Let \(X\) be the quotient space obtained from an 8-sided polygonal region \(P\) by pasting its edges together according to the labelling scheme \(acadbcb^{-1}d\).
Check that all vertices of \(P\) are mapped to the same point of the quotient space \(X\) by the pasting map.
Calculate \(H_1(X)\).
Assuming \(X\) is homeomorphic to one of the surfaces given in Theorem 75.5 (which it is), which surface is it?
Solution
Goal: Analyze the 2-dimensional surface \(X\) given by the polygonal labeling scheme \(w = a c a d b c b^{-1} d\): verify single vertex class, compute first homology \(H_1(X)\), and classify the surface.
1. Part (a): All 8 vertices of \(P\) are identified to a single point.Proof:
1.1. Label the vertices of the 8-gon in counterclockwise order as \(v_0, v_1, v_2, v_3, v_4, v_5, v_6, v_7\), so the oriented boundary edges are:\[a: v_0 \to v_1, \quad c: v_1 \to v_2, \quad a: v_2 \to v_3, \quad d: v_3 \to v_4,\]\[b: v_4 \to v_5, \quad c: v_5 \to v_6, \quad b^{-1}: v_6 \to v_7, \quad d: v_7 \to v_0.\]
1.2. The identifications from identical directed edges yield:- \(e_1(a)\) and \(e_3(a) \implies v_0 \sim v_2\) and \(v_1 \sim v_3\).- \(e_2(c)\) and \(e_6(c) \implies v_1 \sim v_5\) and \(v_2 \sim v_6\).- \(e_4(d)\) and \(e_8(d) \implies v_3 \sim v_7\) and \(v_4 \sim v_0\).- \(e_5(b)\) and \(e_7(b^{-1}) \implies b: v_7 \to v_6\), so \(v_4 \sim v_7\) and \(v_5 \sim v_6\).
1.3. Combining these equivalences:\[v_0 \sim v_2 \sim v_6 \sim v_5 \sim v_1 \sim v_3 \sim v_7 \sim v_4 \sim v_0.\]
1.4. Hence all 8 vertices lie in a single equivalence class, so \(V = 1\).
2. Part (b): Calculation of \(H_1(X)\).\[H_1(X) \cong \mathbb{Z}^3 \oplus \mathbb{Z}/2\mathbb{Z}.\]Proof:
2.1. \(X\) has a standard CW complex structure with one 0-cell \(v\), four 1-cells \(\{a, b, c, d\}\), and one 2-cell \(e^2\) attached along the loop \(w = a c a d b c b^{-1} d\).
2.2. The 1-chains group is \(C_1(X) = \mathbb{Z}a \oplus \mathbb{Z}b \oplus \mathbb{Z}c \oplus \mathbb{Z}d\), and the boundary map \(\partial_1: C_1 \to C_0\) is zero because all 1-cells are closed loops at \(v\).
2.3. The cellular boundary of the 2-cell is:\[\partial_2(e^2) = a + c + a + d + b + c - b + d = 2a + 2c + 2d + 0b = 2(a + c + d).\]
2.4. Thus \(H_1(X) = C_1 / \operatorname{Im}(\partial_2) \cong \langle a, b, c, d \mid 2(a + c + d) = 0 \rangle\).
2.5. Making the change of basis \(u = a + c + d, c' = c, d' = d, b' = b\) for \(\mathbb{Z}^4\) presents the group as:\[\langle u, b', c', d' \mid 2u = 0 \rangle \cong (\mathbb{Z}/2\mathbb{Z}) \oplus \mathbb{Z} \oplus \mathbb{Z} \oplus \mathbb{Z} = \mathbb{Z}^3 \oplus \mathbb{Z}/2\mathbb{Z}.\]
3. Part (c): Topological classification of \(X\).\(X\) is homeomorphic to the connected sum of 4 real projective planes:\[X \cong \#^4 \mathbb{R}P^2 \cong P^2 \# P^2 \# P^2 \# P^2.\]Proof:
3.1. Euler characteristic: \(\chi(X) = V - E + F = 1 - 4 + 1 = -2\).
3.2. Orientability: The edge \(a\) appears twice with the same orientation exponent (\(a\) and \(a\)), so the surface is non-orientable.
3.3. Classification: Any closed, connected non-orientable 2-manifold is classified by its Euler characteristic \(\chi(N_k) = 2 - k\).
3.4. Setting \(2 - k = -2\) gives \(k = 4\).
3.5. Therefore \(X \cong N_4 = \#^4 \mathbb{R}P^2\). Q.E.D.