$\mathbb{R} \to S^1 \times S^1$ defined by $t \mapsto (e^{2\pi i \alpha t}, e^{2 \pi it})$ is not an embedding This is exercise $2.8.2$ from Topology and Groupoids by Ronald Brown.
Let $S^1$ be the unit circle in the complex plane and $\alpha \in \mathbb{R}$ be irrational. I want to show that $f: \mathbb{R} \to S^1 \times S^1, t \mapsto (e^{2\pi i \alpha t}, e^{2 \pi it})$ is not an embedding.
From definitions, there are a few ways to show that $f$ is not am embedding. I can show that an open (closed) set in $\mathbb{R}$ is not taken to an open (closed) set in $S^1 \times S^1$. I can also show that $f$ does not admit a continuous inverse.
I think the key to the problem is understanding how $\alpha$ irrational affects the mapping, but I can't figure out what it's doing. I've tried considering the sets $(0, 1/4)$ and $[0, 1/4]$, but am having trouble showing whether they are closed/open in $S^1 \times S^1$.
 A: The map $f:\Bbb R\to S^1\times S^1$ sending $t$ to 
$$(f_1(t),f_2(t))=\left(e^{2\pi i \alpha t}, e^{2\pi i t}\right)$$ is injective because if $e^{2\pi i s}=e^{2\pi i t}$, then $s-t$ must be an integer, but then $\alpha s-\alpha t$ cannot be an integer, so $e^{2\pi i \alpha s}$ is distinct from $e^{2\pi i\alpha t}$, except when $s=t$.
So we need to show that $f:\Bbb R\to f(\Bbb R)$ is not open. Consider the image of the open interval $(-1,1)$. It contains $(e^0,e^0)$, but no other pair $(a,e^0)$. The aim is to show that $(e^0,e^0)$ is not interior point of $f((-1,1))$. So look at a neighborhood $U\times U$ of $(e^0,e^0)$, where $U$ is the image of $(-\epsilon,\epsilon)$ under the (open) map 
$$\phi:\Bbb R\to S^1,\quad t\mapsto e^{2\pi i t}$$
Then $U=\phi\left(\bigcup_{n\in\Bbb Z}(n-\epsilon,n+\epsilon)\right)$. Note that $f_1$ factors as $\Bbb R\to\Bbb R\xrightarrow\phi S^1$, where the first map is multiplication by $\alpha$. If you can show that the product $\alpha l$ for some integer $l\ne 0$ is in an $\epsilon$-interval around some integer $k$, then $f(l)$ will be in $U\times\{e^0\}\setminus f((-1,1))$, and you are finished (Dirichlet's Approximation Theorem may be useful here).
