# The singular homology groups of a 1-point space are all trivial except the zeroth which is isomorphic to the integers

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## Statement

Let [ilmath](\{x_0\},\{\emptyset,\{x_0\}\})[/ilmath] be the only topological space on the set consisting of just a point, [ilmath]x_0[/ilmath], and let [ilmath]X:\eq\{x\} [/ilmath], then we claim the singular homology groups of [ilmath]X[/ilmath] are as follows:

• [ilmath]H_0(X)\cong\mathbb{Z} [/ilmath]
• [ilmath]H_n(X)\cong 0[/ilmath] for [ilmath]n\in\mathbb{N}_{\ge 1} [/ilmath]

## Proof

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The gist is as follows:

• Notice, for all [ilmath]n\in\mathbb{N}_0[/ilmath] that [ilmath]\phi_n:\Delta_n\rightarrow X[/ilmath] by [ilmath]\phi_n:(t_0,\ldots,t_n)\mapsto x_0[/ilmath] always
• Let [ilmath]\partial_{n,i} [/ilmath] be the [ilmath]i^\text{th} [/ilmath] face map of the [ilmath]n^\text{th} [/ilmath] free singular [ilmath]n[/ilmath]-simplex group, for [ilmath]n\in\mathbb{N}_{\ge 1} [/ilmath]. Then we notice:
• [ilmath]\partial_{n,i}(\phi_n)\eq\phi_{n-1} [/ilmath]
• Proof:
• [ilmath](\partial_{n,i}(\phi_n))(t_0,\ldots,t_n-1)\eq \phi_n(t_0,\ldots,t_{i-1},0,t_i,\ldots,t_{n-1})\eq x_0[/ilmath] and
• [ilmath]\phi_{n-1}(t_0,\ldots,t_n)\eq x_0[/ilmath]
• So we see [ilmath]\partial_{n,i}(\phi_n)\eq\phi_{n-1} [/ilmath] as required.
• Recall that [ilmath]\partial_n(\phi_n):\eq\sum_{i\eq 0}^n(-1)^i\partial_{n,i}(\phi_n)[/ilmath] - for [ilmath]n\in\mathbb{N}_{\ge 1} [/ilmath], as [ilmath]\partial_0[/ilmath] maps everything to [ilmath]0[/ilmath] always.
• Then notice: [ilmath]\partial_n(\phi_n):\eq\sum_{i\eq 0}^n(-1)^i\partial_{n,i}(\phi_n)\eq \sum_{i\eq 0}^n(-1)^i\phi_{n-1} \eq \phi_{n-1}\sum^n_{i\eq 0}(-1)^i [/ilmath]
• Let us consider 2 cases, consider [ilmath]n\eq 2k-1[/ilmath] for [ilmath]k\in\mathbb{N}_{\ge 1} [/ilmath] - for the odd numbers, [ilmath]n\eq 1,3,5,\ldots [/ilmath] and [ilmath]n\eq 2k[/ilmath] for [ilmath]k\in\mathbb{N}_{\ge 1} [/ilmath] for the even numbers, [ilmath]n\eq 2,4,6,\ldots [/ilmath]
1. [ilmath]n\eq 2k-1[/ilmath] for some [ilmath]k[/ilmath].
• [ilmath]\partial_{2k-1}(\lambda \phi_{2k-1})\eq 0[/ilmath] (look at the summation
TODO: Prove it
)
• So [ilmath]\text{Ker}(\partial_{2k-1})\eq S_{2k-1}(X)[/ilmath], and
• [ilmath]\text{Im}(\partial_{2k-1})\eq 0[/ilmath] (the trivial group)
2. [ilmath]n\eq 2k[/ilmath] for some [ilmath]k[/ilmath].
• [ilmath]\partial_{2k}(\lambda \phi_{2k})\eq \lambda\phi_{2k-1} [/ilmath]
• So [ilmath]\text{Ker}(\partial_{2k})\eq 0[/ilmath] (the trivial group), and
• [ilmath]\text{Im}(\partial_{2k})\eq S_{2k-1}(X)[/ilmath]
• Now we can compute the [ilmath]n\in\mathbb{N}_{\ge 1} [/ilmath] homology groups!
• $H_{2k-1}(X):\eq\frac{Z_{2k-1}(X)}{B_{2k-1}(X)}\eq\frac{\text{Ker}(\partial_{2k-1})}{\text{Im}(\partial_{2k})}\eq\frac{S_{2k-1}(X)}{S_{2k-1}(X)}\cong 0$
• $H_{2k}(X):\eq\frac{Z_{2k}(X)}{B_{2k}(X)}\eq \frac{\text{Ker}(\partial_{2k})}{\text{Im}(\partial_{2k+1})} \eq \frac{\text{Ker}(\partial_{2k})}{\text{Im}(\partial_{2(k+1)-1})} \eq\frac{0}{0}\cong 0$ - again the trivial group.
• Lastly let us compute:
• $H_0(X):\eq\frac{Z_0(X)}{B_{-1}(X)}\eq\frac{\text{Ker}(\partial_0)}{\text{Im}(\partial_1)}\eq\frac{S_0(X)}{0}\cong S_0(X)\cong \mathbb{Z}$

[ilmath]S_0(X):\eq\langle\phi_0\rangle [/ilmath] hence that last isomorphism.