Let \(X\) be the quotient space obtained from an 8-sided polygonal region \(P\) by means of the labelling scheme \(abcdad^{-1}cb^{-1}\). Let \(\pi: P \to X\) be the quotient map.
Show that \(\pi\) does not map all the vertices of \(P\) to the same point of \(X\).
Determine the space \(A = \pi(\operatorname{Bd} P)\) and calculate its fundamental group.
Calculate \(\pi_1(X, x_0)\) and \(H_1(X)\).
Assuming \(X\) is homeomorphic to one of the surfaces given in Theorem 75.5, which surface is it?
Solution
Goal: Analyze the quotient surface \(X\) determined by the boundary word \(w = abcdad^{-1}cb^{-1}\) on an octagon \(P\).
1. Part (a): Vertex equivalence classes under \(\pi\).Proof:
1.1. Label the vertices of the octagon \(P\) in cyclic order as \(v_0, v_1, \dots, v_7\), with directed edges \(v_0 \xrightarrow{a} v_1\), \(v_1 \xrightarrow{b} v_2\), \(v_2 \xrightarrow{c} v_3\), \(v_3 \xrightarrow{d} v_4\), \(v_4 \xrightarrow{a} v_5\), \(v_6 \xrightarrow{d} v_5\), \(v_6 \xrightarrow{c} v_7\), \(v_0 \xrightarrow{b} v_7\).
1.2. The identifications induced by the edge labels are:- Label \(a\): \(v_0 \sim v_4\) and \(v_1 \sim v_5\).- Label \(b\): \(v_1 \sim v_0\) and \(v_2 \sim v_7\).- Label \(c\): \(v_2 \sim v_6\) and \(v_3 \sim v_7\).- Label \(d\): \(v_3 \sim v_6\) and \(v_4 \sim v_5\).
1.3. Tracing equivalence classes yields:- Class 1: \(\{v_0, v_1, v_4, v_5\}\) (since \(v_0 \sim v_1 \sim v_5 \sim v_4\)).- Class 2: \(\{v_2, v_3, v_6, v_7\}\) (since \(v_2 \sim v_7 \sim v_3 \sim v_6\)).
1.4. There are exactly 2 equivalence classes. Thus \(\pi\) maps the vertices of \(P\) to two distinct points \(P_1, P_2 \in X\).
2. Part (b): Topology of the 1-skeleton \(A = \pi(\operatorname{Bd} P)\) and \(\pi_1(A)\).Proof:
2.1. The 1-complex \(A\) has 2 vertices \(P_1, P_2\) and 4 directed edges \(a, b, c, d\).
2.2. \(a\) is a self-loop at \(P_1\); \(c\) is a self-loop at \(P_2\); \(b\) is an edge from \(P_1\) to \(P_2\); \(d\) is an edge from \(P_2\) to \(P_1\).
2.3. Choosing the spanning tree \(T = \{b\}\) and collapsing it to a point yields a wedge sum of \(4 - 1 = 3\) circles: \(A \simeq S^1 \vee S^1 \vee S^1\).
2.4. Therefore \(\pi_1(A, P_1) \cong F_3\), the free group of rank 3, with basis \(\alpha = a\), \(\beta = bcb^{-1}\), and \(\gamma = bd\).
3. Part (c): Computation of \(\pi_1(X)\) and \(H_1(X)\).Proof:
3.1. \(X\) is obtained from \(A\) by attaching a 2-cell along the boundary word \(w = abcdad^{-1}cb^{-1}\).
3.2. Rewriting \(w\) in terms of the generators of \(\pi_1(A, P_1)\) based at \(P_1\):\[w = a \cdot (bcb^{-1}) \cdot (bd) \cdot a \cdot (d^{-1}b^{-1}) \cdot (bcb^{-1}) = \alpha \beta \gamma \alpha \gamma^{-1} \beta.\]
3.3. By Seifert-van Kampen theorem,\[\pi_1(X, P_1) \cong \langle \alpha, \beta, \gamma \mid \alpha \beta \gamma \alpha \gamma^{-1} \beta = 1 \rangle.\]
3.4. Abelianizing gives \(H_1(X) \cong \pi_1(X)_{\text{ab}}\): in additive notation,\[\alpha + \beta + \gamma + \alpha - \gamma + \beta = 2\alpha + 2\beta = 2(\alpha + \beta) = 0.\]
3.5. Setting \(u = \alpha + \beta\), the abelian group presentation is \(\langle u, \alpha, \gamma \mid 2u = 0 \rangle \cong \mathbb{Z}_2 \oplus \mathbb{Z} \oplus \mathbb{Z}\).
3.6. Thus \(H_1(X) \cong \mathbb{Z}^2 \oplus \mathbb{Z}_2\).
4. Part (d): Surface classification.Proof:
4.1. \(X\) is a compact, connected 2-manifold without boundary.
4.2. The Euler characteristic is \(\chi(X) = V - E + F = 2 - 4 + 1 = -1\).
4.3. Since \(H_1(X)\) contains non-trivial 2-torsion \(\mathbb{Z}_2\), \(X\) is non-orientable.
4.4. The non-orientable surface of genus \(k\), \(N_k = \#^k \mathbb{R}P^2\), has Euler characteristic \(\chi(N_k) = 2 - k\).
4.5. Setting \(2 - k = -1\) gives \(k = 3\).
4.6. Therefore, \(X\) is homeomorphic to the connected sum of three projective planes:\[X \cong \mathbb{R}P^2 \# \mathbb{R}P^2 \# \mathbb{R}P^2 \quad (\text{or } K \# \mathbb{R}P^2).\]Q.E.D.