Covering maps

hbghlyj

Let $X$ be a path connected cell complex and suppose that there is a map $X\to S^1$ that is not null-homotopic. Let $x$ be a zero-cell of $X$. Prove that $X\vee_x X$ has an irregular cover of degree 3. Does $X\vee_x X$ admit a retract onto a wedge of 2 circles?

hbghlyj

Let \(G = \pi_1(X, x)\). The assumption implies that there exists a surjective homomorphism \(\phi: G \to \mathbb{Z}\). Then \(\pi_1(X \vee_x X, x) = G * G\). Define homomorphisms \(\psi_L: G \to S_3\) and \(\psi_R: G \to S_3\) by composing \(\phi\) with the quotient maps \(\mathbb{Z} \to \mathbb{Z}/3\mathbb{Z} \cong \langle (123) \rangle \subset S_3\) and \(\mathbb{Z} \to \mathbb{Z}/2\mathbb{Z} \cong \langle (12) \rangle \subset S_3\), respectively. These induce a surjective homomorphism \(\rho: G * G \to S_3\). Let \(N = \langle (23) \rangle \subset S_3\), which is not normal in \(S_3\). The preimage \(H = \rho^{-1}(N)\) has index 3 in \(G * G\) and is not normal, corresponding to an irregular covering of degree 3.

---

\(X \vee_x X\) does admit a retract onto a wedge of two circles in the sense that there exist continuous maps \(i: S^1 \vee S^1 \to X \vee_x X\) and \(r: X \vee_x X \to S^1 \vee S^1\) such that \(r \circ i\) is homotopic to the identity on \(S^1 \vee S^1\).

To see why, note that the assumption gives a surjective homomorphism \(\phi: \pi_1(X, x) \to \mathbb{Z}\). Let \(G = \pi_1(X, x)\) so \(\pi_1(X \vee_x X, x) = G * G\). Extend to a surjective homomorphism \(\rho: G * G \to \mathbb{Z} * \mathbb{Z} \cong F_2\) by sending the left copy of \(G\) to the left \(\mathbb{Z}\) via \(\phi\) and the right copy to the right \(\mathbb{Z}\) via \(\phi\). Since \(\phi\) is surjective on each, so is \(\rho\).

Now construct a section \(\sigma: F_2 \to G * G\) for \(\rho\). Let the generators of \(F_2\) be \(l\) and \(r\) (for the left and right \(\mathbb{Z}\)). Choose \(g \in G\) with \(\phi(g) = 1\) (the generator of \(\mathbb{Z}\)), which exists since \(\phi\) is surjective. Set \(\sigma(l)\) to be the copy of \(g\) in the left \(G\) and \(\sigma(r)\) to be the copy of \(g\) in the right \(G\). This defines a homomorphism (as \(F_2\) is free), and \(\rho \circ \sigma = \mathrm{id}_{F_2}\) by construction.

Since \(S^1 \vee S^1\) is a \(K(F_2, 1)\), homotopy classes of maps to and from it are determined by the induced maps on fundamental groups. Thus, there exist continuous maps \(i\) and \(r\) inducing \(\sigma\) and \(\rho\), respectively, on \(\pi_1\). The composition \(r \circ i\) induces \(\mathrm{id}_{F_2}\) on \(\pi_1\), so \(r \circ i \simeq \mathrm{id}\) as maps from \(S^1 \vee S^1\) to itself.