Simplicial homology of a figure-eight graph

1. Figure 8 Here we have: \(C_3 = \emptyset\) \(C_1 = [12], [23], [13], [34], [35], [45] \cong \ZZ^6\) \(C_0 = [1], [2], [3],[4], [5] \cong \ZZ^5\)

So we have \(C_2 \into C_1 \into C_0 \cong 0\xrightarrow{\del_2} \ZZ^6 \xrightarrow{\del_1} \ZZ^5\xrightarrow{\del_0} 0\)

Computing boundary operators, we have

\(\del_1([12]) = [2] - [1]\) \(\del_1([23]) = [3] - [2]\) \(\del_1([13]) = [3] - [1]\) \(\del_1([34]) = [4] - [3]\) \(\del_1([35]) = [5] - [3]\) \(\del_1([45]) = [5] - [4]\)

\(\del_0 = 0\)

And so \(H_0 = \ker \del_0/\im\del_1 = \frac{C_0}{<\del_1([ij])>}\), but from the above calculation we have \([5] = [4] = [3] = [2] = [1]\) in the quotient, so there is just one generator and \(H_0 \cong \ZZ\).

Note that \(\del_2\) is an injection from 0 into \(C_1\), since there are no 2-simplices. Moreover, one can generate two 1-cycles, so we have \(H_1 = \frac{\ker \del_1}{\im \del_2} =\frac{<[23]-[31] + [12],~[45] - [35] + [34]>}{0} \cong \ZZ^2\).

One way to see that these are the generators is to pretend there are two 2-simplices, \([123], [345]\) and compute \(\del_2\) of both of them. Since \(\del_1\del_2 = 0\), anything in the image of \(\del_2\) would have to go to zero anyways, and would thus be in the kernel of \(\del_1\). Since it’s not actually the boundary of any 2-chain, it doesn’t become trivial in homology.

So we have \(H_2 \into H_1 \into H_0 = 0 \into \ZZ^2 \into \ZZ\).