Let \(Y\) be a \(\Delta\)-complex.
Prove that if \(H_4(Y)\ne 0\), then \(Y\) must have a simplex of dimension 4.
Prove that if \(H_4(Y)=\mathbb Z/7\mathbb Z\), then \(Y\) also must have a simplex of dimension 5.
Solution
Let \[ \cdots\longrightarrow C_5(Y) \overset{\partial_5}{\longrightarrow} C_4(Y) \overset{\partial_4}{\longrightarrow} C_3(Y) \longrightarrow\cdots \] be the simplicial chain complex of the \(\Delta\)-complex \(Y\).
1. If \(Y\) has no \(4\)-simplices, then\[ H_4(Y)=0. \]
The simplicial chain group \(C_4(Y)\) is the free abelian group with basis the \(4\)-simplices of \(Y\). If there are no such simplices, then \[ C_4(Y)=0. \] Therefore \[ \ker\partial_4=0, \] and hence \[ H_4(Y) = \ker\partial_4/\operatorname{im}\partial_5 =0. \]
2. Thus, if\[ H_4(Y)\ne0, \]then \(Y\) has a simplex of dimension \(4\).
This is the contrapositive of 1, proving part (a).
3. Suppose \(Y\) has no \(5\)-simplices.Then \(H_4(Y)\) is a subgroup of the free abelian group \(C_4(Y)\) and is therefore torsion-free.
If there are no \(5\)-simplices, then \[ C_5(Y)=0, \] so \[ \operatorname{im}\partial_5=0. \] Consequently \[ H_4(Y) = \ker\partial_4. \] The group \(C_4(Y)\) is free abelian on the \(4\)-simplices, and \[ \ker\partial_4\le C_4(Y). \] Every subgroup of a free abelian group is torsion-free: if \(0\ne m\in\mathbb Z\) and \(mx=0\) in a subgroup, then the same equation holds in the ambient free abelian group, forcing \(x=0\). Hence \(H_4(Y)\) is torsion-free.
4. If\[ H_4(Y)\cong\mathbb Z/7\mathbb Z, \]then \(Y\) has a simplex of dimension \(5\).
The group \(\mathbb Z/7\mathbb Z\) contains nonzero torsion: every nonzero class \(x\) satisfies \[ 7x=0. \] This contradicts the torsion-freeness in 3 if \(Y\) has no \(5\)-simplices. Therefore a \(5\)-simplex must exist, proving part (b).