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I need to prove that $\mathbb{R} \times S^1$ is homeomorphic to $\mathbb{R^2} \setminus \{(0,0)\}$.

I define the map $h:\mathbb{R} \times S^1 \to \mathbb{R^2} \setminus \{(0,0)\}$ by $h(t,\theta)=(e^t\cos \theta,e^t\sin \theta)$, this is easy to show that this is a bijection, and It seems very natural that this function is continuous, and also its inverse $h^{-1}(x,y)=(\ln \sqrt{x^2+y^2},\tan^{-1}\frac{y}{x})$.... But I am not sure what is the best way to prove that these two functions are continuous.

I can use the following theorem:

Theorem: Let $f:A \to X\times Y$ be given by the equation $$f(a)=(f_1(a),f_2(a))$$ Then $f$ is continuous if and only if the functions $$f_1:A \to X \ \ \ \ \ \ \text{and} \ \ \ \ \ \ f_2:A \to Y$$ are continuous.

But I also need to use a theorem like

Theorem(?): Let $g:X\times Y \to \mathbb{R}$ be given by the equation $$g(x,y)=g_1(x)g_2(y)$$ where $$g_1:X \to \mathbb{R} \ \ \ \ \ \ \text{and} \ \ \ \ \ \ g_2:Y \to \mathbb{R}$$ Then $g$ is continuous if (and only if) both $g_1$ and $g_2$ are continuous.

Is that true , and if yes how to prove it ? Thanks for your help in advance.

the8thone
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  • The function in Theorem (?) can be expressed as a composition of continuous functions. The first function is $(x,y) \to (g_1(x),g_2(y))$, and it is followed by the multiplication function $(x,y) \to xy$. – Lee Mosher Jul 29 '14 at 23:47
  • @LeeMosher Thanks! , Actually my problem was how to prove that multiplication function is continuous ? I know it is a dumb one :-| – the8thone Jul 30 '14 at 00:42
  • Using the $\epsilon,\delta$ definition. Very similar to the proof in calculus that if two functions $f(x)$ and $g(x)$ are continuous then the product function $f(x) g(x)$ is continuous. – Lee Mosher Jul 30 '14 at 02:14

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