[Math] How does singular homology H_n capture the number of n-dimensional “holes” in a space

at.algebraic-topologytag-removed

This is a foundational doubt I have. How does singular homology H_n capture the number of n-dimensional holes in a space?

We disregard the case of $H_0$ as it has the very satisfactory explanation that it is the direction sum of $\mathbb Z$ over the path-connected components of the space.

Now, handwaving aside, we consider the most important example of this "detecting hole" phenomenon, viz,, the fact that for $i \geq 1$ $H_i(S^n) = \mathbb(Z)$ if and only if $i = n$. For this we use Mayer-Vietoris and a decomposition of $S_n$ into a union of two open sets which are the complements of the north pole and south pole. And the intersection deformation retracts to $S^{n -1}$ and from the long exact sequence we get the isomorphisms $H_i \cong H_{i -1}$.

Now, by the above computation, it seems that the "hole detection" is achieved via Mayer-Vietoris and going up from the dimension below, using the long exact sequence. Mayer-Vietoris on the other hand depends on the snake lemma, which is very un-geometric and difficult to visualize.

So I would be most grateful for a more intuitive explanation of this hole capturing phenomenon. I can see that it is very natural that boundaries should be cancelled out as the solid simplices can be contracted to the central point. I can also "feel" that a hollow $n$-simplex, there should be a nontrivial $n$-chain which is not a boundary of an $n+1$-chain. But I am still left with a feeling of partial understanding. I hope this fundamental vagueness of understanding of mine can be cleared here.

Best Answer

The "hole detection" is rather in the very definition of homology. Consider, for example, $H_2$: it is morally the set of closed surfaces in you space modulo those that bound a $3$-dimensional body, and if a surface is not the boundary of any $3$-dimensional body then surely there must be a hole entrapped in it, no?

(Morally because when you want to actually implement this, you get a slightly different thing... Although I'd be thrilled to be informed that, in the case of a manifold at least, say, one can somehow construct a free abelian group on the set of maps $\Sigma\to M$ from $k$-manifolds $\Sigma$ to $M$ which, when one mods out the subgroup of those maps that extend to a manifold-with-boundary $N$ such that $\partial N=\Sigma$, gets you $H_k(M)$ or something close)