Suppose $M$ is a topological manifold, triangulated in the sense of CW complexes. If $M$ is closed and connected, I am able to prove (by using the cellular homology and its boundary map) that necessarily $H_n(M)=0$ or $H_n(M)=\mathbb{Z}$.

My definition of orientability is as follows: $M$ is orientable if there exists a class (the fundamental class) $z\in H_n(M)$ such that $\pi(z)$ generates $H_n(M,M-x)\cong \mathbb{Z}$ for all $x\in M$, where $\pi\colon H_n(M) \longrightarrow H_n(M,M-x)$ denotes natural map of the pair $(M,M-x)$.

My question is: in case that $M$ is a triangulated closed connected manifold, is it clear (with my definition of orientability) that $H_n(M)\cong Z$ if and only if $M$ is orientable? I don't see why the map $\pi$ has to be an isomorphism when $H_n(M)\cong \mathbb{Z}$...

  • $\begingroup$ What do you mean by "triangulated in the sense of CW complexes"? $\endgroup$ – Eric Wofsey Jan 6 '18 at 20:10
  • $\begingroup$ I mean that it is a sub CW complex of $\Delta^m$ (the $m$-simplex) for some $m$. Then one can prove that all the cells are of dimension at most $n=\textrm{dim }M$ and each $n-1$ cell is a face of exactly two $n$-cells. $\endgroup$ – J. Karen Jan 6 '18 at 20:20
  • $\begingroup$ Look at Theorem 3.26 in Hatcher $\endgroup$ – Mike Hawk Jan 6 '18 at 20:39

Presumably, your proof that $H_n(M)=0$ or $H_n(M)=\mathbb{Z}$ shows that if $H_n(M)=\mathbb{Z}$, then a generator of $H_n(M)$ can be represented as $z=[\sum c_id_i]$ where the $d_i$ are all the $n$-simplices of some triangulation of $M$ and $c_i=\pm1$ for each $i$. I claim that this $z$ is then a fundamental class of $M$.

Indeed, consider any point $x\in M$. Let us first suppose $x$ is in the interior of one of the $n$-simplices $d_i$, say $d_1$. Then $\pi(z)=[c_1d_1]$, since the simplices $d_i$ for $i\neq 1$ are completely contained in $M-x$. Let us write $D\subset M$ for the closed simplex $d_1$, considered as a subspace of $M$. By excision, the natural map $H_n(D,D-x)\to H_n(M,M-x)$ is an isomorphism. Moreover, $H_n(D,D-x)$ is generated by $[d_1]$ (for instance, because $[\partial d_1]$ is a generator of $H_{n-1}(\partial D)$ and the boundary map $H_n(D,D-x)\cong H_n(D,\partial D)\to H_{n-1}(\partial D)$ is an isomorphism). Since $c_1=\pm 1$, this means $[c_1d_1]$ also generates $H_n(D,D-x)$, and hence $\pi(z)=[c_1d_1]$ generates $H_n(M,M-x)$ as well, as desired.

Unfortunately, this argument does not work if $x$ is not in the interior of an $n$-simplex of the triangulation (a variation would work if $x$ is in the interior of an $(n-1)$-simplex, but it's just hopeless if $x$ is in the $(n-2)$-skeleton). However, if $x$ is not in the interior of an $n$-simplex, you can just perturb your triangulation a little bit near $x$ so that it is (here you have to use the fact that $M$ is a topological manifold, so there is a neighborhood of $x$ which is homeomorphic to a closed ball and then there is a homeomorphism of that ball fixing the boundary which moves $x$ to any other interior point, in particular one which is in the interior of an $n$-simplex). So you can find some triangulation for which $x$ is in the interior of an $n$-simplex, and then apply the argument above for that triangulation.

  • $\begingroup$ Your answer is precisely what I wanted. Thank you! $\endgroup$ – J. Karen Jan 6 '18 at 20:54

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