# fundamental class is the sum of simplices of triangulation of the manifold?

M is an n-dimensional closed orientable manifold. I find in a book "Intuitively,the fundamental class can be thought of as the sum of the (top-dimension) simplices of a suitable triangulation of the manifold". My explanation is we can use the Mayer-vietoris sequence to "glue" the local orientations inductively over a finite oriented atlas together to construct the fundamental class. One open subset of the atlas correspond to one simplex. Am I right?

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Yes, that's exactly how it works. The triangulation need not be "suitable", any triangulation works. You have to use signs if the triangulation is not compatible with the manifold's orientation but that's the only issue. –  Ryan Budney May 21 '13 at 19:54
Not exactly. In general the number of top dimensional simplices has to be higher than the number of open maps of an atlas. First of all you need to have any triangulation. If you do then any triangulation will be fine, and Mayer-Vietoris would not bother with this issue. Fix an orientation in one simplex, and it will spread over the entire connectivity component (over the whole manifold if it is connected; otherwise choose orientation for one simplex per component). –  Włodzimierz Holsztyński May 22 '13 at 8:27
@Wlodzimierz: It's unclear what you are referring to, both with your "not exactly" comment and the "number of open maps of an atlas" comment. –  Ryan Budney May 22 '13 at 13:41
@Ryan: In the Question @jiangsaiyin seems to claim that there always can be a 1-1 correspondence between the open maps of an atlas and the simplices. That's not quite true. Also the mentioning of Mayer-Vietoris sounds not clear, more like unnecessary. On the other hand IF a triangulation of a compact manifold is given then one may hope, somewhat naively but why not, that each top simplex can be expanded to a map (just make them small enough to get different maps for different simplices), and indeed we would get a 1-1 correspondence between maps and the top simplices. –  Włodzimierz Holsztyński May 22 '13 at 20:25
@jiangsaiyin: sphere $S^n$ can be covered by an atlas consisting of 2 maps, but for every positive $n$ the minimal number of $n$-simplices of a triangulation of $S^n$ is higher, is $n+2$. –  Włodzimierz Holsztyński May 22 '13 at 20:29

Here is how I would prove it. You need to know two things. 1. If $M$ is connected, then $H_n (M) \to H_n (M,M-x)$ is an isomorphism (proven in any appropriate textbook). 2. If $\sigma:\Delta^n \to R^n$ is an embedding that has $o$ as an interior point of its image, then it represents a generator of $H_n (R^n,R^n - o)$. This is easily reduced to showing that the identity map on $\Delta^n$ represents a generator of $H_n (\Delta^n, \Delta^n - o)=H_n (\Delta^n, \partial \Delta^n)$, where $o$ is an interior point. By long exact sequence and excision, $H_n (\Delta^n, \partial \Delta^n) \to H_{n-1} (\partial \Delta^n) = H_{n-1}(\partial \Delta^n, \Lambda^n) \cong H_{n-1} (\Delta^{n-1}, \partial \Delta^{n-1})$, where $\Lambda^n$ is the horn, i.e. the union of all but one boundary simplices. By induction, the claim follows.
To get back to the manifold: pick a point $x$ in the interior of $\sigma$, one of the top-dimensional simplices in your triangulation. All that remains to be proven is that the sum of all top-dimensional simplices maps to $\sigma$ under $H_n (M) \to H_n (M,M-x)$, but that follows directly from what I wrote.