Let \(X\) be a hexagon in \(\mathbb R^2\). Define an equivalence relation on \(X\) corresponding to labeling the 6 edges in the boundary of \(X\) in a counterclockwise fashion in order by: counterclockwise \(a\), counterclockwise \(b\), counterclockwise \(a\), counterclockwise \(b\), counterclockwise \(a\), counterclockwise \(b\). Let \(M\) be the corresponding quotient space. Compute \(H_n(M)\) for all \(n\ge 0\).
Solution
Label the vertices of the hexagon counterclockwise by \(v_0,\ldots,v_5\), so the boundary edges are \[ a:v_0\to v_1, \quad b:v_1\to v_2, \quad a:v_2\to v_3, \quad b:v_3\to v_4, \quad a:v_4\to v_5, \quad b:v_5\to v_0. \]
1. The quotient \(M\) has two \(0\)-cells\[ u=[v_0]=[v_2]=[v_4], \qquad v=[v_1]=[v_3]=[v_5], \]two \(1\)-cells \(a,b\), and one \(2\)-cell \(e^2\).The edge \(a\) is oriented from \(u\) to \(v\), while \(b\) is oriented from \(v\) to \(u\).
Identifying all edges carrying the same label and orientation identifies their initial vertices with one another and their terminal vertices with one another. For the three \(a\)-edges this identifies \(v_0,v_2,v_4\) and separately \(v_1,v_3,v_5\). The \(b\)-edge identifications give the same two classes, with the opposite direction. The interior of the hexagon is unaffected and gives one open \(2\)-cell.
2. The cellular chain groups are\[ C_2(M)=\mathbb Z\langle e^2\rangle, \qquad C_1(M)=\mathbb Z\langle a,b\rangle, \qquad C_0(M)=\mathbb Z\langle u,v\rangle, \]and\[ \partial_1(a)=v-u, \qquad \partial_1(b)=u-v. \]Consequently\[ \ker\partial_1=\mathbb Z\langle a+b\rangle. \]
The formulas for \(\partial_1\) are terminal vertex minus initial vertex. For a chain \(ra+sb\), \[ \partial_1(ra+sb)=(r-s)(v-u), \] which vanishes exactly when \(r=s\). Thus the kernel is generated by \(a+b\).
3. The cellular boundary of the \(2\)-cell is\[ \partial_2(e^2)=3a+3b=3(a+b). \]
The attaching loop of the hexagonal \(2\)-cell traverses the boundary word \[ ababab \] with every occurrence oriented in the positive direction of the corresponding \(1\)-cell. Hence the degree of the attaching map on each of the two \(1\)-cells is \(3\). The cellular boundary formula therefore gives \[ \partial_2(e^2)=3a+3b. \]
4. The zeroth and first homology groups are\[ H_0(M)\cong\mathbb Z, \qquad H_1(M)\cong\mathbb Z/3\mathbb Z. \]
The \(1\)-skeleton contains both vertices joined by edges, so \(M\) is connected and hence \(H_0(M)\cong\mathbb Z\). Also, by <1>2–<1>3, \[ H_1(M) =\ker\partial_1/\operatorname{im}\partial_2 =\mathbb Z\langle a+b\rangle/3\mathbb Z\langle a+b\rangle \cong\mathbb Z/3\mathbb Z. \]
5. We have\[ H_n(M)=0 \qquad\text{for every }n\ge2. \]
Since \(C_2(M)\cong\mathbb Z\) and \[ \partial_2(e^2)=3(a+b)\ne0, \] the map \(\partial_2\) is injective, so \[ H_2(M)=\ker\partial_2=0. \] There are no cells in dimensions greater than \(2\), hence \(H_n(M)=0\) for \(n\ge3\).
6. Therefore\[ H_n(M)\cong \begin{cases} \mathbb Z, & n=0,\\ \mathbb Z/3\mathbb Z, & n=1,\\ 0, & n\ge2. \end{cases} \]
Combine 4 and 5.