Universal Coefficient Theorem for Homology

We present a direct proof of the universal coefficient theorem for homology that is simpler and shorter than the standard proof.

Theorem 1 Given a chain complex C in which each C_{n} is free abelian, and a coefficient group G, we have for each n the natural short exact sequence

0 − ; − ; → H_{n}(C) ⊗ G − ; − ; → H_{n}(C ⊗ G) − ; − ; → Tor(H_{n−1}(C), G) − ; − ; → 0,  (2)

which splits (but not naturally).

In particular, this applies immediately to singular homology.

Theorem 3 Given a pair of spaces (X, A) and a coefficient group G, we have for each n the natural short exact sequence

0 − ; − ; → H_{n}(X, A) ⊗ G − ; − ; → H_{n}(X, A; G) − ; − ; → Tor(H_{n−1}(X, A), G) − ; − ; → 0,

which splits (but not naturally).

We shall derive diagram (2) as an instance of the following elementary result.

Lemma 4 Given homomorphisms f: K → L and g: L → M of abelian groups, with a splitting homomorphism s: L → K such that f ^{∘} s = id_{L}, we have the split short exact sequence

0 − ; − ; → Ker f − ; − ; →^{⊂} Ker(g^{∘}f) − ; − ; →^{f′} Ker g − ; − ; → 0,  (5)

where f′ = f| Ker(g^{∘}f), with the splitting s′ = s| Ker g: Ker g → Ker(g^{∘}f).

Proof We note that f′ and s′ are defined, as f(Ker(g^{∘}f)) ⊂ Ker g and s(Ker g) ⊂ Ker(g^{∘}f). (In detail, if l ∈ Ker g, (g^{∘}f)sl = gfsl = gl = 0 shows that sl ∈ Ker(g^{∘}f).) Then f ^{∘} s = id_{L} restricts to f′ ^{∘} s′ = id. Since Ker f ⊂ Ker(g^{∘}f), we have Ker f′ = Ker f ∩ Ker(g ^{∘} f) = Ker f. This completes the proof.

Preliminaries We consider a chain complex C as in Theorem 1. We adopt the usual notation: Z_{n} for the group of n-cycles, B_{n} for the group of n-boundaries, and H_{n} = H_{n}(C) = Z_{n}/B_{n}; also Z_{n}(C ⊗ G) etc. for the chain complex C ⊗ G. The key idea of our proof of Theorem 1 is to express ∂: C_{n} → C_{n−1} as the composite

∂: C_{n} − ; − ; → \frac{C_{n}}{B_{n}} − ; − ; → \frac{C_{n}}{Z_{n}} − ; − ; →^{≅} B_{n−1} − ; − ; →^{⊂} Z_{n−1} − ; − ; →^{⊂} C_{n−1}.  (6)

By definition, we have the short exact sequence

0 − ; − ; → B_{n} − ; − ; →^{⊂} Z_{n} − ; − ; → H_{n} − ; − ; → 0  (7)

for any n. As B_{n} and Z_{n} are subgroups of the free abelian group C_{n} and therefore free abelian, we recognize (7) as a free resolution of H_{n}.

Also by definition, from diagram (6) we have the short exact sequence

0 − ; − ; → Z_{n} − ; − ; →^{⊂} C_{n} − ; − ; →^{∂} B_{n−1} − ; − ; → 0,  (8)

which splits because B_{n−1} is free abelian. Using (7) and (8), we may rewrite the canonical short exact sequence

0 − ; − ; → Z_{n}/B_{n} − ; − ; →^{⊂} C_{n}/B_{n} − ; − ; → C_{n}/Z_{n} − ; − ; → 0

as

0 − ; − ; → H_{n} − ; − ; → C_{n}/B_{n} − ; − ; →^{∂} B_{n−1} − ; − ; → 0,  (9)

where ∂ is a quotient of ∂. As B_{n−1} is free abelian, this short exact sequence also splits, and we may choose a splitting homomorphism s: B_{n−1} → C_{n}/B_{n}. Proof of Theorem 1 We tensor diagram (6) and the above short exact sequences with G, to form the following diagram of exact sequences.

\xymatrix{ &  & 0 \ar[d] \\  &  & H_{n} ⊗ G \ar[d] \\ B_{n} ⊗ G \ar[r] & C_{n} ⊗ G \ar[r] & \frac{C_{n}}{B_{n}} ⊗ G \ar[r] \ar[d]^{∂⊗id} & 0 & 0 \ar[d] \\ 0 \ar[r] & Tor(H_{n−1}, G) \ar[r] & B_{n−1} ⊗ G \ar[u]^{s⊗id} \ar[rr] \ar[drr] \ar[d] &  & Z_{n−1} ⊗ G \ar[d] \\  &  & 0 &  & C_{n−1} ⊗ G}  (10)

Both vertical sequences are exact, since (9) and (8) split. The upper horizontal sequence is exact, because − ⊗ G is right exact. The lower horizontal sequence is exact, as Tor(H_{n−1}, G) = Ker[B_{n−1} ⊗ G → Z_{n−1} ⊗ G] by definition.

We deduce that B_{n−1}(C ⊗ G) = Im(∂ ⊗ id) = Im[B_{n−1} ⊗ G → C_{n−1} ⊗ G]. On replacing n − 1 by n, we see that

B_{n}(C ⊗ G) = Im[B_{n} ⊗ G − ; − ; → C_{n} ⊗ G] = Ker[C_{n} ⊗ G − ; − ; → (C_{n}/B_{n}) ⊗ G].

Then (C_{n}⊗G)/B_{n}(C⊗G) ≅ (C_{n}/B_{n})⊗G. Also, Z_{n}(C⊗G) = Ker[C_{n}⊗G → Z_{n−1}⊗G], and hence (C_{n} ⊗ G)/Z_{n}(C ⊗ G) ≅ Im[C_{n} ⊗ G → Z_{n−1} ⊗ G] ⊂ Z_{n−1} ⊗ G.

We next use the canonical short exact sequence

0 − ; − ; → \frac{Z_{n}(C ⊗ G)}{B_{n}(C ⊗ G)} − ; − ; →^{⊂} \frac{C_{n} ⊗ G}{B_{n}(C ⊗ G)} − ; − ; → \frac{C_{n} ⊗ G}{Z_{n}(C ⊗ G)} − ; − ; → 0

to rewrite our target group as

H_{n}(C ⊗ G) = \frac{Z_{n}(C ⊗ G)}{B_{n}(C ⊗ G)} ≅ Ker [\frac{C_{n}}{B_{n}} ⊗ G − ; − ; → Z_{n−1} ⊗ G] .

We now apply Lemma 4 with K = (C_{n}/B_{n}) ⊗ G, L = B_{n−1} ⊗ G and M = Z_{n−1} ⊗ G, and use the splitting s ⊗ id: L → K. Diagram (10) identifies Ker[K → L] with H_{n} ⊗ G and Ker[L → M] with Tor(H_{n−1}, G). These identifications reduce the split short exact sequence (5) to the desired (2).