Zeroth homology group of a singleton

I'm reading the chapter "Homology" on Topology and Geometry by Bredon, and I tried to calculate the group $H_0(*)$ by the definition but I can't do it. Here's my attempt:

We shall calculate the $0$th homology group of the singleton $X=\{*\}$. Observe that there exists a single map $\sigma_0 \colon \Delta_0 \to X$. Hence the singular $0$-chain group $\Delta_0(X)$ is the group $\{0 \sigma_0, 1 \sigma_0, -1 \sigma_0,...\}$. The $0$th homology group of $\{*\}$ is $(\ker \partial_0)/(\mbox{im } \partial_{1})$. Note that $\partial_0 \colon \Delta_0(X) \to \Delta_{-1}(X)$, and, by definition, $\Delta_{-1}(X)=0$. Hence \begin{align*} \partial_0 \colon \Delta_0(X) &\to \Delta_{-1}(X) \\ n \sigma_0 &\mapsto 0, \end{align*} which implies $\ker \partial_0 = \Delta_0(X)$. Analogously there exists a single map $\sigma_1 \colon \Delta_1 \to X$. Hence $\Delta_1(X) = \{0 \sigma_1, 1 \sigma_1, -1 \sigma_1,...\}$. We have, by definition, \begin{align*} \partial_1 \colon \Delta_1(X) &\to \Delta_{0}(X) \\ n \sigma_1 &\mapsto n (\sigma_1 \circ F_0^1 - \sigma_1 \circ F_1^1). \end{align*} Since $X$ is a singleton, we have $\sigma_1 \circ F_0^1 - \sigma_1 \circ F_1^1 = \sigma_0$, which implies $\partial_1$ is a bijection, hence $\mbox{im } \partial_{1} = \Delta_0(X)$. Finally, we have $H_0(X)= \Delta_0(X)/ \Delta_0(X) \cong 0$.

I know that $H_0(X) \cong \mathbb{Z}$, but I don't understand the explanation given by Bredon.

• What don't you understand? If you are more specific, we can help you. Jul 7 '17 at 17:34

$$\newcommand{\si}{\sigma}$$ You are right that $H_0(X)$ and $H_1(X)$ are free Abelian groups with generators $[\si_0]$ and $[\sigma_1]$ respectively (I prefer this notation with brackets). The error is that the boundary $\partial_1[\si_1]$ is not $[\si_0]$ as you have it, but is zero. Thus the cycle group is $Z_0=\Delta_0(X)=\Bbb Z[\si_0]$ and the boundary group is $B_0=0$. Then $H_0(X)=Z_0/B_0\cong\Bbb Z$.
• I'm failing to see that $\partial_1 [\sigma_1]=0$. If $\partial_1 (n \sigma_1) = n \partial_1 ( \sigma_1)$ (by definition), then it can't "change" the coefficient $n$ to $0$, right? Jul 7 '17 at 18:12
• @HilarioFernandes $\partial_1$ of a path is the difference between its endpoints considered as elements of $Z_0(X)$. In our example the path $\sigma_1$ has both endpoints the same, so $\partial_1[\sigma_1]=[\sigma_1]-[\sigma_1]=0$. Jul 7 '17 at 18:52
• Oh, I see. By realizing that $\sigma_1 \circ F_0^1 = \sigma_0 = \sigma_1 \circ F_1^1$, we have $n(\sigma_1 \circ F_0^1 - \sigma_1 \circ F_1^1) = n(\sigma_0-\sigma_0)=(n-n)\sigma_0 = 0 \sigma_0$. Jul 7 '17 at 19:26