Showing that $f(x, y)=xy$ is continuous on $S^1$ This is question from studying general topology.
Let $f : S^1 \times S^1 \rightarrow S^1$ defined by $f(x, y)=xy$
How can I show that $f$ is continuous?
My attempt : Let $U : $ {$e^{i\theta}$ | $\alpha < \theta < \beta $ } : open set in $S^1$.
Then, $f^{-1}(U)$={$(e^{i \theta _1}, e^{i \theta _2})$ | $\alpha + 2n\pi < \theta_1 + \theta_2 < \beta + 2n\pi$} , which is homeomorphic to $V$ = {$(x, y)$ | $\alpha + 2n\pi < x+y < \beta + 2n\pi$} .
Since $V$ is open, $f$ is continuous.
 A: $f(e^{it}, e^{ir})  = e^{i (t+r)} $ is a composition of the two continuous functions $(r,t) \mapsto t+r$ and $e^{i x}$.
A: The notation
$$f^{-1}(U)=\{(e^{i \theta _1}, e^{i \theta _2}) \mid \alpha + 2n\pi < \theta_1 + \theta_2 < \beta + 2n\pi\}$$ is fuzzy. What is $n$ here? Also when you state that it is homeomorphic to
$$V= \{(x, y) \mid \alpha + 2n\pi < x+y < \beta + 2n\pi\}$$
this deserves some argument as well as to precise to which set the $(x,y)$ belong to.
This question looks simple... but formulating a precise answer is not so easy (at least for me!).
I would suggest the following:

*

*Take $(x,y) = (e^{i\alpha}, e^{i\beta})\in S^1 \times S^1$. We have to prove that $f$ is continuous at $(x,y)$.

*By considering the continuous map $(u,v) \mapsto (ue^{-i\alpha},ve^{-i\beta})$ defined on $S^1 \times S^1$, we can without loss of generality suppose that $(x,y) = (1,1)$.

*The open $S^1_+=\{z \in S^1 \mid \Re z >0\}$ is homeomorphic to $(-\frac{\pi}{2},\frac{\pi}{2})$ under the standard homeomorphism $\varphi$ and $(1,1) \in S^1_+ \times S^1_+$.

*$f = \overline{\varphi} \circ S \circ \Phi$ where

$$\begin{array}{l|rcl}
\Phi : & S^1_+  \times S^1_+& \longrightarrow & (-\frac{\pi}{2},\frac{\pi}{2}) \times (-\frac{\pi}{2},\frac{\pi}{2}) \\
    & (u, v) & \longmapsto & (\varphi^{-1}(u),\varphi^{-1}(v)) \end{array}$$
$$\begin{array}{l|rcl}
S: &  (-\frac{\pi}{2},\frac{\pi}{2}) \times (-\frac{\pi}{2},\frac{\pi}{2})& \longrightarrow & (-\pi,\pi) \\
    & (r,s) & \longmapsto & r+s \end{array}$$
$$\begin{array}{l|rcl}
\overline{\varphi}: &  (-\pi,\pi)& \longrightarrow & S^1 \\
    & t & \longmapsto & e^{it} \end{array}$$
are all continuous map.
A: Let $\iota\colon S^1 \to \Bbb R$ be the inclusion. By definition of subspace topology, $\iota$ is continuous. So $\iota\times \iota\colon S^1\times S^1 \to \Bbb R \times \Bbb R = \Bbb R^2$ is also continuous. Define $\widetilde{f}\colon \Bbb R^2 \to \Bbb R$ by $\widetilde{f}(x,y) = xy$. This $\widetilde{f}$ is obviously continuous, as it is a polynomial. Then $$f = \widetilde{f}\circ (\iota\times \iota)$$is the composition of continuous maps, hence continuous.
