Complex argument is continuous in quotient topology

Question

Consider $$\operatorname{Arg}:\mathbb{C}^*\rightarrow\mathbb{R}/2\pi\mathbb{Z}$$ $$\operatorname{Arg}(z)=\arg(z)+2\pi\mathbb{Z}$$

where $\arg(z)$ is the principal argument. Show that such function is continuous in the quotient topology. I never encountered such topology before so I don't know how to do that, however I have a hint (that actually seems to me harder):

Show that for every convergent sequence of non-zero complex numbers $z_n\rightarrow z$ and $\theta \in \operatorname{Arg}(z)$ we can choose for every $n\in \mathbb{N}$ a $\theta_n \in \text{Arg}(z_n)$ s.t. $\theta_n\rightarrow \theta$

Answer 1

Define $arg:\Bbb C-\{(0,0)\}\rightarrow \Bbb R$ as follows: $$arg(x,y)=tan^{-1}(|\frac{y}{x}|); \ y≥0,x>0$$ $$arg(0,y)=\frac{\pi}{2};\ y>0$$ $$arg(x,y)=\pi-tan^{-1}(|\frac{y}{x}|);\ y≥0,x<0$$ $$arg(x,y)=\pi+tan^{-1}(|\frac{y}{x}|);\ y≤0,x<0$$ $$arg(0,y)=\frac{3\pi}{2};\ y<0$$ $$arg(x,y)=2\pi-tan^{-1}(|\frac{y}{x}|);\ y<0,x>0$$

Where $tan^{-1}:\Bbb R\rightarrow (-\frac{\pi}{2},\frac{\pi}{2})$ is the inverse of $tan_{|(-\frac{\pi}{2},\frac{\pi}{2})}\rightarrow \Bbb R$.

Now $\Bbb C^*$ is a metric space. Therefore to show the continuity of the map $$\operatorname{Arg}:\mathbb{C}^*\rightarrow\mathbb{R}/2\pi\mathbb{Z}$$ $${Arg}(z)=\arg(z)+2\pi\mathbb{Z}$$ we may use sequential criteria. But note that $arg $ is continuous at each point of $\Bbb C-\{(x,0):x≥0\}$ , hence $Arg$ is also continuous on $\Bbb C-\{(x,0):x≥0\}$ as the quotient map $\pi:\Bbb R\rightarrow \frac{\Bbb R}{2\pi \Bbb Z}$ is continuous.

Now to check the continuity of $Arg$ on $\{(x,0)\in \Bbb C:x>0\}$ notice that if a sequence $\{(x_n,y_n)\}_n$ in $\Bbb C ^*$ converges to $(x,0)$ with $x>0$, then either $arg$ converges to $0$ or converges to $2\pi$ i.e. $Arg$ converges to either $0\ +\ 2\pi \Bbb Z$ or $2\pi\ +\ 2\pi \Bbb Z$ but $0\ +\ 2\pi \Bbb Z\ =\ 2\pi\ +\ 2\pi \Bbb Z$. Hence $Arg$ is also continuous.

Answer 2

If you have a topological space $X$, a set $M$ and a surjective function $f : X \to M$ , you can give $M$ the finest topology $\tau_f$ such that $f$ becomes a continuous map. Explicitly, this topology consists of all $U \subset M$ such that $f^{-1}(U)$ is open in $X$. It is called the quotient topology induced by $f$. It has the following universal property: If $Y$ is any topological space and $g : M \to Y$ is any function such that $g \circ f$ is continuous, then $g : (M,\tau_f) \to Y$ is continuous. This is true because for any open $V \subset Y$ we have $f^{-1}(g^{-1}(V)) = (g \circ f)^{-1}(V)$ open in $X$, hence $g^{-1}(V) \in \tau_f$.

A typical application is this: You have a space $X$ and an equivalence relation $\sim$ on $X$. If $X/\sim$ denotes the set of equivalence classes, then you get a canonical surjection $p : X \to X/\sim, p(x) = [x] =$ equivalence class of $x$ and endow $X/\sim$ with the quotient topology.

In the present case we have the following equivalence relation on $\mathbb{R}$: $x \sim y$ iff $x - y \in 2\pi\mathbb{Z}$. We have $\mathbb{R}/\sim \phantom{} = \mathbb{R}/2\pi\mathbb{Z}$, the latter being the quotient group of the abelian group $\mathbb{R}$ modulo the subgroup $2\pi\mathbb{Z}$. Its elements have the form $x + 2\pi\mathbb{Z}$ with $x \in \mathbb{R}$, and it is endowed with the quotient topology induced by $p : \mathbb{R} \to \mathbb{R}/2\pi\mathbb{Z}$.

Now let $S^1 = \{ z \in \mathbb{C} \mid \lvert z \rvert = 1 \}$. This is a subgroup of $\mathbb{C} \setminus \{ 0 \}$ with respect to complex multiplication. Define $$\phi : \mathbb{R} \to S^1, \phi(x) = e^{ix} .$$ This is a continuous surjective group homomorphism with kernel $ker\phi = 2\pi\mathbb{Z}$. Thus we obtain a group isomorphism $$\Phi : \mathbb{R}/2\pi\mathbb{Z} \to S^1, \Phi([x]) = \phi(x) .$$ By definition $\Phi \circ p = \phi$, therefore $\Phi$ is continuous. Moreover, $\Phi^{-1}$ is readily seen to be continuous (so that $\Phi$ is a homeomorphism): Let $z_0 \in S^1$ and $V$ be an open neigborhood of $[x_0] = \Phi^{-1}(z)$ in $\mathbb{R}/2\pi\mathbb{Z}$. This means that $p^{-1}(V)$ is an open neigborhhood of $x_0$ in $\mathbb{R}$. Choose $\epsilon \in (0,\pi)$ such that $W = (x_0 - \epsilon, x_0 + \epsilon) \subset p^{-1}(V)$. Then $p^{-1}(p(W)) = W$, thus $p(W)$ is an open neigborhood of $[x_0]$ in $\mathbb{R}/2\pi\mathbb{Z}$ which is contained in $V$. For sufficiently small $\delta >0$ we see that $\lvert z - z_0 \rvert < \delta$ implies $\Phi^{-1}(z) \in p(W)$. This means that $\Phi^{-1}$ is continuous.

Let us consider the continuous map $$\psi : \mathbb{C} \setminus \{ 0 \} \to S^1, \psi(z) = \frac{z}{\lvert z \rvert} .$$ Then obviously $\Phi \circ \text{Arg} = \psi$, i.e. $$\text{Arg} = \Phi^{-1} \circ \psi.$$ This proves the continuity of $\text{Arg}$ and morever explains that $\text{Arg}$ is noting else than $\psi$ interpreted by identifying $S^1$ with $\mathbb{R}/2\pi\mathbb{Z}$.