Queston on the definition of singular homology From the Hatcher's 

can someone told me why $\sigma$ has singularities ?
Thank you 
 A: The author is emphasising the difference between a singular simplex and a simplicial simplex. Among other things, a simplicial simplex has to be an injective map when restricted to its interior. This simply isn't the case for singular simplexes - they can be as non-injective (or singular) as they like - we only require that they are continuous maps.
It's helpful to give an example of how this affects computations. Given that the singular chain complex of a space is almost always incalculable, I'll use a single point as the example space.
When calculating the simplicial complex associated to $\{\ast\}$ we note that there exists exactly one simplicial simplex $\sigma_0\colon\Delta_0\to\{\ast\}$ and there exist no higher simplicial simplexes because any such simplex would fail to be injective on their interior. It follows that the associated chain complex is given by $$\cdots\to 0\to 0\to \mathbb{Z}_{ \langle \sigma_0 \rangle}.$$
Now, what about the associated singular chain complex for $\{\ast\}$? Well, there is still exactly one simplex $\sigma_0\colon\Delta_0\to\{\ast\}$ but now we also have exactly one continuous map (and so singular simplex) $\sigma_n\colon\Delta_n\to\{\ast\}$ for each $n\geq 1$. It follows that the associated singular chain complex for $\{\ast\}$ is given by $$\cdots\to \mathbb{Z}_{ \langle \sigma_2 \rangle}\to \mathbb{Z}_{ \langle \sigma_1 \rangle}\to \mathbb{Z}_{ \langle \sigma_0 \rangle}.$$
