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i have this example :

The homology of the space $X=\lbrace x \rbrace$ .

for all $p\geq 0$, there is a unique singular p-simplex $\sigma_p:\Delta_p\rightarrow X$, and for $p>0$ we have $\partial_i\sigma_p=\sigma_{p-1}$ so

$\partial\sigma_p=(\sum_{i=0}^p (-1)^i)\sigma_{p-1}= 0 ~~\text{if $p$ is odd and }~~\sigma_{p-1} ~~\text{if not}$

so the singular chains complexe is $\mathbb{A} \stackrel{0}\longleftarrow \mathbb{A}\stackrel{id}\longleftarrow \mathbb{A}...$

I dont understand why $C_0(X,\mathbb{A})=C_1(X,\mathbb{A})=...=\mathbb{A}$

Please help me

Thank you.

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What is $\Bbb A$? –  Stefan Hamcke Oct 27 '13 at 17:41
    
I am slightly confused - for an $m$-dimensional simplicial complex shouldn't $C_n(X,\mathbb{A})=0$ if $n>m$? –  levitopher Oct 27 '13 at 17:43
2  
The chain group is generated by the unique simplex, so it is isomorphic to $\Bbb A$. –  Stefan Hamcke Oct 27 '13 at 17:46
1  
Yes, there can be only one. –  Stefan Hamcke Oct 27 '13 at 17:49
1  
@levitopher: Perhaps you are confusing simplicial and singular homology. –  Stefan Hamcke Oct 27 '13 at 17:56

1 Answer 1

For $p\geqslant 0$ the group $C_p(X,A)$ is a free $A$-module generated by one element $\sigma_p$; it follows that $C_p(X,A)\cong A$ by mapping $\sigma_p$ to $1$ and extending linearly.

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