Cellular homology of the one-polygon surface model

By [F1], it is enough to compute on the indicated polygonal model; the homeomorphism transfers the resulting singular homology groups to $X$. In the commutator polygon, let $v_0,\dots ,v_{4g}=v_0$ be the successive corners. For each handle block $a_i{b}_i{a}_i^{-1}{b}_i^{-1}$, its side identifications join the four corners in that block successively: the $a_i$ pair identifies the first with the fourth and the second with the third, while the $b_i$ pair identifies the second with the next block's first and the third with the fourth. Thus every corner lies in one class. There are consequently one $0$-cell, $2g$ paired $1$-cells, and one $2$-cell.

For $g=0$, the sphere digon has two corner classes: its paired sides identify each corner with itself, leaving the two distinct corners $v_0,v_1$. Orient the single paired edge from $v_0$ to $v_1$. The $1$-cell incidence formula gives ${\mathrm{\partial }}_{1}e=v_1-v_0$. The attaching word $a{a}^{-1}$ has total exponent zero, so its $2$-cell has ${\mathrm{\partial }}_2=0$. Therefore $\ker {\mathrm{\partial }}_1=0$ and $\mathrm{coker}{\mathrm{\partial }}_1\cong \mathbb{Z}$.

When $g\ge 1$, every $1$-cell is a loop at the unique vertex, so ${\mathrm{\partial }}_1=0$ by [F3]. The coefficient of each $1$-cell in ${\mathrm{\partial }}_2$ is the degree of the attaching word after the other edges are collapsed; each label occurs once with each exponent, so that degree is $1-1=0$. Hence ${\mathrm{\partial }}_2=0$. The chain groups are $C_2=\mathbb{Z}$, $C_1={\mathbb{Z}}^{2g}$, $C_0=\mathbb{Z}$, and $C_q=0$ for $q\ge 3$.

Taking kernels modulo images in the chain complexes of steps 2.1 and 1.2 gives the stated cellular homology groups in both cases. By [F4], these are the singular homology groups, and the cellular generators in the commutator model are exactly its side-loop classes. Counting cells gives $1-2g+1=2-2g$ for $g\ge 1$ and $2-1+1=2$ for the digon, proving the Euler-characteristic claim by [F5].

To verify the subdivision claim, let the new vertex be $w$ and orient the two replacement edges from the old vertex $v$ to $w$ and from $w$ to $v$. Their boundaries are $w-v$ and $v-w$; all other loop edges still have zero boundary. In the attaching word the subdivided side occurs once in each direction, so each new edge also has total exponent zero and ${\mathrm{\partial }}_2=0$. The kernel of ${\mathrm{\partial }}_1$ is generated by the sum of the two replacement edges and the other $2g-1$ loop edges. This gives the asserted basis, with the original side class represented by that sum.