How to calculate the integral homology of a cell complex?

Question

Consider the following 2-dimensional hexagon, with its edges oriented as indicated by the arrows.

Let $X$ be the cell complex obtained by making the following three identifications on the edges of the hexagon: $a_1 = a_2, b_1 = b_2, c_1 = c_2$.

I am new in algebraic topology, so I am having trouble in understanding the construction of cell complexes. I have to find integral homology of $X$. Any help regarding this will be appreciated

Answer 1

Let $X$ be your space, and let $B$ be a ball in the interior of your hexagon, and hence in your space. Choose $b\in B$, and the pair of open sets $(B,X\smallsetminus b)$ that intersect at $B\smallsetminus b$. Note that $X\smallsetminus b$ deformation retracts onto three circles, one for each pair of edges in your hexagon. Both the open sets and their intersection have trivial $H_i$ for $i>1$, and the Mayer-Vietoris gives you that $H_i(X) = 0$ for $i>2$ and an exact sequence

$$0\to H_2(X) \to H_1(B\smallsetminus b)\to H_1(X\smallsetminus b)\to H_1(X) \to \widetilde{H}_0(B\smallsetminus b)=0$$

The image of the map $H_1(X\smallsetminus b)\to H_1(X)$ is zero since the generators of $H_1(X\smallsetminus b)$ can be taken to be a loop traversing an edge once and going around $b$, which is nullhomologous in $X$. This means that we have a short exact sequence

$$0\to H_2(X) \to H_1(B\smallsetminus b)\to H_1(X\smallsetminus b)\to 0$$

The last terms is $\mathbb Z^3$, and you can check that the arrow $\mathbb Z\to\mathbb Z^3$ is given by $(2,2,2)^t$ under the identification above. This means that $H_2(X) = 0$, since this map is injective.

Since the kernel of $H_1(X) \to 0$ is the cokernel of $(2,2,2)^t$, we get that $H_1(X)= \mathbb Z/2 \oplus \mathbb Z\oplus\mathbb Z$, and of course $X$ is connected. To see this last claim, you may compute the Smith normal form of the corresponding matrix, or note that there is an automorphism of $\mathbb Z^3$ that sends $e_1\to e_1+e_2+e_3$, $e_2\to e_2$ and $e_3\to e_3$, and the subgroup generated by $(2,0,0)$ to that generated by $(2,2,2)$.

Answer 2

Compute the cellular homology, there exists one two cell $t$, three 1-cels $a,b,c$ and 1 $0$ cells $x$.

$d_2:H_2(T_2,T_1)\rightarrow H_1(T_1,T_0)$ is defined by $d_2(t)=2a+2b+2c$,

$d_1:H_1(T_1,T_0)\rightarrow H_1(T_0,T_{-1})$ $d_1(a)=0, d_1(b)=0, d_2(c)=0$ since the vertices at the end of the 1-cell are identified.

so $H_2(T)=0$, $H_1(T)$ is the quotient of $\mathbb{Z}\oplus\mathbb{Z}\oplus\mathbb{Z}$ by the subgroup generated by $(2,2,2))$ and $H_0(T)$ is $\mathbb{Z}$.