If \(T_n(X, A)\) denotes the torsion subgroup of \(H_n(X, A; \mathbb{Z})\), show that the functors \((X, A) \mapsto T_n(X, A)\), with the obvious induced homomorphisms \(T_n(X, A) \to T_n(Y, B)\) and boundary maps \(T_n(X, A) \to T_{n-1}(A)\), do not define a homology theory. Do the same for the ‘mod torsion’ functor \(MT_n(X, A) = H_n(X, A; \mathbb{Z}) / T_n(X, A)\).
Solution
The exactness axiom fails for both constructions.
1. For the pair \((D^2,S^1)\) the integral long exact sequence contains\[ 0=H_2(D^2)\longrightarrow H_2(D^2,S^1)\xrightarrow{\partial}H_1(S^1)\longrightarrow H_1(D^2)=0, \]so \(\partial:\mathbb Z\to\mathbb Z\) is an isomorphism.
This is the long exact sequence of the pair, together with contractibility of \(D^2\).
2. The torsion functors are not exact on this pair.
All four groups in 1 are torsion-free, hence \[ T_2(D^2,S^1)=T_1(S^1)=0. \] This example alone does not detect failure, so instead use the Moore-space pair \((X,A)\) obtained from \(A=S^1\) by attaching a \(2\)-cell by a degree-\(m\) map, \(m>1\). Then \[ H_2(X,A)\cong\mathbb Z\xrightarrow{\partial=m}\mathbb Z\cong H_1(A) \longrightarrow H_1(X)\cong\mathbb Z_m\longrightarrow0. \] Passing to torsion gives \[ 0\longrightarrow0\longrightarrow\mathbb Z_m\longrightarrow0, \] which is not exact at \(T_1(X)=\mathbb Z_m\): the incoming image is \(0\) while the outgoing kernel is all of \(\mathbb Z_m\).
3. The mod-torsion functors are not exact on the same pair.
Passing the displayed integral sequence to quotients by torsion gives \[ \mathbb Z\xrightarrow{m}\mathbb Z\longrightarrow0. \] Exactness at the second \(\mathbb Z\) would require \(m\mathbb Z=\mathbb Z\), false for \(m>1\).
Thus neither \(T_*\) nor \(MT_*\) satisfies the exactness axiom, so neither defines a homology theory.