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Suppose $X$ is endowed with the trivial topology, e.g. $X$ and $\emptyset$ are the only open sets. For the sake of simplicity, I'll assume that $X$ is finite, $|X| = m$.

Now, the $n$-th module of the singular chain complex of $X$ should be a free $\mathbb{Z}$-module generated by the basis of every map from the n-simplex to $X$, since all those maps are continuous.

Looking back at the example of a single point space, the boundary operators could be described explicitly in terms of $n$ odd or even since the cardinality of the basis was always one. In this case, however, I fail to see what those maps are supposed to look like. I'd appreciate some hints on how to tackle this problem.

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Do you want to implicitly use the definitions to compute homology, or is it sufficient to notice that any non-empty $X$ with trivial topology is necessarily contractible to a single point? –  Thomas Andrews Dec 6 '11 at 16:38
    
I don't know how contractibility of a space relates to the associated singular chain complex since I pretty much just have the definitions at my disposal, but I look forward to any solution that doesn't require a vast machinery. I can always look up the details. –  athanasios Dec 6 '11 at 16:56
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Expanding on Thomas's comment, the map $H\colon X\times I\to X$ given by $H(x,t)=x$ for $0\leq t<1$ and $H(x,1)=x_0$ for some fixed $x_0\in X$ gives a homotopy from the identity map to the constant map (check that it's continuous!). This means that $X$ is homotopy equivalent to the single point space $*$, and because singular homology is a homotopy invariant, it follows that $X$ has the same homology as $*$. Hence, $H_0(X;\mathbb{Z})=\mathbb Z$ and $H_n(X;\mathbb{Z})=0$ for $n>0$.

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Thanks. I looked up homotopy invariance of homology and this is precisely what I was looking for - a rather elementary way to get around the cumbersome definition of singular homology! –  athanasios Dec 6 '11 at 17:51
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