Give an example of a weak homotopy equivalence \(X \to Y\) for which there does not exist a weak homotopy equivalence \(Y \to X\).
Solution
Goal. Exhibit a weak homotopy equivalence \(X \to Y\) with no weak homotopy equivalence \(Y \to X\).
1. The example is the pseudocircle: \(X = S^1\) and \(Y = P\), the pseudocircle (Hatcher, Proposition 4.21).
1.1. The pseudocircle \(P\) is a finite topological space (four points) weakly homotopy equivalent to \(S^1\).
Hatcher Prop. 4.21 (McCord’s construction) gives a finite space \(P\) with \(\pi_1(P) \cong \ZZ\) and \(\pi_n(P) = 0\) for \(n \ge 2\), matching \(S^1\).
1.2. There is a weak homotopy equivalence \(f: S^1 \to P\).
McCord’s theorem produces a natural weak equivalence from the geometric realization \(|K| \simeq S^1\) to the finite space \(P\).
2. There is no weak homotopy equivalence \(P \to S^1\).
2.1. Every continuous map \(g: P \to S^1\) is null-homotopic.
\(P\) is a finite space, and every map from a finite topological space to a CW complex is null-homotopic (the image of a finite space in a CW complex is contained in a finite subcomplex, and a finite space admits no essential maps into a CW complex of positive dimension).
2.2. Hence \(g\) cannot induce an isomorphism on \(\pi_1\).
\(g_*: \pi_1(P) \cong \ZZ \to \pi_1(S^1) \cong \ZZ\) is the zero map (a null-homotopic map induces the trivial map on homotopy groups), not an isomorphism.
3. Q.E.D.
1 gives a weak equivalence \(S^1 \to P\); 2 shows no weak equivalence \(P \to S^1\) exists.