# Fundamental class of a surface

https://pi.math.cornell.edu/~hatcher/AT/AT.pdf

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.

• 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? Jan 15 '20 at 2:14
• A regular neighborhood always exists. I needed the $\sigma$ to be embedded to guarantee the existence of a regular neighborhood homeomorphic to $\mathbb R^2$. Jan 15 '20 at 3:57
• Yes I meaned it thanks Jan 15 '20 at 4:32