Let $\mathbb{R}$ act in $\mathbb{R}^2$ by $\lambda \cdot (x,y) := (x + \lambda, y), \ \forall \lambda \in \mathbb{R}, (x,y) \in \mathbb{R}^2$. If we define an equivalence relation $\sim$ by $$(x,y) \sim (x',y') \iff (x,y) = \lambda\cdot(x',y'), \ \text{for some } \lambda \in \mathbb{R}$$ i.e., we have $(x,y) \sim (x',y')$ iff $x = x' + \lambda$ and $y = y'$, for some $\lambda$. Now, we may define \begin{align} R & = \{((x,y), (x',y')) \in \mathbb{R}^2 \times \mathbb{R}^2 : (x,y) \sim (x',y')\} \\ & \cong \{((x,y), (x + \lambda, y) ) \in \mathbb{R}^2 \times \mathbb{R}^2: \lambda \in \mathbb{R}\}. \end{align} How can I show that $R$ is actually an embedded submanifold of $\mathbb{R}^2 \times \mathbb{R}^2$?
I've noticed that this problem is suspiciously similar to Example 10.3 of Lee's Introduction to Smooth Manifolds (2nd edition), although (seemingly) harder.
Edit: Moreover, I know that:
The equivalence classes are the lines parallel to the $x$ axis. Indeed, if $(x,y) \sim (x',y')$, then $y' = y$ and $x' = x + \lambda$, for some $\lambda \in \mathbb{R}$, and thus $(x',y')$ lies in such a line. The other way around is analogous.
The main tool I use to show that some subset $S$ of a manifold $M$ is an embedded submanifold is to show that $S$ can be seen as a preimage of some regular value under some smooth map $f$, which is pretty much what I've tried on this particular problem. I did not get far with it, as I didn't find any map $f$ that did the job. I still suspect this strategy will work on this question, though.