Let \(p: E \to X\) be the map constructed in the proof of Lemma 60.5. Let \(E'\) be the subspace of \(E\) that is the union of the \(x\)-axis and the \(y\)-axis. Show that \(p \mid E'\) is not a covering map.
Solution
Goal: Prove that the restriction \(q = p|_{E'}: E' \to X = A \vee B\) of the grid covering map to the union of the coordinate axes \(E' = (\mathbb{R} \times \{0\}) \cup (\{0\} \times \mathbb{R})\) is not a covering map.
1. Setting and structure of the map \(p\):Proof:
1.1. The total space \(E = (\mathbb{R} \times \mathbb{Z}) \cup (\mathbb{Z} \times \mathbb{R}) \subset \mathbb{R}^2\) is the infinite grid in the plane.
1.2. The base space \(X = A \vee B\) is the wedge of two circles \(A\) and \(B\) intersecting at a basepoint \(x_0\).
1.3. The covering map \(p: E \to X\) wraps each horizontal grid line \(\mathbb{R} \times \{n\}\) around circle \(A\) and each vertical grid line \(\{m\} \times \mathbb{R}\) around circle \(B\), sending all integer lattice points \(\mathbb{Z} \times \mathbb{Z}\) to \(x_0\).
1.4. The subspace \(E' = (\mathbb{R} \times \{0\}) \cup (\{0\} \times \mathbb{R})\) is the cross formed by the \(x\)-axis and \(y\)-axis.
2. Preimage of the basepoint in \(E'\):Proof:
2.1. The preimage \(q^{-1}(x_0) = E' \cap (\mathbb{Z} \times \mathbb{Z}) = (\mathbb{Z} \times \{0\}) \cup (\{0\} \times \mathbb{Z})\).
2.2. At the origin \((0, 0) \in E'\), both the horizontal and vertical axes meet, forming a 4-branch cross.
2.3. For any non-zero point \((n, 0) \in E'\) (\(n \in \mathbb{Z} \setminus \{0\}\)), \(E'\) consists purely of the horizontal segment \((n - \varepsilon, n + \varepsilon) \times \{0\}\), which has 2 branches (homeomorphic to an open interval).
3. Failure of local homeomorphy and even covering:Proof:
3.1. Let \(n \in \mathbb{Z} \setminus \{0\}\) and consider \(e = (n, 0) \in q^{-1}(x_0)\).
3.2. Any sufficiently small open neighborhood \(V\) of \(e\) in \(E'\) is a 1-dimensional open interval \((n - \varepsilon, n + \varepsilon) \times \{0\}\) along the \(x\)-axis.
3.3. Under \(q\), the image \(q(V)\) lies entirely within the circle \(A\), so \(q(V) \cap (B \setminus \{x_0\}) = \varnothing\).
3.4. However, every open neighborhood \(U\) of \(x_0\) in the wedge space \(X = A \vee B\) contains points of both \(A \setminus \{x_0\}\) and \(B \setminus \{x_0\}\).
3.5. Therefore, \(q(V)\) cannot contain any open neighborhood of \(x_0\) in \(X\), so \(q\) is not an open map and fails to be a local homeomorphism at \((n, 0)\).
3.6. Consequently, no open neighborhood of \(x_0\) in \(X\) is evenly covered by \(q\).
4. Conclusion:\(q = p|_{E'}\) is not a covering map. Q.E.D.