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

{{Stub page|grade=A*|msg=Link to stuff, like the singular homology groups or group. Make this embedded in the wiki}}
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==Statement==
Let {{M|(\{x_0\},\{\emptyset,\{x_0\}\})}} be the only [[topological space]] on the set consisting of just a point, {{M|x_0}}, and let {{M|X:\eq\{x\} }}, then we claim the singular homology groups of {{M|X}} are as follows{{rHTJWV}}:  
* {{M|H_0(X)\cong\mathbb{Z} }}
* {{M|H_n(X)\cong 0}} for {{M|n\in\mathbb{N}_{\ge 1} }}
==Proof{{rHTJWV}}==
{{Requires work|grade=B|msg=Fill it out}}
The gist is as follows:
* Notice, for all {{M|n\in\mathbb{N}_0}} that {{M|\phi_n:\Delta_n\rightarrow X}} by {{M|\phi_n:(t_0,\ldots,t_n)\mapsto x_0}} always
* Let {{M|\partial_{n,i} }} be the {{M|i^\text{th} }} face map of the {{M|n^\text{th} }} free singular {{n|simplex}} group, for {{M|n\in\mathbb{N}_{\ge 1} }}. Then we notice:
** {{M|\partial_{n,i}(\phi_n)\eq\phi_{n-1} }}
*** '''Proof:'''
**** {{M|(\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}} and
**** {{M|\phi_{n-1}(t_0,\ldots,t_n)\eq x_0}}
*** So we see {{M|\partial_{n,i}(\phi_n)\eq\phi_{n-1} }} as required.
* Recall that {{M|\partial_n(\phi_n):\eq\sum_{i\eq 0}^n(-1)^i\partial_{n,i}(\phi_n)}} - for {{M|n\in\mathbb{N}_{\ge 1} }}, as {{M|\partial_0}} maps everything to {{M|0}} always.
** Then notice: {{M|\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 }}
*** Let us consider 2 cases, consider {{M|n\eq 2k-1}} for {{M|k\in\mathbb{N}_{\ge 1} }} - for the odd numbers, {{M|n\eq 1,3,5,\ldots }} and {{M|n\eq 2k}} for {{M|k\in\mathbb{N}_{\ge 1} }} for the even numbers, {{M|n\eq 2,4,6,\ldots }}
***# {{M|n\eq 2k-1}} for some {{M|k}}. 
***#* {{M|\partial_{2k-1}(\lambda \phi_{2k-1})\eq 0}} (look at the summation {{XXX|Prove it}})
***#** So {{M|\text{Ker}(\partial_{2k-1})\eq S_{2k-1}(X)}}, and
***#** {{M|\text{Im}(\partial_{2k-1})\eq 0}} (the trivial group)
***# {{M|n\eq 2k}} for some {{M|k}}.
***#* {{M|\partial_{2k}(\lambda \phi_{2k})\eq \lambda\phi_{2k-1} }}
***#** So {{M|\text{Ker}(\partial_{2k})\eq 0}} (the trivial group), and
***#** {{M|\text{Im}(\partial_{2k})\eq S_{2k-1}(X)}}
* Now we can compute the {{M|n\in\mathbb{N}_{\ge 1} }} homology groups!
** {{MM|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 }}
** {{MM|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:
** {{MM|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}  }}

{{M|S_0(X):\eq\langle\phi_0\rangle }} hence that last isomorphism.
==References==
<references/>
{{Theorem Of|Algebraic Topology|Homology|Singular Homology}}