Show that this map is an immersion but not an embbedding Define $C=\mathbb{R}^2/\sim$,  where "$\sim$" is the equivalence relation: $(x,y) \sim (x+k,y)$ and $k \in \mathbb{Z}$.
Show that $\gamma:(0,\pi/3)\to C$ given by $$\gamma(t)=(1/2+ \cot(t),\sin(t))$$ is not an embedding.
This map is an immersion (I did this). But I don´t have an idea how to show that is not an embedding.
 A: Let us observe that it would be better to write
$$\gamma(t) = [1/2 + \cot(t),\sin(t)]$$
where $[-]$ denotes equivalence class.
$\gamma$ is an embedding of smooth manifolds. Since you know that it is an immersion, it suffices to show that it is a topological embedding, i.e. that the map
$$\bar \gamma : (0,\pi/3) \to G = \gamma((0,\pi/3)), \bar \gamma(t) = \gamma(t)$$
is a homeomorphism.
Let $p_2 : \mathbb R^2 \to \mathbb R$ be the projection onto the second coordinate. Clearly $p_2(x,y) = p_2(x+k,y)$, thus $p_2$ induces a continuous map $\pi : C \to \mathbb R$. We have $\pi \circ \gamma = \sin$. Since $\sin$ maps $(0,\pi/3)$ homeomorphically onto $(0,\sqrt 3/2)$, we see that $\bar \gamma$ has a continuous right inverse $\phi : G  \to (0,\pi/3)$. In fact, take $\phi =\arcsin(\pi(w))$ for $w \in G$.
From $\phi \circ \bar \gamma = id$ we conclude that $\bar \gamma$ is injective. Thus it maps $(0,\pi/3)$ bijectively onto $G$. Therefore $\phi$ is the inverse of $\bar \gamma$. Since it is continuous, we see that $\bar \gamma$ is a homeomorphism.
