I read that both the zeroth homology and cohomology both count the connected components of a finite simplicial complex. (I am restricting to finite simplicial complexes so that connected=path connected and direct sum=direct product.)

However, by Poincare duality, $H^k\cong H_{n-k}$.

Hence, does that mean that actually $H^n$ and $H_n$ also count the connected components of a simplicial complex? (Since $H^0\cong H_{n}$ and $H^n\cong H_{0}$ by Poincare duality).

However that doesn't seem right? I recall that $H_n$ counts the number of $n$-dimensional "holes", which can be different from the number of connected components.

Thanks for any enlightenment.

Update: I realize Poincare duality only works for manifold without boundary. Is this the reason?

  • $\begingroup$ A form of Poincare duality does indeed hold for manifolds with boundary. A triangulation of a given manifold actually gives rise to a dual cell structure. In this way counting the 0-dimensional holes is the same as counting the n-dimensional holes of the dual structure. I'd suggest the wiki article en.wikipedia.org/wiki/Poincar%C3%A9_duality for some background. $\endgroup$
    – Tyrone
    Dec 7, 2017 at 14:49

1 Answer 1


A few preliminary notes:

  • To simplify things I will assume the coefficient ring $R = \mathbb Z.$ The same arguments also go through for general $R,$ provided we restrict to $\mathbb Z$-orientable surfaces.

  • The usual formulation of Poincaré duality holds for compact, oriented manifolds without boundary (to see where this can fail, consider the unit ball in $\mathbb R^n$). There is a relative version for manifolds with boundary, but I won't go into that here.

Recall that if $M$ is a (closed oriented) manifold, we can decompose it into its path components as $M = \bigsqcup_{\alpha \in A} M_{\alpha}.$ Then we get an isomorphism, $$ H_k(M) \cong \bigoplus_{\alpha \in A} H_k(M_{\alpha}) $$ and similarly for cohomology. The important point is that if $M$ is orientable, then so are all of its path components. So assuming $M$ has dimension $n$ we get,

$$ H_n(M_{\alpha}) \cong \mathbb Z. $$

From this you see that for each path component, each path component gives rise to a (co)homology class in grading $n.$

Now what you're probably wondering is why the last claim I stated holds. To simplify things I will further assume $M$ is connected and explain why $H_n(M) \cong \mathbb Z.$

Formally we can prove this assuming $M$ is triangulable ($M$ is homeomorphic to a simplicial complex, where all the top dimensional simplicies have dimension $n$), which is always true albeit annoying to prove. One can check that if you sum the $n$-simplicies arising from the triangulation (with appropriate orientations), you get a cycle and hence a non-trivial class in the top homology. The idea here is we can partition $M$ into these $n$-dimensional solid triangles, which fit together nicely in the sense that each $(n-1)$-simplex is a face of two $n$-simplicies. This already gives you some idea of the non-triviality of the top homology,since these simplicies somewhat loop back together.

The idea of $n$-dimensional holes does somewhat work to give some intuition also, where we restrict to the case where $M$ is smooth and embedded in $\mathbb R^{n+1}.$ I quite like this explanation which basically considers inserting an $(n+1)$-dimensional solid object inside the space; in this simplified setting it will be an open subset of $\mathbb R^{n+1}.$ Of course I don't think this will hold in general, but by considering the compact orientable surfaces (as illustrated in the link above) you can convince yourself that it works for $n=2.$ It's instructive to think about how this relates to the more formal viewpoint of adding simplicies.

The case of cohomology is a very different matter, since it doesn't measure $n$-dimensional holes. In that case I think the best way to get intuition is via the de Rham cohomology for smooth manifolds.


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