Let \(X\) be the subspace of \(\mathbb{R}^2\) consisting of the four sides of the square \([0,1] \times [0,1]\) together with the segments of the vertical lines \(x = 1/2, 1/3, 1/4, \ldots\) inside the square. Show that for every covering space \(\tilde{X} \to X\) there is some neighborhood of the left edge of \(X\) that lifts homeomorphically to \(\tilde{X}\). Deduce that \(X\) has no simply-connected covering space.
Solution
Let \[ L=\{0\}\times[0,1] \] be the left edge.
1. We first prove a compact local-injectivity lemma.Let \(f:E\to M\) be continuous, where \(M\) is metric.Suppose \(K\subseteq E\) is compact, \(f|_K\) is injective, and every \(k\in K\) has a neighborhood on which \(f\) is injective.Then there is an open neighborhood \(N\) of \(K\) on which \(f\) is injective.
Let \(d\) be a metric on \(M\) and define \[ g:E\times E\to\mathbb R, \qquad g(e,e')=d(f(e),f(e')). \] Put \[ Z=g^{-1}(0)=\{(e,e'):f(e)=f(e')\}. \]
For each \(k\in K\), choose an open neighborhood \(V_k\) on which \(f\) is injective. Then \[ (V_k\times V_k)\cap Z \subseteq \Delta_E. \] Let \[ V=\bigcup_{k\in K}(V_k\times V_k), \qquad O=(E\times E\setminus Z)\cup V. \] The set \(O\) is open.
We claim \[ K\times K\subseteq O. \] Indeed, if \(k\ne k'\) lie in \(K\), injectivity of \(f|_K\) gives \[ f(k)\ne f(k'), \] so \((k,k')\notin Z\). If \(k=k'\), then \((k,k)\in V_k\times V_k\subseteq V\).
Since \(K\times K\) is compact and lies in the open set \(O\), the product-neighborhood lemma for compact sets gives open neighborhoods \(W,W'\) of \(K\) such that \[ W\times W'\subseteq O. \] For completeness, this product-neighborhood lemma follows by covering each slice \(\{k\}\times K\) by finitely many product neighborhoods inside \(O\), intersecting their first factors to obtain a product neighborhood of \(\{k\}\times K\), then using compactness of the first copy of \(K\) and intersecting finitely many corresponding second factors.
Set \[ N=W\cap W'. \] If \(e,e'\in N\) and \(f(e)=f(e')\), then \[ (e,e')\in (N\times N)\cap Z\subseteq O\cap Z\subseteq V. \] Hence \(e,e'\) lie together in some \(V_k\), where \(f\) is injective, so \(e=e'\). Thus \(f|_N\) is injective.
2. Let\[ p:\widetilde X\to X \]be any covering space.Choose a point \(\widetilde \ell_0\) above the lower-left corner of \(L\).The inclusion path of \(L\) has a unique lift beginning at \(\widetilde \ell_0\); denote its image by \(\widetilde L\).Then\[ p|_{\widetilde L}:\widetilde L\to L \]is a homeomorphism.
The lifted path is injective. If two distinct parameters had the same lifted point, their images under \(p\) would be the same point of the injectively parametrized left edge \(L\). Thus the lifted path is a continuous bijection from the compact interval \(I\) onto \(\widetilde L\). Since \(\widetilde X\) is Hausdorff locally over the metric space \(X\) along each covering sheet, this map is a homeomorphism onto its image. Equivalently, its inverse is simply the restriction of \(p\) followed by the inverse parametrization of \(L\).
3. There is an open neighborhood \(N\) of \(\widetilde L\) such that\[ p|_N:N\longrightarrow p(N) \]is a homeomorphism, and \(p(N)\) is an open neighborhood of \(L\) in \(X\).
2.A covering map is locally a homeomorphism, hence locally injective near every point of \(\widetilde L\).Also \(p\) is injective on \(\widetilde L\) by
2.
Apply 1 with \[ E=\widetilde X, \qquad M=X\subseteq\mathbb R^2, \qquad K=\widetilde L. \] We obtain an open neighborhood \(N\) of \(\widetilde L\) on which \(p\) is injective.
Every covering map is open: on each covering sheet it is a homeomorphism onto an open subset of the base, and an arbitrary open set is the union of its intersections with such sheets. Therefore \(p(N)\) is open. The continuous bijection \[ p|_N:N\to p(N) \] is open, hence is a homeomorphism. Since \(p(\widetilde L)=L\), the set \(p(N)\) contains \(L\).
4. Every open neighborhood \(U\) of \(L\) in \(X\) contains, for all sufficiently large \(n\), the entire rectangular loop\[ R_n = \partial\bigl([0,1/n]\times[0,1]\bigr). \]
The complement \[ F=X\setminus U \] is closed in the compact metric space \(X\), hence compact. It is disjoint from the compact set \(L\). Therefore their Euclidean distance is positive: \[ \delta=d(L,F)>0. \] Choose \(n\) with \[ 1/n<\delta. \] Every point of \(R_n\) has Euclidean distance at most \(1/n\) from \(L\), so no point of \(R_n\) lies in \(F\). Thus \[ R_n\subseteq U. \]
5. The loop \(R_n\) is nontrivial in \(\pi_1(X)\).
Define \[ \varphi_n:X\to S^1 \] to be constant equal to \(1\) on all of \(X\) except the vertical segment \[ V_n=\{1/n\}\times[0,1], \] and on that segment set \[ \varphi_n(1/n,t)=e^{2\pi i t}. \] This is continuous. The segment \(V_n\) is isolated from all the other interior vertical segments in the horizontal direction, and at its two endpoints the displayed formula equals \(1\), agreeing with the constant definition on the top and bottom edges.
Orient \(R_n\) so that it traverses \(V_n\) from bottom to top. All other portions of \(R_n\) map constantly to \(1\), while \(V_n\) maps once around \(S^1\). Hence \[ (\varphi_n)_*([R_n])=1\in\pi_1(S^1)\cong\mathbb Z. \] Therefore \([R_n]\ne1\) in \(\pi_1(X)\).
6. The space \(X\) has no simply connected covering space.
Suppose \(p:\widetilde X\to X\) were a simply connected covering. By 3, choose an open neighborhood \(U=p(N)\) of \(L\) that lifts homeomorphically to \(N\). By 4, choose \(n\) with \[ R_n\subseteq U. \]
The loop \(R_n\) therefore has the closed lift \[ \widetilde R_n=(p|_N)^{-1}\circ R_n \] in \(\widetilde X\). Since \(\widetilde X\) is simply connected, \(\widetilde R_n\) is nullhomotopic. Composing a nullhomotopy with \(p\) shows that \(R_n\) is nullhomotopic in \(X\). This contradicts 5.
Hence no simply connected covering of \(X\) exists.