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I apologize in advance if this question is inappropriate for this forum.

I am reading Hatcher's book and have the following problem: Let $X$ be a space, $G$ an abelian group and $R$ a ring. In his proof of the universal coefficient theorem in $\S$ 3.1, he considers a natural homomorphism $h\colon H^k(X;G)\to \text{Hom}(H_k(X),G)$. However, when computing cup product structures in $\S$ 3.2, specifically in examples 3.8 and 3.9, he seems to be using a homomorphism $\tilde{h}\colon H^k(X;R)\to \text{Hom}(H_k(X;R),R)$.

I think I can see that such a homomorphism always exists when $R$ is a ring. More precisely, given a cocyle $\phi\colon C_k(X)\to R$, we can define $\tilde{\phi}\colon Z_k(X)\otimes R\to R$ by $\tilde{\phi}(\alpha \otimes r)=\phi(\alpha)r$, and the fact that $\delta\phi=\phi\partial=0$ means that this gives rise to a homomorphism $H_k(X;R)\to R$. In other words, the construction yields a map $\tilde{h}\colon H^k(X;R)\to \text{Hom}(H_k(X;R),R)$.

What I would like to know is:

(a) Is there a way to define something similar to $\tilde{h}$ with $G$ in place of $R$?

(b) Is this map $\tilde{h}$ surjective as $h$ is?

(c) What is its kernel?

Thank you.

share|cite|improve this question
A ring $R$, in Hatcher's context, already has the structure of an abelian group with respect to addition, so everything in 3.1 applies. It is true that a ring has additional structure besides addition --- namely multiplication --- but your questions (a,b,c) are answered by ignoring multiplication. The whole point of cup product is to exploit the additional multiplicative structure on $R$. – Lee Mosher Apr 12 '12 at 16:54
@Lee: Note that homology on the right hand side of $h$ has coefficients in $\mathbb Z$ while $\tilde{h}$ has coefficients in $R$. Hence it's a bit more subtil than just arguing that a ring is a group with some additional properties. – Ralph Apr 12 '12 at 17:30
@Ralph: ah, right. Got it. – Lee Mosher Apr 12 '12 at 17:44
Here's some pure speculation... given $G$ there should always be a group homomorphism from $\mathbb{Z}\rightarrow G$, right? This induces a group homomorphism $Hom(H_k(X),\mathbb{Z})\righarrow Hom(H_k(X),G)$. Maybe this can help you use the $\tilde{h}$ from $R=\mathbb{Z}$ to construct the map you're after. Of course, it can only be a group homomorphism, since $Hom(H_k(X),G)$ is only a group, so there is no guarantee it'll have the nice properties that $\tilde{h}$ has. Now, here's a question: what do you need $\tilde{h}_G$ for? Why isn't $h$ good enough? – David White Apr 12 '12 at 19:16
It seems to me that you get a homomorphism $H^k(X;A)\to\text{Hom}(H_k(X;B),C)$ when (and probably only when) you have a pairing $A\otimes B\to C$. The $h$ in the question comes from the fact that an abelian group $G$ is a $\mathbb Z$-module, i.e., a pairing $G\otimes\mathbb Z\to G$. The $\tilde h$ comes from the ring multiplication $R\otimes R\to R$. – Andreas Blass Apr 12 '12 at 21:01

Claim 1: If $R$ is a principal ideal domain, then there is a short exact sequence $$0 \to Ext^1_R(H_{n-1}(X;R),R) \to H^n(X;R) \xrightarrow[]{\tilde{h}} Hom_R(H_n(X;R),R) \to 0\hspace{10pt}(\ast)$$

This answers b) and c).

Claim 2: If $R$ or $H_\ast(X)$ is torsion-free, then $\tilde{h} = h$ (for $G=R$).

Proof: Denote the simplicial complex of $X$ by $S(X)$. Then $C := S(X) \otimes_{\mathbb Z} R$ is a complex of free $R$-modules. Hence, by the universal coefficient theorem (cf. Hatcher, after Corollary 3.4) we have the short exact sequence $$0 \to Ext^1_R(H_{n-1}(C),R) \to H^n Hom_R(C,R) \xrightarrow[]{\tilde{h}} Hom_R(H_n(C),R)\to 0.$$ But $Hom_R(C,R) = Hom_R(S(X) \otimes_{\mathbb Z} R,R) \cong Hom(S(X),R)$ (cf. III (3.3) in Brown: Cohomology of Groups) and $H_n(C) = H_n(X;R)$. Hence the short exact sequence above transforms into $(\ast)$.

Moreover, by universal coefficients, $$H_n(X;R) = H_n(X) \otimes_{\mathbb Z}R \oplus Tor_1^{\mathbb Z}(H_{n-1}(X),R) = H_n(X) \otimes_{\mathbb Z}R$$ if $R$ or $H_{n-1}(X)$ is torsion-free. Hence $$Hom_R(H_n(X;R),R) = Hom_R(H_n(X) \otimes_{\mathbb Z} R,R) \cong Hom(H_n(X),R).$$

This shows claim 2.

Remark: a) Example 3.8 from Hatcher uses $R=\mathbb{Z}/2$.

b) Example 3.9 uses $R=\mathbb{Z}/m$. Hence $(\ast)$ doesn't apply (for general $m$). In particular it can't be used to show that $\tilde{h}$ is surjective. This may be the reason for Hatcher's alternative argumentation in this example.

Added: I want to add an application of the short exact sequence $(\ast)$ that is of particular importance: Let $R=k$ be a field. Thus $H_\ast(X;k)$ is a projective $k$-module and hence $Ext_k^1(H_{n-1}(X;k),k)=0$ for all $n$. Now $(\ast)$ yields an isomorphism $$H^n(X;k) \xrightarrow[]{\sim}Hom_k(H_n(X;k),k),$$ i.e. $H^n(X;k)$ is just the dual space of $H_n(X;k)$. This property is frequently used in the literature.

share|cite|improve this answer
I thought he wanted to take the construction of $\tilde{h}$, which he understands, and try to make it work to get a map like $h$, i.e. a map $H^k(X;G)\rightarrow Hom(H_k(X),G)$. That's how I read the ``with $G$ in place of $R$'' part of his question. If I'm right, then this answer is probably not what he's looking for, though I still think it's cool. – David White Apr 13 '12 at 0:39
@David: The OP defines a map $\tilde{h}$ and asks in b), c) about the image and kernel of $\tilde{h}$. And that are the questions I answer. Where's the problem ? – Ralph Apr 13 '12 at 0:57
$\mathbb Z/m$ is hereditary iff $m$ is square free (and then it is semisimple) – Mariano Suárez-Alvarez Apr 13 '12 at 1:05
@Mariano: You beat me by a few seconds. – Ralph Apr 13 '12 at 1:11
:) Also, a semi-hereditary ring has $Tor$ dimension at most $1$, but over $\mathbb Z/P^r$ with $r>1$ we have $\mathrm{Tor}_n(\mathbb Z/p,\mathbb Z/p)\neq0$ for all $n$. – Mariano Suárez-Alvarez Apr 13 '12 at 1:22

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