In Example 3.31 in Hatcher's Algebraic Topology(p.241), there is a figure of a $\Delta$-complex structure of the closed orientable surface $M$ of genus $g$ ($g=2$ in the figure). Hatcher says that, the $2$-cycle formed by the sum of all $4g$ $2$-simplices with the signs indicated in the figure, represents a fundamental class $[M]$ of $M$. I want to understand this.

It suffices to show that $[M]$ corresponds to the generator of $H_1(S^1)$ under the following isomorphisms, for each $x \in M$:

$$H_2(M) \to H_2(M,M-x) \leftarrow H_2(U,U-x) \to H_2(\Bbb R^2,\Bbb R^2-0)\to H_1(\Bbb R^2-0)\to H_1(S^1)$$

where $U$ is an open neighborhood of $x$ in $M$ homeomorphic to $\Bbb R^2$, and the second isomorphism is excision.

It is easy to examine the maps except for the second one. The generator of $H_1(S^1)$ (the loop wrapping once the circle) corresponds to the generator of $H_2(U,U-x)$ which is represented by, say a relative cycle $\sigma: \Delta^2 \to M$ with $x \in \text{int} (\sigma(\Delta^2))$. But how can I know that $[M]$ corresponds to $[\sigma]$ under the second isomorphism?


This is really just a matter of understanding the actual form of the excision isomorphism.

But we can pick $U$, $\sigma$, and $x$ somewhat carefully, to make this easier.

Instead of working with the given $\Delta$-complex structure, let's work with its 2nd barycentric subdivision, which is guaranteed to be an actual simplicial complex whose individual simplices are actually embedded. You probably know that a homology class is invariant under subdivision, so the two classes formed by summing the simplices of the 2nd barycentric subdivision and by summing the 2-simplices of the original $\Delta$-complex structure are equal. That gives us the freedom to work with teh 2nd barycentric subdivision.

Choose $\sigma$ to be a 2-simplex of the 2nd barycentric subdivision. And now choose $U$ to be a regular neighborhod of $\sigma$, chosen so small that $\sigma$ is the unique 2-simplex of the 2nd barycentric subdivision that is contained in $U$; because it's a regular neighborhood of a closed disc in a manifold, $U$ is homeomorphic to $\mathbb R^2$.

Finally, pick $x$ to be a point in the interior of $\sigma$.

What we need to show is that if $c$ is the sum of all simplices of the 2nd barycentric subdivision then the class $[c] \in H_2(M,M-x)$ is the equal to the image, under excision, of the class $[\sigma] \in H_2(U,U-x)$. And this is really just a matter of understanding a concrete description of the excision homomorphism.

Excision can be described like this:

If you have a $k$-cycle $c = \sum a_i \tau_i$ of $(M,M-x)$, and if each $\tau_i$ is contained in either $M-x$ or $U$, and if $c'$ is obtained from $c$ by discarding all terms $a_i\tau_i$ such that $\tau_i$ is contained in $M-x$, then $c'$ is a $k$-cycle of $(U,U-x)$, and the excision map $H_k(U,U-x) \to H_k(M,M-x)$ takes $[c']$ to $[c]$.

Now let's apply this. Certainly the given $c$ is a cycle of $M$, and it is in fact a fundamental cycle, representing the fundamental class $[M]$. So $c$ is certainly also a cycle of $(M,M-x)$. Applying the above description of excision, the terms that we remove from $c$ are all of the terms except for $\sigma$, which is the only term contained in $U$. So all that's left is $c'=\sigma$, and we're done.

  • $\begingroup$ Thanks I clearly see that $[\sigma]$ and $[c]$ is the same in $H_2(M,M-x)$. One last question: Is there a reason that you specified the "2nd" subdivision? Is it just for the guarantee of a regular neighborhood? $\endgroup$
    – blancket
    Jan 15 '20 at 2:14
  • $\begingroup$ A regular neighborhood always exists. I needed the $\sigma$ to be embedded to guarantee the existence of a regular neighborhood homeomorphic to $\mathbb R^2$. $\endgroup$
    – Lee Mosher
    Jan 15 '20 at 3:57
  • $\begingroup$ Yes I meaned it thanks $\endgroup$
    – blancket
    Jan 15 '20 at 4:32

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