My old book gives the following proof of the Universal Coefficient Theorem for Homology:


Then, in the case of the Universal Coefficient Theorem for Cohomology, iy just says:

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with no proof. I'm trying to adapt the proof of the version from Homology to the version of Cohomology. It seems that since the result is analogous, the proof might be analogous. For example, I should start picking an exact sequence just like in the first proof. Should I take them in the reverse order, like this:

$$0 \to Z_{n-1}(C)\to C_{n-1}\to Z_n(C)\to H_n(C)\to 0\ ?$$

Or, in order to prove it, I should use the theorem for Homology and arrive at the one for Cohomology?


I've told to consider:

$$\cdots \to Hom(C_{n-1},G)\to^{\delta^n} \delta^n Hom(C_n,G)\to^{\delta^{n+1}}\to Hom(C_{n+1},G)\to \cdots$$

where $\delta^n = Hom(\partial_n, i)$ is the Hom of $\partial_n:C_n\to C_{n-1}$ and the identity homomorphism $i:G\to G$.

Then, the $n-dimensional cohomology module

$$H^n[Hom(C,G)] = Z^n[Hom(C,G)]/B^n[Hom(C,G)]$$

of $Hom(C,G)$ is called the $n$-dimensional cohomology oh $C$ over the coefficient module $G$ and will be denoted by $H^n(C;G)$. Our objective is to determine $H^n(C;G)$ in terms of $H_n(C)$ and $H_{n-1}(C)$. For this purpose, let us define a homomorphism

$$h:H^n(C;G)\to Hom[H_n(C),G)]$$

for every integer $n$ as follows: Let

$$p:Z^n[Hom(C,G)]\to H^n(C;G)$$

denote the natural projection. To define $h$, let $x$ denote an arbitrary element of $H^n(C;G)$. Selec an element

$$z\in Z^n[Hom(C,G)]\subset Hom(C_n,G)$$

satisfying $p(z) = x$. Since $z\in Hom(C_n,G)$, $z$ is by definition a homomorphism:

$$z:C_n\to G$$

Since $\delta^{n+1}(z) = z\circ \partial_{n+1} = 0$, $z$ sends the submodule $B_n(C)$ into the element $0$ of $G$. Hence $z$ induces a homomorphism

$$z_*:H_n(C)\to G$$

Thus $z_*$ is an element of $Hom[H_n(C),G]$. Since the choice is completely determined by $x$, we may define:

$$h:H^n(C;G)\to Hom[H_n(C),G]$$

by taking $h(x) = z$. It is straightfoward to verify that $h$ is an homomorphism.

Dualizing the proof for the homology version, we have the theorem for cohomology

But this cohomology thing is so different. Do I need to use the proof for the homology or just use the theorem for homology and prove the theorem for cohomology? How should I prove everything?

  • 1
    $\begingroup$ So, what happens if you do start with that sequence? $\endgroup$ Commented Jun 18, 2017 at 21:56

1 Answer 1


Let $Z_n$ be the kernel of $\partial_n$ and $B_{n-1}$ the image of $\partial_n$. Then we have a short exact sequence

$$ 0 \to Z_n \to C_n \stackrel{\partial_n}{\to} B_{n-1} \to 0 $$

Now $B_{n-1}$ is an $R$-submodule of $C_{n-1}$ which is a free $R$-module. Since $R$ is a principal ideal domain, $B_{n-1}$ is also free. Therefore the sequence splits and so taking $\text{Hom}_R(-,G)$ we obtain a short exact sequence

$$ 0 \to \text{Hom}_R(B_{n-1},G) \stackrel{\delta_{n-1}}{\to} \text{Hom}_R(C_n,G) \to \text{Hom}_R(Z_n,G) \to 0 $$

I am calling $\delta_{n-1}$ the dual of $\partial_n$, that is what you call $\text{Hom}(\partial_n,i)$.

Let that rest for now and notice that we have a short exact sequence of complexes (each column is a complex)

$$ \require{AMScd} \begin{CD} 0 @>>> Z_n @>>> C_n @>{\partial_n}>> B_{n-1} @>>> 0 \\ @. @V0VV @V{\partial_n}VV @V0VV \\ 0 @>>> Z_{n-1} @>>> C_{n-1} @>{\partial_{n-1}}>> B_{n-2} @>>> 0 \end{CD} $$

When you take $\text{Hom}(-,G)$, you get again a short exact sequence of co-complexes (this is just complexes with the index going upwards)

$$ \begin{CD} 0 @>>> \text{Hom}_R(B_{n-2},G) @>{\delta_{n-2}}>> \text{Hom}_R(C_{n-1},G) @>>> \text{Hom}_R(Z_{n-1},G) @>>> 0 \\ @. @V0VV @V{\delta_{n-1}}VV @V0VV \\ 0 @>>> \text{Hom}_R(B_{n-1},G) @>{\delta_{n-1}}>> \text{Hom}_R(C_n,G) @>>> \text{Hom}_R(Z_n,G) @>>> 0 \end{CD} $$

A short exact sequence of co-complexes induces a long exact sequence in cohomology. The cohomology of the middle column is just the cohomology of the complex $C$. The cohomology of each of the other two columns is just the co-complex itself, since the differentials are zero. Therefore we have a long exact sequence

$$ \text{Hom}_R(Z_{n-1},G) \to \text{Hom}_R(B_{n-1},G) \to H^n(C;G) \to \text{Hom}_R(Z_n,G) \to \text{Hom}_R(B_n,G) $$

and so there will be a short exact sequence

$$ 0 \to \text{(Cokernel of first map)} \to H^n(C;G) \to \text{(Kernel of last map)} \to 0 $$

Let us find out what is the map $\text{Hom}_R(Z_{n-1},G) \to \text{Hom}_R(B_{n-1},G)$. This map is the connecting homomorphism $\delta$. Recall that this map was defined as follows. Take an element $f \in \text{Hom}_R(Z_{n-1},G)$. It must come from an element $g$ of $\text{Hom}_R(C_{n-1},G)$. Compute $\partial_{n-1} (g)$ and this must be in the image of the other $\partial_{n-1}$, the horizontal one. That is, $\partial_{n-1}(g) = \partial_{n-1}(h)$ for some $h \in \text{Hom}(B_{n-1},G)$. And then $\delta(f)=h$. Recall also that choices did not matter.

Now that $f$ comes from $g$ means that $f$ is the restriction of $g$ to $Z_{n-1}$. And so we can take $h$ to be the restriction of $f$ to $B_{n-1}$. So the map $\delta$ sends a map to its restriction to $B_{n-1}$.

Now that we have identified these maps, consider the short exact sequence

$$ 0 \to Z_n \to B_n \to H_n(C) \to 0 $$

and apply $\text{Hom}_R(-,G)$ to it. We get a long exact sequence

$$ 0 \to \text{Hom}_R(H_n(C),G) \to \text{Hom}_R(B_n,G) \to \text{Hom}_R(Z_n,G) \to \text{Ext}_R^1(H_n(C),G) \to 0 $$

The last term should really be $\text{Ext}_R^1(B_n,G)$, but since $B_n$ is free, it vanishes. The middle map is given by restriction and so this exact sequence tells us the kernel and the cokernel of the restriction maps, they are the first and the last term, respectively (without counting the zeros, of course).

Now putting those back in our short exact sequence (for $n-1$ in the first and $n$ in the second)

$$ 0 \to \text{(Cokernel of restriction)} \to H^n(C;G) \to \text{(Kernel of restriction)} \to 0 $$

we obtain

$$ 0 \to \text{Ext}_R^1(H_{n-1}(C),G) \to H^n(C;G) \to \text{Hom}_R(H_n(C),G) \to 0 $$

The second part, about this sequence being split, comes from the beginning. Our original short exact sequence was split. If you follow the splitting along all this process, it will give you a splitting of this last short exact sequence. Now that you have seen how to do the main thing, doing the second part should not be a problem.


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