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I'm following Tu's Introduction to Manifolds to build an atlas on the unit circle $S^1$. In order to build charts, Tu defines the following maps:

$$ \phi_1(e^{it})=t, \; - \pi< t < \pi , \\ \phi_2(e^{it})=t, \; 0< t < 2\pi. $$ He says that both maps are homeomorphisms onto their images. I don't understand this conclusion: an homeomorphism is supposed to be a continuous bijective map, and Tu himself says that $\phi_1$ and $\phi_2$ are branches of the complex logarithm. Isn't there a problem with the cut in the complex logarithm?

glS
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    There would be a problem if you're trying to define Log on the whole of $\mathbb{C}\setminus{0}$, but he is only defining $\phi_1$ for a circle minus the point ${-1}$. The point is that $\phi_1$ and $\phi_2$ together cover the whole circle. – Teddy38 Jan 04 '21 at 12:05
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    An homeomorphism is more than just a continuous bijective map. – Thorgott Jan 04 '21 at 14:53
  • @Teddy38 I do not have a solid grasp of complex analysis, but here is what I don't understand precisely: log is not defined on the real negative axis, so we must be careful with $\phi_1$ and $\phi_2$. On the circle $e^{it}$, $t=\pi$ should then be a problematic value, shouldn't it? I hear that for $\phi_1$, we do not include -1, but what about $\phi_2$? From $0$ to $2\pi$, we go straight through that real negative axis. Thank you for helping! –  Jan 05 '21 at 11:44

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Usually one defines a branch of the logarithm on a sliced plane $\mathbb C \setminus L(c)$, where $L(c) =\{\lambda c \mid \lambda \ge 0\}$ is a ray through some $c \ne 0$.

Of course we may assume $c \in S^1$. Writing $c = e^{i\gamma}$ with $\gamma \in \mathbb R$, for each $w \in \mathbb C \setminus L(c)$ there exists a unique $\varphi(w) = \varphi_\gamma(w) \in (\gamma,\gamma+2\pi)$ such that $e^{i\varphi(w)} = w/\lvert w \rvert$. Then a logarithm branch is given by

$$\ln = \ln_\gamma : \mathbb C \setminus L(c) \to \mathbb C, \ln w = \ln \lvert w \rvert + i\varphi(w)$$ The image of $\ln_\gamma$ is the open strip $\{x+iy \mid x > 0, \gamma < y < \gamma + 2\pi \} =(0,\infty) \times (\gamma,\gamma+2\pi)$.

If we pick another $\gamma'$ with $e^{i\gamma'} = c$, then we have $\gamma' - \gamma = 2k\pi$ for some $k \in \mathbb Z$ which gives another function $\varphi_{\gamma'} : \mathbb C \setminus L(c) \to (\gamma+2k\pi,\gamma+ 2(k+1)\pi), \varphi_{\gamma'}(w) = \varphi_\gamma(w) + 2k\pi$, and another logarithm branch $\ln_{\gamma'}$ on $\mathbb C \setminus L(c)$.

Tu's maps are obtained as follows:

  1. $c = 1$. Take $\gamma = 0$ and get $\phi_1(w) = \varphi_0(e^{it}) = t$ for $w = e^{it}$ with $t \in (0,2\pi)$.

  2. $c = -1$. Takte $\gamma = -\pi$ and get $\phi_2(w) = \varphi_{-\pi}(e^{it}) = t$ for $w = e^{it}$ with $t \in (-\pi,\pi)$.

To see that these maps form an atlas on $S^1$, it is essential to know that $\varphi_\gamma \mid_{S^1 \setminus \{c\}} : S^1 \setminus \{c\} \to (\gamma,\gamma+2\pi)$ is a homeomorphism. Have a look at Why is $\sigma(u,v)=(\cos u,\sin u,v)$ a homeomorphism on $0< u <2\pi$? and Is $f(t)=(\cos(2\pi t),\sin(2\pi t))$ an open function?

Paul Frost
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