From Hatcher's Algebraic Topology:
Cycles in singular homology are defined algebraically, but they can be given a somewhat more geometric interpretation in terms of maps from finite $\Delta$-complexes. To see this, note that a singular $n$-chain $\xi$ can always be written in the form $\sum_i \varepsilon_i \sigma_i$ with $\varepsilon_i = \pm 1$, allowing repetitions of the singular $n$-simplices $σ_i$. Given such an $n$-chain $\sum_i \varepsilon_i \sigma_i$, when we compute $\partial \xi$ as a sum of singular $(n−1)$-simplices with signs $\pm 1$, there may be some canceling pairs consisting of two identical singular $(n−1)$-simplices with opposite signs. Choosing a maximal collection of canceling pairs, construct an $n$-dimensional $\Delta$ complex $K_\xi$ from a disjoint union of $n$-simplices, one for each $\sigma_i$, by identifying the pairs of $(n−1)$-dimensional faces corresponding to the chosen canceling pairs. The $\sigma_i$'s then induce a map $K_\xi→X$. If $\xi$ is a cycle, all the $(n−1)$-simplices of $K_\xi$ come from canceling pairs, hence are faces of exactly two $n$-simplices of $K_\xi$. Thus $K_\xi$ is a manifold, locally homeomorphic to $\mathbb R^n$, except at a subcomplex of dimension at most $n−2$.
I don't really understand this. Does this mean that if the $n$-chain is a chain of $2$-simplices with two nonzero terms that cancel each other after applying the boundary operator (they have the same boundary oriented in opposite directions), then after taking the quotient space of their disjoint union identifying the boundaries, we get a $2$-space that's a manifold except near the $3$-vertices? (This space will be homeomorphic to $S^2$?) Is this is the intended meaning of the paragraph?