It is a fact that if \(X\) is a single point then \(H_1(X)=0\). One of the following is the correct justification of this fact in terms of the singular chain complex. Which one is correct and why is it correct?
\(C_1(X)=0\).
\(C_1(X)\neq0\) but \(\ker\partial_1=0\), with \(\partial_1:C_1(X)\to C_0(X)\).
\(\ker\partial_1\neq0\) but \(\ker\partial_1=\operatorname{im}\partial_2\), with \(\partial_2:C_2(X)\to C_1(X)\).
Solution
The correct answer is (c).
1. For every \(n\ge0\), the singular chain group of the one-point space \(X=\{*\}\) is\[ C_n(X)\cong\ZZ. \]
There is exactly one continuous map \[ \sigma_n:\Delta^n\longrightarrow\{*\}, \] so there is exactly one singular \(n\)-simplex. By definition, \(C_n(X)\) is the free abelian group on the singular \(n\)-simplices, hence \[ C_n(X)=\ZZ\langle\sigma_n\rangle\cong\ZZ. \] In particular, \(C_1(X)\neq0\), so alternative (a) is false.
2. The boundary map\[ \partial_1:C_1(X)\longrightarrow C_0(X) \]is the zero map.
The two faces of the unique singular \(1\)-simplex are both the unique singular \(0\)-simplex \(\sigma_0\). Therefore \[ \partial_1\sigma_1 =\sigma_0-\sigma_0 =0. \] Hence \[ \ker\partial_1=C_1(X)\cong\ZZ\neq0, \] so alternative (b) is false.
3. The boundary map\[ \partial_2:C_2(X)\longrightarrow C_1(X) \]is surjective.
All three faces of the unique singular \(2\)-simplex are the unique singular \(1\)-simplex \(\sigma_1\). Thus \[ \partial_2\sigma_2 =\sigma_1-\sigma_1+\sigma_1 =\sigma_1. \] Since \(\sigma_1\) generates \(C_1(X)\), it follows that \[ \operatorname{im}\partial_2=C_1(X). \]
4. Therefore\[ H_1(X) =\ker\partial_1/\operatorname{im}\partial_2 =0. \]
By 2 and 3, \[ \ker\partial_1 =C_1(X) =\operatorname{im}\partial_2, \] and this common group is nonzero. This is exactly alternative (c).