This seemingly simple question stands unanswered on math.stackexchange.com for a couple of days (https://math.stackexchange.com/questions/680211/a-question-about-the-universal-coefficient-theorem), so I decided to repost it here, although it is hardly research related.

Let $X$ be some topological space, $R$ be a (unital) PID and $G$ be an $R$-module. I'm trying to see how the fact that the cohomology groups $H^*(X;G)$ may be found as the cohomology of either $\mathrm{Hom}_\mathbb{Z}(C_*(X;\mathbb{Z}),G)$ or $\mathrm{Hom}_R(C_*(X;R),G)$ relates to the universal coeffients theorems.

One has \begin{align} H^n(X;G)&\cong \mathrm{Hom}_\mathbb{Z}(H_n(X;\mathbb{Z}),G)\oplus \mathrm{Ext}^1_\mathbb{Z}(H_{n-1}(X;\mathbb{Z}),G)\\ &\cong\mathrm{Hom}_R(H_n(X;R),G)\oplus \mathrm{Ext}^1_R(H_{n-1}(X;R),G). \end{align} Now let us try to derive the latter isomorphism from the universal coefficient theorem for homology. Replacing the homology groups in the second line with the corresponding expressions of the form $*\otimes_\mathbb{Z} R\oplus \mathrm{Tor}_1^\mathbb{Z}(*,R)$ we obtain quite the isomorphism. I can see that $\mathrm{Hom}_\mathbb{Z}(H_n(X;\mathbb{Z}),G)\cong \mathrm{Hom}_R(H_n(X;\mathbb{Z})\otimes_\mathbb{Z}R,G)$ via tensor-hom adjunction, i.e. the very first summands on both sides coincide. However, the isomorphism still seems dubious to me, since, if I'm not mistaken, any of the four summands on the right may actually be nonzero.

I do understand that the most suspicious part is the exact sequence in the UCT for homology being a split exact sequence of $R$-modules. However, in the stackexchange question I provide an argument seemingly showing that it actually is such.

I would be truly grateful for a detailed explanation of what is what here.

MAJOR UPDATE. Thanks to Neil Strickland's comment the question has boiled down to the following.

As I see it, the UCTs imply the following isomorphism of $R$-modules: $$\mathrm{Hom}_\mathbb{Z}(H_n(X;\mathbb{Z}),G)\oplus \mathrm{Ext}_\mathbb{Z}(H_{n-1}(X;\mathbb{Z}),G)=\\\mathrm{Hom}_R(H_n(X;\mathbb{Z})\otimes_\mathbb{Z} R,G)\oplus \mathrm{Hom}_R(\mathrm{Tor}^\mathbb{Z}(H_{n-1}(X;\mathbb{Z}),R),G)\oplus\\\mathrm{Ext}_R(H_{n-1}(X;\mathbb{Z})\otimes_\mathbb{Z} R,G)\oplus\mathrm{Ext}_R(\mathrm{Tor}^\mathbb{Z}(H_{n-2}(X;\mathbb{Z}),R),G).$$ I would like to see a direct proof.

Let us refer to the isomorphism as $A\oplus B=C\oplus D\oplus E\oplus F$. Tensor-hom adjunction shows that $A=C$. Thus we are to show $B=D\oplus E\oplus F$.

Now, as Neil pointed out, $R$ is either a free $\mathbb{Z}$-module or a $\mathbb{Z}/p\mathbb{Z}$-algebra for some prime $p$. In the first case it isn't hard to see that $D=F=0$ and $B=E$, achieving a proof. I'm now interested in the second case, what is the situation here?

  • $\begingroup$ I don't understand the question. Why do you want to derive the second one from the first one? Just mimic the standard textbook proof over $\mathbb{Z}$: it works over any PID. (I guess the only thing that you need is the fact that a submodule of a free module is free.) $\endgroup$ Feb 19, 2014 at 14:10
  • $\begingroup$ I'm trying to understand the way the two statements interact and the correct way of looking at it all. More specifically, I have some sort of argument leading to dubious results. I would like to pinpoint the mistake. $\endgroup$
    – imakhlin
    Feb 19, 2014 at 14:15
  • 5
    $\begingroup$ As $R$ is a domain, it is not hard to see that it is either torsion-free over $\mathbb{Z}$, or an algebra over $\mathbb{Z}/p$ for some prime $p$. In the first case the homological UCT has only one term, and in the second case everything is straightforwardly determined by mod $p$ (co)homology. $\endgroup$ Feb 19, 2014 at 15:31
  • $\begingroup$ @NeilStrickland. Wow, thanks! This goes to show that if $R$ is torsion-free, then $$\mathrm{Ext}_\mathbb{Z}(H_{n-1}(X;\mathbb{Z}),G)=\mathrm{Ext}_R(H_{n-1}(X; \mathbb{Z})\otimes R,G)$$ and the other two summands vanish. Thus the (2 summands)=(4 summands) isomorphism holds indeed. Am I correct? $\endgroup$
    – imakhlin
    Feb 19, 2014 at 18:37
  • 1
    $\begingroup$ @Igor: $H_{n-1}(X;\mathbb{Z}/p)$ is free over the field $\mathbb{Z}/p$, so the module $H_{n-1}(X;R)=H_{n-1}(X;\mathbb{Z}/p)\otimes R$ is free over $R$, so both of the Ext groups that you mention are zero. $\endgroup$ Feb 19, 2014 at 18:57

1 Answer 1


Well, I seem to have sorted it all out, might as well write it all up.

Statement. For arbitrary abelian groups $A$, $B$ and $C$, a PID $R$ and an $R$-module G one has an isomorphism of $R$-modules $$\mathrm{Hom}_\mathbb{Z}(A,G)\oplus\mathrm{Ext}_\mathbb{Z}(B,G)=\\\mathrm{Hom}_R(A\otimes R,G)\oplus\mathrm{Hom}_R(\mathrm{Tor}(B,R),G)\oplus\mathrm{Ext}_R(B\otimes R,G)\oplus\mathrm{Ext}_R(\mathrm{Tor}(C,G)).$$ (My "dubious" isomorphism is then obtained by setting $A=H_n(X;\mathbb{Z})$, $B=H_{n-1}(X;\mathbb{Z})$, $C=H_{n-2}(X;\mathbb{Z})$.)

Proof. Let our isomorphism be $U\oplus V=W\oplus X\oplus Y\oplus Z$. Tensor-hom adjunction provides $U=W$.

The free resolution $$0\rightarrow\mathbb{Z}^P\rightarrow\mathbb{Z}^Q\rightarrow C\rightarrow 0$$ gives rise to the exact sequence $$0\rightarrow\mathrm{Tor}(C,G)\rightarrow R^P\rightarrow R^Q\rightarrow C\otimes R\rightarrow 0.$$ Hence $\mathrm{Tor}(C,G)$ is free ($R$ is a PID) and $Z=0$. (Which was our only chance, by the way.)

Similarly, we have a free $R$-module resolution $$ 0\rightarrow\mathrm{Tor}(B,G)\rightarrow R^I\rightarrow R^J\rightarrow B\otimes R\rightarrow 0. $$ $R^I/\mathrm{Tor}(B,G)$ is free which allows us to write $R^I=\mathrm{Tor}(B,G)\oplus R^K$. Applying the functor $\mathrm{Hom}_R(-,G)$ to this resolution we obtain the complex $$X\leftarrow X\oplus G^K\overset{f}{\leftarrow} G^J.$$ Its homology is equal to $G^K/\mathrm{im} f$ on one hand and $Y$ (by construction) on the other.

Further, applying $\mathrm{Hom}_\mathbb{Z}(-,G)$ to $$0\rightarrow\mathbb{Z}^I\rightarrow\mathbb{Z}^J\rightarrow B\rightarrow 0$$ gives the complex $$0\leftarrow G^I=X\oplus G^K\overset{f}{\leftarrow}G^J.$$ The former complex is an augmentation of the latter. The latter's homology is $X\oplus (G^K/\mathrm{im} f)=X\oplus Y$, but also $V$ by construction. Thus we arrive at the sought-for $V=X\oplus Y$. Q.E.D.

Now it would be interesting to put the exact sequence $$0\rightarrow\mathrm{Ext}_R(B\otimes R,G)\rightarrow\mathrm{Ext}_\mathbb{Z}(B,G)\rightarrow\mathrm{Hom}_R(\mathrm{Tor}(B,R),G)\rightarrow 0$$ in a broader context. What is relevant spectral sequence here? What are other ways of understanding it?

  • $\begingroup$ By the way, does anyone actually know how to use the label-ref mechanism properly on MO? I'm pretty sure there is a way, although I wasn't able to find a single example of someone successfully doing so. $\endgroup$
    – imakhlin
    Feb 20, 2014 at 16:05

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