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I have an exercise in topology and I am struggling with it:

a) Proof that a homeomorphism $S^1 \times S^1\to \mathbb T ^2$ exists.

b) For which $p,q \in \mathbb R$ is the curve $$\gamma(p,q):=\mathbb{R}\to S^1 \times S^1 , t\mapsto (\exp(2* \pi*t*p),\exp(2*\pi*t*q) $$ injective?

c) Draw a sketch of the curve $\gamma(2,3)$ for $r=1$ and $R=2$.

I think the answer for a) is that $$f((a,b),(c,d))=((c+R)*a,(c+R)*b,d*r)$$ is the requested homeomorphism with the inverse $$f^{-1}(a,b,c)= (\frac{a}{\sqrt{a^2+b^2}}, \frac{y}{\sqrt{a^2+b^2}},\frac{\sqrt{a^2+b^2}-R}{r},\frac{c}{r})$$.

My Problem is b) and c). In b) I think that I have to choose $p$ and $q$ so that the $\exp$ is not periodic, is this right? If it is right how can I find this $p$ and $q$ and if it is nit true, what I have to do? And at the moment im completley confused which effect the parameter have for the sketch in c).

Thank you :)

Definition: $\mathbb T^2$ is the surface of revolution generated by revolving the circle $\{(x,0,z):(x-R)^2+z^2=r^2\}$ around the $z$-axis.

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  • $\begingroup$ Your question is not complete unless you tell us what definition of $\mathbb T^2$ you are using. This is important not just to understand your equations, which seem somewhat random, because also because one of the most common definitions, $\mathbb T^2$ is defined to be $S^1 \times S^1$. $\endgroup$ – Lee Mosher Nov 4 '19 at 22:44
  • $\begingroup$ I added the definition. $\endgroup$ – alpaka123 Nov 4 '19 at 23:03
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For b. As for $q=0$ or $p=0$ is clear that the map is not injective, we may assume $pq\neq 0$. Note that if $\gamma$ is not inyective there should exist $t_1,t_2$ such that (I added an $i$ that I think you forgot).

$$ (e^{2\pi it_1 p},e^{2\pi it_1 q})=(e^{2\pi it_2 p},e^{2\pi it_2 q})\quad \Rightarrow \quad e^{2\pi i(t_1-t_2) p}=1=e^{2\pi i(t_1-t_2) q} $$

Therefore $(t_1-t_2) p=n\in \mathbb Z$ and $(t_1-t_2) q=m\in \mathbb Z$. Therefore $p/q=(n/m)$ is rational. Which proves that $\gamma$ is NOT injective if $p/q$ is rational. I leave to you the proof of the reverse implication but the final result is $\gamma$ is injective if and only if $p/q$ is irrational.

That gives a fairly good idea for c. As (2/3) is rational $\gamma$ is not injective. Moreover you have to look for the smallest $t$ such that both $2t$ and $3t$ are integers. 2,3 are co-prime then the smallest such $t$ is 1 and the image $\gamma(2,3)(\mathbb R)$ is equal to $\gamma(2,3)([0,1])$. Lastly noting than $e^{2\pi i 2t},e^{2\pi i 3t}$ covers the circle two and three times for $0\leq t\leq 1$ respectively. you see the plot is a line that winds around one direction two times and around the other three times.

I guess that's enough to draw it but if you have troubles let me know :)

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