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I want to check that my understanding is correct about cohomology.

Let $X$ be a topological space $G$ be an abelian group. The universal coefficients theorem, as stated in hatcher, says that the following sequence of $\mathbb{Z}$-modules is split exact $$0\to \text{Ext}(H_{n-1}(X),G)\to H^n(X,G)\to \hom(H_n(X),G)\to 0,$$ where $H_i(X)$ is the singular homology group with $\mathbb{Z}$-coefficients.

Later in a comment, Hatcher mentioned that we take the chain complex of free $R$-modules $$\cdots\to C_{n-1}(X,R)\to C_n(X,R)\to C_{n+1}(X,R)\to \cdots$$ and take its homology modules $H_n(X,R)$, as well as the homology modules after applying $\hom_R(-,R)$, denoted $H^n(X,R)$. Then the universal coefficients theorem says that the following is a split short exact sequence of $R$-modules $$0\to \text{Ext}_R(H_{n-1}(X,R),R)\to H^n(X,R)\to \hom_R(H_n(X,R),R)\to 0.$$


However I feel that Hatcher did not give the more general case, so the following is my speculation:

Since $\mathbb{Z}$ is initial in the category of unital rings, let $R$ be any unital ring, and $\varphi:\mathbb{Z}\to R$ be the unique homomorphism, making $R$ a $\mathbb{Z}-R$ left bimodule. Consider the chain complex with $\mathbb{Z}$ coefficients $$\cdots\to C_{n-1}(X)\to C_n(X)\to C_{n+1}(X)\to \cdots$$ and apply $\hom_\mathbb{Z}(-,R)$ to obtain a complex of $R$-modules $$\cdots\to \hom_\mathbb{Z}(C_{n+1}(X),R)\to \hom_\mathbb{Z}(C_{n}(X),R)\to \hom_\mathbb{Z}(C_{n-1}(X),R)\to \cdots,$$ whose homology yields $h^n(X,R)$.

My two questions are:

  1. $h^n(X,R) \cong H^n(X,R)$? Which one is the 'correct' definition of cohomology?
  2. Is there a split exact sequence of $R$-modules $$0\to \text{Ext}_\mathbb{Z}(H_{n-1}(X),R)\to h^n(X,R)\to \hom_\mathbb{Z}(H_n(X),R)\to 0?$$
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2 Answers 2

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EDIT: The version of (2) using homology with coefficients in R is not true. Consider $X = \Bbb{RP}^2$, and $R = \Bbb{Z}/4$.

The homology groups are $$ \Bbb Z, \Bbb Z/2, 0, \ldots $$ The universal coefficient theorem tells you that the homology and cohomology with mod-4 coefficients is $$ \Bbb Z/4, \Bbb Z/2, \Bbb Z/2, 0, \ldots $$ but your formula predicts that the cohomology with mod-4 coefficients is $$ \Bbb Z/4, \Bbb Z/2, \Bbb Z/2^2, \Bbb Z/2, 0, \ldots $$

However, (2) is true as you've stated it. Sorry; I got mixed up as to the two displayed equations.

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  1. Remember that $C_n(X;R) \cong C_n(X) \otimes_\mathbb{Z} R$, and $\hom_R(M \otimes_\mathbb{Z} R, N) \cong \hom_\mathbb{Z}(M,N)$.

  2. Yes, apply your first exact sequence with $R$ instead of $G$.

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    $\begingroup$ For 2 you need to observe,too, that since the universal coeffecient exact sequence is natural in the coefficients the maps given by multiplication by elements of $R$ commutes with the maps in the exact sequence; also, the splitting of the exact sequece is natural in the coefficients (and not on the space) so the splitting is also $R$-linear. $\endgroup$ Nov 9, 2014 at 19:17
  • $\begingroup$ Indeed, it could well be the case that the maps in the UCT sequence for a ring $R$ be not $R$-linear, or that the splitting is only $\mathbb Z$-linear. One has to check that this is not so. $\endgroup$ Nov 9, 2014 at 19:19
  • $\begingroup$ @MarianoSuárez-Alvarez Good point, I should have mentioned it; I didn't notice that the OP didn't state that the first exact sequence was just said to be of $\mathbb{Z}$-modules. $\endgroup$ Nov 9, 2014 at 19:43
  • $\begingroup$ @MarianoSuárez-Alvarez It appears to me that the counterexample that Tyler provided is correct, could you please help me confirm that? $\endgroup$
    – mez
    Nov 9, 2014 at 23:07
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    $\begingroup$ Well, the exact sequence (with homology with coefficients in R) exists if R is a principal ideal domain. That is the standard hypothesis in this context. (For a general ring there is a universal coefficients spectral sequence, not short exact sequences) $\endgroup$ Nov 10, 2014 at 1:35

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