Inducing an embedding from a continuous and free group action Let $G$ be a topological group and $X$ be a space with a continuous and free group action $G\times X\to X$, $(g,x)\mapsto g\cdot x$. The map
$$\phi:G\times X\to X\times X,\quad \phi(g,x)=(g\cdot x,x)$$
is continuous by continuity of the action and injective by freeness of the action.
My question is: must $\phi$ be a topological embedding?
 A: This does not have to be a topological embedding.
Here is a counter example:
Let $G=\mathbb{Z}$ and $X=S^1$, given some irrational $\alpha\notin \mathbb{Q}$ we have the action:
$$m.e^{2\pi i t} = e^{2 \pi i (t+ \alpha \cdot m)} $$
The invective map $\phi: \mathbb{Z} \times S^1 \to S^1\times S^1$, $(m,e^{2\pi i t}) \mapsto (e^{2\pi i (t+\alpha m)},e^{2\pi i t})$
is not a topological embedding.
To see this we observe that if a group $G$ is complete and topologically embedded in to a group $Y$ via a homomorphism $f:G\to Y$ then the image $f(G)$ is complete as well.
To see this, take a Cauchy sequence $\{y_i\}\in f(G) $ i.e. for any neighborhood $V$ of the identity $e_Y$ there is $N\in \mathbb{N} $ such that $\forall m_1,m_2>N \ \ y_{m_1}y_{m_2}^{-1}\in U$
Its pre-image $\{g_i\}$ ($f(g_i)=y_i$) is a Cauchy sequence as well since for any neighborhood $U$ of the identity $e_G$ we look at the $N$ that corresponds to the neighborhood $f(U)$ (this is open since we assume topological embedding) and we have that  $\forall m_1,m_2>N \ \ y_{m_1}y_{m_2}^{-1}\in f(U) \Rightarrow  g_{m_1}g_{m_2}^{-1}\in U$
Now since $G$ is complete there exist a limit $g$ of the sequence $\{g_i\}$, we get that $f(g)$ is a limit of $\{y_i\}$ hence the image is complete.
It is a well known fact that a complete subspace is in fact closed, hence the image $f(G)$ must be closed in $Y$.
Returning to the counter example, we can see that $\phi $ is a homomorphism and that the image is not closed, since is dense but to the whole space.
The image is elements of the form $(e^{2\pi i (t+\alpha m)},e^{2\pi i t})$ and for example the element $(e^{2\pi i 0.5},e^{2\pi i 0.25})$ is not of this form since the second coordinate would imply  $t=0.25+k$ and from the first we get $m=\frac{0.25+k}{\alpha}+k'$ ($k,k'\in \mathbb{Z}$) in contradiction to $m\in \mathbb{Z}$. So the image is not the whole space.
The image is dense since for any $(e^{2\pi i s},e^{2\pi i t})=(e^{2\pi i t}\cdot e^{2\pi i (s-t)},e^{2\pi i t}) $ since the $\mathbb{Z}$ orbit $e^{2\pi i \alpha \cdot z }$ is dense in $S^1$ there is $m\in \mathbb{Z}$ s.t $e^{2\pi i \alpha \cdot m }$ is arbitrary close to $e^{2\pi i (s-t)}$, hence we can always find an element in the image that is close to $ (e^{2\pi i s},e^{2\pi i t}) $, so the orbit is dense.
Having the image dense but not the whole space implies it is not closed, hence $\phi$ is not a topological embedding.
