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.

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.
