Compute the singular homology groups of $S^1$ How do I go about computing the singular homology groups of $S^1$?
Anything to get me started or a full answer is appreciated.
EDIT: I've realised that while this is a simple question, there are various ways to get the answer, as Johannes Hahn has suggested. In particular, I'm looking into attempting to do calculations on singular homology groups. How will that look like? (Is it very similar to doing the calculations to find the simplicial homology groups and then do a slight change of notation?)
 A: If you want to compute the singular homology without using an isomorphism to some other homology theory (such as simplicial, Cech, ...), then the standard way is to use Mayer--Vietoris.  You decompose $S^1$ into the union of two intervals, each overlapping near their two endpoints, and use the fact that intervals are contractible (and hence have the homology of a point).
A: The standard way of doing it would be to first prove standard theorems about singular homology (Eilenberg-Steenrod-Axioms, Mayer-Vietoris, suspension isomorphism, ...). Then one can easily (that is easy with all these theorems) prove that $H_n(S^m;A) = \begin{cases} A & n=0 \vee n=m \\ 0 & \text{otherwise}\end{cases}$.
Oooor: One gets ones hands dirty and actually calculates. This is clearly the less preferable way and heavily depends on what is already known and what is not.
Oooor: One knows about the several other ways to define homology (cellular, simplicial, Cech, ...) and that they all give the same result. In that case one can use the easier to calculate homologies.
Johannes Hahn.
A: You could triangulate $S^1$ (use the border of a 2-simplex), then compute the simplicial homology of that simplex, and then use that simplicial and singular homology agree.
Or, you could calculate the $\pi_{1}$, and use Hurewicz theorem.
