Let \(Y\) be the quasi-circle, a closed subspace of \(\mathbb{R}^2\) consisting of a portion of the graph of \(y = \sin(1/x)\), the segment \([-1, 1]\) in the \(y\)-axis, and an arc connecting these two pieces. Collapsing the segment of \(Y\) in the \(y\)-axis to a point gives a quotient map \(f: Y \to S^1\). Show that \(f\) does not lift to the covering space \(\mathbb{R} \to S^1\), even though \(\pi_1(Y) = 0\). Thus local path-connectedness of \(Y\) is a necessary hypothesis in the lifting criterion.
Solution
Let \[ L=\{0\}\times[-1,1] \] be the vertical segment and let \(G\) be the oscillating graph portion of \(y=\sin(1/x)\) used in the quasi-circle. Before the extra connecting arc is added, \(L\) and \(G\) are the two path components of the topologist sine curve.
1. The image of every path in \(Y\) meets only a finite truncation of the oscillating graph \(G\) near the \(y\)-axis.
Let \[ \alpha:I\to Y \] be a path. The only way for a path to pass between \(L\) and \(G\) is through the added connecting arc, since \(L\) and \(G\) are distinct path components of the topologist sine curve itself.
Let \(b\in G\) be the endpoint of the added arc on the graph, and write its positive \(x\)-coordinate as \(x_b\). On \(G\), projection to the \(x\)-coordinate is a homeomorphism onto an interval \((0,x_b]\).
Consider the components of the set of parameter values for which \(\alpha\) lies in \(G\setminus\{b\}\). Except possibly for the two components meeting the endpoints of \(I\), each such component is an open interval whose two endpoint values under \(\alpha\) are \(b\): leaving the graph can occur only through the attachment point \(b\).
The path \(\alpha\) is uniformly continuous. Choose \(\delta>0\) such that \[ |s-t|<\delta \quad\Longrightarrow\quad \|\alpha(s)-\alpha(t)\|<x_b/2. \] Any graph excursion that reaches a point with \(x<x_b/2\) has Euclidean distance at least \(x_b/2\) from \(b\) in the horizontal coordinate alone. Hence its parameter interval has length at least \(\delta\). There can be only finitely many pairwise disjoint excursion intervals of length at least \(\delta\).
On each of these finitely many intervals, the continuous positive \(x\)-coordinate attains a positive minimum on the closed interval obtained by adding its endpoints. The same is true for the at most two components meeting \(0\) or \(1\). All remaining excursions stay in \(x\ge x_b/2\). Taking the minimum of these finitely many positive lower bounds and \(x_b/2\) gives some \(\varepsilon>0\) such that every point of \(\alpha(I)\cap G\) has \[ x\ge\varepsilon. \]
2. Every loop in \(Y\) is contained in a contractible subspace of \(Y\).
Let \(\gamma:S^1\to Y\) be a loop. By 1, choose \(\varepsilon>0\) so that the part of its image in \(G\) lies in the compact graph segment \[ G_\varepsilon=G\cap\{x\ge\varepsilon\}. \]
The union of
- the vertical interval \(L\),
- the added connecting arc,
- the finite graph segment from the arc endpoint down to \(G_\varepsilon\),
is a finite tree: these three arc pieces are attached without forming a cycle. It contains the whole image of \(\gamma\) and is contractible. Therefore \(\gamma\) is nullhomotopic in \(Y\).
3. Consequently\[ \boxed{\pi_1(Y)=0.} \]
Every loop is nullhomotopic by 2.
4. Let\[ p:\mathbb R\to S^1, \qquad p(t)=e^{2\pi i t}, \]be the standard covering.Suppose for contradiction that the quotient map\[ f:Y\to S^1 \]has a lift\[ \widetilde f:Y\to\mathbb R \]with\[ p\widetilde f=f. \]
This is the negation of the desired nonlifting statement.
5. The map \(\widetilde f\) is constant on the collapsed segment \(L\).
The quotient map \(f\) sends all of \(L\) to one point \(z_0\in S^1\). Therefore \[ \widetilde f(L)\subseteq p^{-1}(z_0). \] The fiber \(p^{-1}(z_0)\) is a discrete subset of \(\mathbb R\), while \(L\) is connected. The continuous image of a connected space in a discrete space is a single point. Hence \(\widetilde f|_L\) is constant.
6. The lift \(\widetilde f\) factors through the quotient \(f:Y\to S^1\):there is a continuous map\[ s:S^1\to\mathbb R \]such that\[ \widetilde f=s\circ f. \]
The map \(f\) is precisely the quotient obtained by collapsing \(L\) to one point, and it is one-to-one on the remaining quotient classes. By 5, \(\widetilde f\) is constant on every fiber of this quotient map. The universal property of quotient maps therefore gives a unique continuous factor \(s\) with \[ \widetilde f=s f. \]
7. The map \(s\) would be a section of \(p\):\[ p\circ s=\operatorname{id}_{S^1}. \]
Using 4 and 6, \[ f=p\widetilde f=p s f. \] Since \(f:Y\to S^1\) is surjective, equality after composition with \(f\) implies \[ p s=\operatorname{id}_{S^1}. \]
8. No such section \(s\) exists.
On fundamental groups, the identity \[ p s=\operatorname{id}_{S^1} \] would give \[ p_*s_*=\operatorname{id}_{\pi_1(S^1)}. \] But \[ \pi_1(\mathbb R)=0, \] so \[ s_*:\pi_1(S^1)\to\pi_1(\mathbb R) \] is the zero map, making \(p_*s_*\) zero rather than the identity on \[ \pi_1(S^1)\cong\mathbb Z. \] Contradiction.
9. Therefore \(f\) has no lift to \(\mathbb R\to S^1\), despite satisfying\[ f_*(\pi_1(Y))=0. \]
Nonexistence of the lift follows from <1>4–<1>8, while 3 gives the trivial subgroup condition. Thus the local path-connectedness hypothesis in the usual lifting criterion cannot simply be omitted.