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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.

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    To show the equivalence between simplicial and singular homology we need to first prove that the identity map is a generator, as in Hatcher Example 2.23 used in Theorem 2.27, which uses tools such as excision and long exact sequence. So if what you want to calculate is *singular* homology of the sphere, I would suspect there is no easier way without excision, or, barycentric subdivision.2017-06-26

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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.

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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.

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    That's possible!2011-01-27