A map that has no lift

Question

I am trying to understand an example given in Chapter 8 of Lee's Introduction to Topological Manifold.

If $B$ is a topological space and $\varphi:B\rightarrow\mathbb{S}^1$ is a continuous map, we define a lift of $\varphi$ to be a continuous map $\tilde\varphi:B\rightarrow\mathbb{R}$ such that $\varphi=\varepsilon\tilde\varphi$, where $\varepsilon:\mathbb{R}\rightarrow\mathbb{S}^1$ is given by $\varepsilon(r)=e^{2\pi ir}$.

The example claims that there is no lift of the identity map on $\mathbb{S}^1$. The reasoning is that if a lift $\sigma:\mathbb{S}^1\rightarrow \mathbb{R}$ did exist, then $\varepsilon\circ\sigma=Id$ would mean that $2\pi\sigma$ is a continuous choice of angle function on the circle, and that "any choice of angle function would have to change by $2\pi$ as one goes around the whole circle, and thus cannot be continuous on the whole circle".

I do not understand this reasoning. Could someone please explain what is going on in more detail?

Specifically, why would the continuous choice of angle function changing by $2\pi$ imply that it cannot be continuous on the whole circle? It might be that I'm just not connecting the dots of something obvious.

Answer 1

Ok, so you agree that in this case $2\pi\sigma: S^1 \to \mathbb{R}$ would be an angle function, sending a point on the unit circle to its argument as a complex number. The trouble is that the argument is only defined up to multiples of $2\pi$. To see this trouble, suppose we start with $2\pi\sigma(1)=0$. Then I move the input around the circle, and the output has to change continuously. So I get $2\pi\sigma(i)=\pi/2$, then $2\pi\sigma(-1) = \pi$, then $2\pi\sigma(-i) = 3\pi/2$, and finally when I come full circle I arrive at $2\pi\sigma(1) = 2\pi$. But that contradicts the value I started with, $2\pi\sigma(1)=0$. Because the output had to track the input continuously, going around the circle had to cause the angle to increase by $2\pi$. There's no way for this function to be self-consistent and continuous.

Answer 2

Let's suppose for each $z\in S^1$ we have $z=\epsilon\circ\sigma(z)=e^{2\pi i\sigma(z)}$. It means $\sigma(z)$ is actually an argument of the complex number $z$. The reasoning is that such a function can't be continuous on all $S^1$. Why? Well, first of all note that each complex number has infinitely many arguments and the difference between each two is a multiple of $2\pi$. Now, if you want the argument function to be continuous then you must choose a branch: for example for all $z\in S^1$ you choose $\sigma(z)$ to be the argument which is between $0$ and $2\pi$. Another option: for each $z\in S^1$ you choose $\sigma(z)$ to be the argument which is between $-\pi$ and $\pi$. But I say it will still not be continuous everywhere.

Let's assume you chose $\sigma(z)$ to be the argument in $(-\pi,\pi]$. What happens near the point $z=-1$ in that case? On the points on the circle which are below $-1$ the value of $\sigma(z)$ will be very close to $-\pi$. But on the points on the circle which are above $-1$ the value of $\sigma(z)$ will be very close to $\pi$. So $\sigma$ can't be continuous at the point $-1$. And you will get a similar problem with any branch you choose.

Answer 3

Here is an alternative argument.

Assume there exists a lift $l: S^1 \to \mathbb{R}$ of the identity map on $S^1$. It is injective since $l(a) = l(b)$ implies $a = \varepsilon(l(a)) = \varepsilon(l(b)) = b$.

Let $S^1_+$ denote the closed upper half circle and $S^1_-$ the closed lower half circle. Then $J_\pm = l(S^1_\pm)$ are compact connected subset of $\mathbb{R}$ (recall the continuous images of connected sets are connected). Thus $J_\pm$ are closed intervals containing the two distinct points $r_\pm = l(\pm1)$. We conclude that $J_+ \cap J_-$ contains the closed interval $[a,b]$ between $r_+$ and $r_-$. Let $r \in (a,b)$. We have $r = l(z_+) = l(z_-)$ with $z_+ \in S^1_+$ and $z_- \in S^1_-$. Injectivity of $l$ implies $z_+ = z_- = z$. But then $z \in S^1_+ \cap S^1_- = \{ 1, -1 \}$. Hence $r = l(z) \in \{ r_+, r_- \} = \{ a, b \}$ which is a contradiction.

Note that we have more generally shown that there does not exist a continuous injection $ S^1 \to \mathbb{R}$.

Remark.

We used some well-known facts:

1) $S^1_\pm$ is connected.

2) Continuous images of connected sets are connected.

3) The compact connected subsets of $\mathbb{R}$ are precisely the closed intervals.

If you don't like to invoke these facts, you can proceeed as follows.

Let $I_+ = [0,1], I_- = [-1,0]$. The restrictions $l_\pm = (l \circ \varepsilon) \mid_{I_\pm} : I_\pm \to \mathbb{R}$ are injective. We have $l_+(0) = l_-(0) = l(1) = a$ and $l_+(1) = l_-(-1) = l(-1) = b$. W.l.o.g. we may assume $a < b$. The mean value theorem implies that $(a,b) \subset l_\pm(I_\pm)$. Let $r \in (a,b)$. We have $r = l_+(t_+) = l_-(t_-)$ with $t_+ \in I_+$ and $t_- \in I_-$. Setting $z_\pm = \varepsilon(t_\pm)$ we see that $l(z_+) = l(z_-)$. Injectivity of $l$ implies $z_+ = z_- = z$. But then $z \in l_+(I_+) \cap l_-(I_-) = \{ 1, -1 \}$. Hence $r = l(z) \in \{ a, b \}$ which is a contradiction.