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I have been studying a course in algebraic topology that follows Hatcher's textbook on the subject. I have some queries as to how certain things are defined.

The first part of the text defines the cochain groups $S_n (X)$ to be the free abelian groups generated by singular n simplices on $X$. Then the cochain groups are $S^n (X;G):=Hom(S_n (X),G)$.

Then in chapter 3.2 we consider cohomology with coefficients in a commutative ring $R$. So now we appear to have $S^n (X;R):=Hom(S_n (X),R)$.

In order that this is well defined I think we must treat $R$ as a group when considering these homomorphisms. So we would have to view $R$ as a group under it's addition or multiplication and since we don't necessarily have $1 \in R$ we must choose the addition for this. None of this is discussed in Hatcher and is glossed over in every text I can find.

Things then get more complicated when discussing the cap product. At this stage it is shown in Hatcher (page 249 "connection with cup product") that for $\psi \in S^l(X;R), \phi \in S^k(X;R)$ and $\alpha \in S_{k+l}(X;R)$: \begin{equation} \psi (\alpha \cap \phi) = (\phi \cup \psi) (\alpha) \end{equation} But the proof of this requires that $\psi (r \sigma) = r \psi(\sigma)$ i.e. $\psi$ is an $R$-homomorphism. This suggests that the definitions we are working with are in fact: $S_n (X) = S_n (X;R)$ (coefficients in $R$) and $S^n (X;R):=Hom_R(S_n (X;R),R)$.

So we come to my questions:

  1. How are the chain groups really meant to be defined when we have coefficients in a ring $R$?
  2. If we do have several definitions floating about how do we know which one is being used.
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But what is the problem? ${\rm Hom}_R$ simply means $R$-homomorhisms, and $R$ is certainly a module over itself. –  M.B. Mar 6 '14 at 21:24
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Since $S_n(X;R)\cong S_n(X)\otimes R$, one gets $Hom_R(S_n(X;R),R)\cong Hom(S_n(X),R)$ canonically. –  archipelago Mar 6 '14 at 23:33
    
Thank you for your replies. My problem was that this definition was never given in the text and so it was not clear that the co-chains were in fact R-homomorphisms. I see now that the isomorphism given by @archipelago means that the two definitions I was thinking of are in fact the same. However I still think that this should have been discussed in the text. –  Edward Hart Mar 7 '14 at 8:59

1 Answer 1

  1. Take the underlying additive group of $R$. The multiplicative structure only enters when discussing the cup product.

  2. The two definitions you've given are equivalent.

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