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I'm trying to solve exercise 1.3.7 in Hatcher's Algebraic Topology:

Let $Y$ be the quasi-circle that is the union of a portion of the graph $y = \sin(1/x)$, the line segment $[-1,1]$ in the $y$-axis, and an arc connecting the two pieces. Collapsing the segment 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 $p:\mathbb{R} \to S^1$, even though $\pi_1(Y) = 0$.
Topologist's sine curve with loop

I was able to show that $\pi_1(Y)$ is simply-connected by decomposing it into 2 pieces and showing that a path in $Y$ must be contained in a subspace that is homeomorphic to $\mathbb{R}$. What I'm having difficulty with is showing that $f$ doesn't have a lift. I've shown that if it has a lift $\tilde f$, $\tilde{f}([-1,1])$ is one point. I feel that this contradicts with the continuity of $\tilde f$, but I cannot rigorously prove it.

Any help would be appreciated. This is self-study by the way.

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2 Answers 2

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I'll use the convention that $S^1 = \{z\in\mathbb C : \lvert z \rvert = 1\}$ and $p(t)=e^{2\pi i t}$, to avoid the clash of notations between ordered pairs in $\mathbb R^2$ and open intervals in $\mathbb R$.

Let $L$ be the segment on the $y$-axis. WLOG assume that $f(L)=\{1\}$. Suppose $\tilde f\colon Y\rightarrow R$ is a lift. Since $Y\setminus L$ is connected, $\tilde f(Y\setminus L)$ is also connected, so it must lie in one of the components of $p^{-1}(f(Y\setminus L))=\mathbb R \setminus 2\pi \mathbb Z$, say the interval $(0,2\pi)$. It follows by the surjectivity of $f$ that $\tilde f(Y\setminus L)$ must be precisely $(0,2\pi)$. Now $Y$ is compact, so $\tilde f(Y) \supset [0,2\pi]$. Thus, $\{0,2\pi\}\subset \tilde f(L)$, contradicting your observation that $\tilde f(L)$ is one point.

Alternatively for the last step, observe that $\tilde f(L)$ should be connected, but $\tilde f(L) \subset 2\pi\mathbb Z$ is a discrete set containing at least two points.

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  • $\begingroup$ I'm not convinced by your answer - what will go wrong with your proof if $Y=S^1$? $\endgroup$
    – Shaked
    Jul 14, 2017 at 20:54
  • $\begingroup$ Nothing I would think. Thus $Y$ from the problem is circle-like in a certain sense, even though $\pi_1(Y) = 0$! $\endgroup$
    – epimorphic
    Dec 28, 2017 at 16:35
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    $\begingroup$ The point of the exercise, per Hatcher, is to illustrate the fact that lifts can fail to exist if the hypotheses of the lifting criterion are violated. The identity map on $S^1$ fails the lifting criterion because $\operatorname{id}_{*}(\pi_1(S^1)) \not\subset p_*(\pi_1(\mathbb R))$. The quotient map $Y \to S^1$ from the problem does satisfy that part, but fails the stipulation that $Y$ should be locally path-connected. $\endgroup$
    – epimorphic
    Dec 28, 2017 at 17:21
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Consider the injective path $g:[0,\infty)\to Y$ starting at $(0,0)$, running through the arc connecting $(0,0)$ and $(1/\pi,0)$, and sending $x$ to $(1/x,\sin x)$ for $x\ge\pi$. Assume that $\tilde f:Y\to R$ is a lift for $f:Y\twoheadrightarrow S^1$. As $f$ is injective except for the vertical segment, and $g$ avoids this segment, $\tilde fg$ must be injective, and thus strictly increasing. Denote the point $\tilde fg(0)=\tilde f(0,0)$ by $y_0$. Consider the points $x_n=\left(\frac1{n\pi},0\right)$. We have $x_n=g(n\pi)$, so the sequence $y_n=\tilde f(x_n)$ increases. But $(x_n)_n\to (0,0)$, so $y_n\to y_0$, which is impossible as $y_0=\tilde fg(0)$ is less than $y_1=\tilde fg(\pi)$.

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  • $\begingroup$ (+1) Thank you. Your answer works too but I had to accept dunce cap's answer since he/she continued what I started. $\endgroup$
    – PeterM
    Oct 22, 2013 at 22:04
  • $\begingroup$ What's the definition of g from 0 to $\pi$? $\endgroup$
    – 6666
    Feb 21, 2016 at 0:06
  • $\begingroup$ @6666: $g$ restricts to a homeomorphism from $[0,\pi]$ to the arc connecting $(0,0)$ and $(1/\pi,0)$. I've edited my answer. $\endgroup$ Feb 21, 2016 at 17:56
  • $\begingroup$ Thx! That makes sense $\endgroup$
    – 6666
    Feb 21, 2016 at 18:03

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