TITLE: Singular homology of spaces with trivial topology
QUESTION [5 upvotes]: 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.

REPLY [4 votes]: 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 $n>0$.