4

For $X$ a path-connected abelian CW complex with finitely many cells in each dimension, and $X_{(p)}$ be the localization of $X$ at prime $p$ (so that $\tilde{H}_*(X_{(p)}) = \tilde{H}_*(X)\otimes \mathbb{Z}_{(p)}$ where $\mathbb{Z}_{(p)}$ means the localization of $\mathbb{Z}$ with $\mathbb{Z}-(p)$), I am trying to prove that $H^*(X_{(p)}; \mathbb{Z}_{(p)}) = H^*(X; \mathbb{Z}) \otimes \mathbb{Z}_{(p)}$ as graded rings.

My idea is to prove: (1): $H^*(X_{(p)}; \mathbb{Z}_{(p)}) = H^*(X; \mathbb{Z}_{(p)})$; (2): $H^*(X; \mathbb{Z}_{(p)}) = H^*(X; \mathbb{Z}) \otimes \mathbb{Z}_{(p)}$. For step (1) I ran into trouble trying to prove the isomorphisms using universal coefficient theorem, where $Ext_{\mathbb{Z}}(\mathbb{Z}_{(p)}; \mathbb{Z}_{(p)})$ needs to be computed. I just want to ask if this idea works and any hints for the details(of either steps) would help.

I am not completely sure if my idea works out, and a more straightforward idea for the proof would certainly help.

Kevin
  • 53

1 Answers1

1

I recommend changing the ring you are using. The universal coefficient theorem works for any PID. A nontrivial localization of a PID is a PID, hence, $\mathbb{Z}_{(p)}$ is a PID. Hence, we have a map of exact sequences from

$0 \rightarrow \operatorname{Ext}(H_{n-1}(X),\mathbb{Z}) \rightarrow H^n(X) \rightarrow \operatorname{Hom}(H_n(X),\mathbb{Z}))\rightarrow 0$

to

$0 \rightarrow \operatorname{Ext}_{\mathbb{Z}_{(p)}}(H_{n-1}(X_{(p)})_,\mathbb{Z}_{(p)}) \rightarrow H^n(X_{(p)},\mathbb{Z}_{(p)}) \rightarrow \operatorname{Hom}_{\mathbb{Z}_{(p)}}(H_n(X_{(p)}),\mathbb{Z}_{(p)}) \rightarrow 0$

induced by the inclusion of $\mathbb{Z}$ into $\mathbb{Z}_{(p)}$ and the inclusion $X \rightarrow X_{(p)}$.

The right most map is a localization, see here. Since localization is exact and localization commutes with Hom in this case, $\operatorname{Ext}_{\mathbb{Z}_{(p)}}$ can be computed by localizing, hence leftmost map is a localization.

If we have a map of short exact sequences that is a localization of the two outermost terms, then by the 5 lemma it is a localization of the middle term. Hence, the map $H^*(X) \rightarrow H^*(X_{(p)})$ is a localization thinking of these as modules over $\mathbb{Z}$. This is also a ring map, hence it is a localization of the ring $H^*(X)$ at $p$ as well.

Since in this case localization is given by tensoring with $\mathbb{Z}_{(p)}$, we are done.

Connor Malin
  • 11,661
  • In the upper exact sequence shouldn't it be $H_n(X; \mathbb{Z})$ to make the rightmost map be a localization? – Kevin Jun 21 '21 at 14:47
  • @Kevin Good catch, otherwise it wouldn't be finitely generated. I've fixed. – Connor Malin Jun 21 '21 at 14:53
  • Thanks! Also in the lower exact sequence should it be $H^*(X_{(p)}; \mathbb{Z}_{(p)})$? Also could you explain a bit more on how to get the lower exact sequence? I couldn't find a reference for this version of universal coefficient theorem. – Kevin Jun 21 '21 at 15:02
  • @Kevin So the usual universal coefficient theorem comes from examining the map $H^n(X; A) \rightarrow \operatorname{Hom}(H_n(X),A)$ for some abelian group $A$. However, if $A$ is a module over the ring $R$, then $H_n (X;R)$ has an R-module structure and it makes sense to make the comparison $H^n(X;A) \rightarrow \operatorname{Hom}_R(H_n(X;R),A)$. Now if $R$ is a PID the same proof of the UCT holds but with Hom and Ext now taken over $R$. – Connor Malin Jun 21 '21 at 15:14
  • So the second SES is coming from taking the coefficient ring$\mathbb{Z}{(p)}$, but since we are local, $\mathbb{Z}$ coefficients give the same answer as $\mathbb{Z}{(p)}$ coefficients. – Connor Malin Jun 21 '21 at 15:15
  • Ah I see, thank you very much! – Kevin Jun 21 '21 at 15:51
  • @ConnorMalin Would you mind elaborating on how you obtain the map of short exact sequences induced by the inclusions $\mathbb{Z} \rightarrow \mathbb{Z}{(p)}$ and $X \rightarrow X{(p)}$? I assume you use naturality of $0 \rightarrow \text{Ext}R^1(H{n-1}(X;R),M) \rightarrow H^n(X;M) \rightarrow \text{Hom}_R(H_n(X;R),M) \rightarrow 0$ in $X$ and $M$. Either way I use naturality, I cannot seem to find a map as the one you wrote. – deeppurp Jun 22 '21 at 14:51
  • @deeppurp Unless I am very mistaken it is really just applying those two natural maps. – Connor Malin Jun 22 '21 at 17:00
  • @ConnorMalin It might be, I am absolutely no expert in naturality of short exact sequences. My problem is that singular cohomology as functor is contravariant in $X$, but covariant in $M$. Hence the two inclusions induce maps $H^n(X, \mathbb{Z}) \rightarrow H^n(X, \mathbb{Z}{(p)}) \leftarrow H^n(X{(p)}, \mathbb{Z}{(p)})$, which is why I am unable to see any natural map $H^n(X,\mathbb{Z}) \rightarrow H^n(X{(p)}, \mathbb{Z}_{(p)})$. – deeppurp Jun 25 '21 at 13:13
  • 1
    @deeppurp That is a very good point. The key is that one of these maps is an isomorphism, so we can turn it around. Consider the (contravariant) map on SES with coefficients in $\mathbb{Z}_{(p)}$ and use the previous arguments to deduce all the maps are p local equivalences. Since both the domain and the codomain are p local (by our choice of coefficients), all these maps are isomorphisms. – Connor Malin Jun 25 '21 at 15:26