been looking for a way to prove a result for general abelian groups, how do you know that $h_n(X;G)$ is singular homology with $\mathrm{Hom}(G,\mathbb{Z})$ coefficients? – Javi Jan 31 '19 at 09:15