Why $\Delta_1(S^1)=\mathbb{Z}$ and $\Delta_n(S^1)=0$ for $n\geq 2$?

Question

I find it difficult to understand it when I read Hatcher's Algebraic Topology. In Example 2.2, I can understand $\Delta_0(S^1)=\mathbb{Z}$. But how to illustrate why $\Delta_1(S^1)=\mathbb{Z}$ and $\Delta_n(S^1)=0$ for $n\geq 2$?

Answer 1

Hatcher chose a $\Delta$-complex structure on $S^1$ with exactly one $0$-cell $v$, one $1$-cell $e$, and zero $n$-cells for $n \geq 2$. But $\Delta_i(S^1)$ is the free abelian group generated by the open $n$-simplices of $S^1$. In other words, we count the number of $i$-cells to find the rank of $\Delta_i(S^1)$ and obtain $$ \Delta_i(S^1) = \begin{cases} \mathbb{Z}v \cong \mathbb{Z} \quad & i = 0, \\ \mathbb{Z}e \cong \mathbb{Z} & i = 1, \\ 0 & i \geq 2. \end{cases}$$