Universal Coefficient Theorem

In Projective Resolution of Modules, we’ve introduced enough homological algebras. Now we can state and prove the Universal Coefficient Theorem, which relates homology and cohomology via the $\mathrm{Ext}$ and $\mathrm{Hom}$ functors.

Universal Coefficient Theorem (for Cohomology)

Let $C$ be a chain complex of free abelian groups and $G$ be an abelian group. Then for each $n$, there is a short exact sequence: $0\to \mathrm{Ext}(H_{n-1}(C),G)\to H^n(C;G)\to \mathrm{Hom}(H_n(C),G)\to 0$ This sequence splits, but the splitting is not natural.

Proof Sketch

1. For the chain complex $C=\{C_n,d_n\}$, we have the cycles $Z_n=\ker d_n$ and boundaries $B_n=\text{im }{d}_{n+1}$. By definition, the homology is $H_n=Z_n/B_n$.
2. These groups fit into two short exact sequences for each $n$:
   • $0\to B_n\to Z_n\to H_n\to 0$ (from the definition of homology)
   • $0\to Z_n\to C_n\to B_{n-1}\to 0$ (from the definition of cycles and boundaries)
3. Since $C_n$ is a free abelian group, its subgroup $B_{n-1}$ must also be free. Because $B_{n-1}$ is a free (and thus projective) module, the second sequence $0\to Z_n\to C_n\to B_{n-1}\to 0$ splits.
4. Applying the functor $\text{Hom}(-,G)$ is contravariant and left-exact. Applying it to the split sequence from step 3 gives a split short exact sequence of cochain complexes: $0⟵\text{Hom}(Z_n,G)⟵\text{Hom}(C_n,G)⟵\text{Hom}(B_{n-1},G)⟵0$
5. A short exact sequence of cochain complexes induces a long exact sequence in cohomology. This gives: $\cdots \to H^{n-1}(\text{Hom}(B,G))\to H^n(\text{Hom}(Z,G))\to H^n(C;G)\to H^n(\text{Hom}(B,G))\to \dots$
6. By analyzing the differentials and applying the long exact sequence for Ext derived from the sequence $0\to B_n\to Z_n\to H_n\to 0$, one can identify the terms in the long exact sequence from step 5. This process yields the short exact sequence stated in the theorem.
7. To show the sequence splits, one constructs a splitting map $\alpha :\text{Hom}(H_n(C),G)\to H^n(C;G)$. This is done by choosing a splitting $s_n:C_n\to Z_n$ of the sequence in step 3. For any homomorphism $\phi :H_n(C)\to G$, the composition $\phi \circ q\circ s_n:C_n\to G$ (where $q:Z_n\to H_n(C)$ is the quotient map) defines a cocycle whose cohomology class gives the desired splitting.

$\square$

Properties of the Ext Functor

Lemma

The Ext functor has the following properties:

1. $\text{Ext}(M_1\oplus M_2,N)\cong \text{Ext}(M_1,N)\oplus \text{Ext}(M_2,N)$
2. $\text{Ext}(M,N_1\oplus N_2)\cong \text{Ext}(M,N_1)\oplus \text{Ext}(M,N_2)$
3. $\text{Ext}(\mathbb{Z},N)=0$ for any abelian group $N$.
4. $\text{Ext}(\mathbb{Z}/n\mathbb{Z},N)\cong N/nN$.

e.g. We compute the cohomology groups $H^n(\mathbb{R}{\mathrm{P}}^5;\mathbb{Z}/2)$ using the Universal Coefficient Theorem. Since the sequence splits, $H^n({\text{RP}}^5;{\mathbb{Z}}_2)\cong \text{Ext}(H_{n-1}({\text{RP}}^5),{\mathbb{Z}}_2)\oplus \text{Hom}(H_n({\text{RP}}^5),{\mathbb{Z}}_2)$. We compute the two terms for each $n$. The calculation is summarized in the following table:

$n$	$H_n({\text{RP}}^5)$	$\text{Hom}(H_n,\mathbb{Z}/2)$	$\text{Ext}(H_{n-1},\mathbb{Z}/2)$	$H^n({\text{RP}}^5;\mathbb{Z}/2)$
0	$\mathbb{Z}$	$\mathbb{Z}/2$	0	$\mathbb{Z}/2$
1	$\mathbb{Z}/2$	$\mathbb{Z}/2$	$\text{Ext}(\mathbb{Z},\mathbb{Z}/2)=0$	$\mathbb{Z}/2$
2	$0$	$0$	$\text{Ext}(\mathbb{Z}/2,\mathbb{Z}/2)=\mathbb{Z}/2$	$\mathbb{Z}/2$
3	$\mathbb{Z}/2$	$\mathbb{Z}/2$	$\text{Ext}(0,\mathbb{Z}/2)=0$	$\mathbb{Z}/2$
4	$0$	$0$	$\text{Ext}(\mathbb{Z}/2,\mathbb{Z}/2)=\mathbb{Z}/2$	$\mathbb{Z}/2$
5	$\mathbb{Z}$	$\mathbb{Z}/2$	$\text{Ext}(0,\mathbb{Z}/2)=0$	$\mathbb{Z}/2$

Universal Coefficient Theorem for Homology

If $C$ is a chain complex of free abelian group $G$, then there are natural short exact sequences $0\to H_n(C)\otimes G\to H_n(C;G)\to \mathrm{Tor}(H_{n-1}(C),G)\to 0$ for all n and all $G$, and these sequences split, though not naturally.