TITLE: Homology of the Sphere
QUESTION [6 upvotes]: In my algebraic topology class we showed how to calculate the homology groups of $S^n$, using the tools of singular homology, however we did not discuss other ways of doing it; my question - is there any relatively simple way of doing this, using simplicial homology? I tried thinking about this for a bit, but couldn't see any obvious direction.
Thanks.

REPLY [7 votes]: Using simplicial homology you can triangulate the sphere as the boundary of an $(n+1)$-simplex, and work out the chain complex by hand. 
With cellular homology it is even easier since $S^n$ is the union of an $n$-cell and a $0$-cell. The chain complex has a single $\mathbb Z$ in degree $0$ and a single $\mathbb Z$ in degree $n$. In all other degrees it is zero.

REPLY [4 votes]: Another rather quick way to compute the groups is with cellular homology.  Here the $n$th chain group is generated by the $n$-cells of your CW-complex.  The boundary map has to do with degree maps; but in your case it is simple.  An $n$-sphere is a 0-cell with an $n$-cell attached by mapping the boundary $S^{n-1}$ to the 0-cell.  If $n>1$ then all the maps in the chain complex must be 0 because the chain groups are trivial except for the $n$-chains and 0-chains.  The homology of such a complex is easy to compute.