Let \(X\) be the figure eight and let \(Y\) be the theta space. Describe maps \(f: X \to Y\) and \(g: Y \to X\) that are homotopy inverse to each other.
Solution
Goal: Explicitly construct continuous maps \(f: X \to Y\) and \(g: Y \to X\) between the figure-eight space \(X = S^1 \vee S^1\) and the theta space \(Y = S^1 \cup ([-1, 1] \times \{0\})\), and prove they are homotopy inverses.
1. Geometric models for \(X\) and \(Y\):1. Figure eight \(X = C_1 \cup C_2 \subset \mathbb{R}^2\), where \(C_1\) and \(C_2\) are circles intersecting at the single basepoint \(p = (0, 0)\).2. Theta space \(Y = U \cup L \cup I_0 \subset \mathbb{R}^2\), where:- \(U = \{(x, y) \in S^1 : y \ge 0\}\) is the upper semicircle from \(a = (-1, 0)\) to \(b = (1, 0)\),- \(L = \{(x, y) \in S^1 : y \le 0\}\) is the lower semicircle from \(a\) to \(b\),- \(I_0 = \{(x, 0) \in \mathbb{R}^2 : -1 \le x \le 1\}\) is the horizontal diameter from \(a\) to \(b\).
2. Construction of \(g: Y \to X\) (collapsing the central segment):Define \(g: Y \to X\) by the quotient map that collapses the contractible segment \(I_0\) to the basepoint \(p\):- For \(z \in I_0\): \(g(z) = p\).- For \(z \in U\): \(g(z)\) maps \(U\) homeomorphically onto \(C_1\) with \(g(a) = g(b) = p\).- For \(z \in L\): \(g(z)\) maps \(L\) homeomorphically onto \(C_2\) with \(g(a) = g(b) = p\).By the Pasting Lemma, \(g\) is continuous.
3. Construction of \(f: X \to Y\) (attaching loops):Define \(f: X \to Y\) by sending the basepoint \(p \in X\) to \(a \in Y\):- On \(C_1\): \(f\) traces the loop formed by following \(U\) from \(a\) to \(b\), then returning along \(I_0\) from \(b\) to \(a\).- On \(C_2\): \(f\) traces the loop formed by following \(L\) from \(a\) to \(b\), then returning along \(I_0\) from \(b\) to \(a\).By the Pasting Lemma on \(X = C_1 \cup C_2\), \(f\) is continuous with \(f(p) = a\).
4. Homotopy \(g \circ f \simeq \operatorname{id}_X\):Proof:
4.1. The composition \((g \circ f)|_{C_1}\) traverses \(g(U) = C_1\) followed by the constant path \(g(I_0) = c_p\).
4.2. Since concatenation with the constant path is homotopic to the original loop (\(C_1 \ast c_p \simeq C_1\)), \((g \circ f)|_{C_1} \simeq \operatorname{id}_{C_1}\).
4.3. Similarly, \((g \circ f)|_{C_2}\) traverses \(g(L) \ast g(I_0) = C_2 \ast c_p \simeq C_2\), so \((g \circ f)|_{C_2} \simeq \operatorname{id}_{C_2}\).
4.4. Combining these homotopies fixes the basepoint \(p\), yielding \(g \circ f \simeq \operatorname{id}_X\).
5. Homotopy \(f \circ g \simeq \operatorname{id}_Y\):Proof:
5.1. The segment \(I_0\) is a deformation retract of \(Y\) onto the tree segment \(I_0\) connecting \(a\) and \(b\), which itself deformation retracts to \(a\).
5.2. Explicitly, define \(H: Y \times [0, 1] \to Y\) by linearly sliding the segment \(I_0\) onto the point \(a\):\[H((x, 0), t) = ((1-t)x - t, 0) \quad \text{for } (x, 0) \in I_0,\]while keeping points on \(U\) and \(L\) stationary up to reparametrization.
5.3. At \(t = 0\), \(H_0 = \operatorname{id}_Y\).
5.4. At \(t = 1\), \(H_1 = f \circ g\).
5.5. Thus \(f \circ g \simeq \operatorname{id}_Y\).
6. Conclusion:\(f\) and \(g\) are homotopy inverse to each other, establishing that the figure eight and the theta space are homotopy equivalent: \(X \simeq Y\). Q.E.D.