Is $y^2=x(x-1)^2$ an immersed submanifold?

Question

Is the curve $y^2=x(x-1)^2$ an immersed submanifold in $\mathbb{R}^2$?

It's certainly not embedded since it intersects itself at $(1,0)$. I'm aware of techniques to prove a subset is not a immersed submanifold, but how can you find an appropriate topology and smooth structure to show this is immersed? It seems hard to pick one out of the blue.

Answer 1

Let $c$ denote the curve in question, i.e. $c=\{y^2=x(x-1)^2\}\subset\mathbb{R}^2$. It is easy to verify that $c\setminus\{(1,0)\}$ is a smooth curve (around every point there is a neighborhood with a smooth parametrization). Furthermore, each one of the two "branches" at $(1,0)$ has a smooth parametrization. For example:

$$t\mapsto (t,\sqrt{t}(t-1))$$ and $$t\mapsto(t,-\sqrt{t}(t-1)).$$ Hence the answer is yes.

Answer 2

Yes, your curve is the image $\gamma (\mathbb R)$ of the global immersion $$ \gamma:\mathbb R \to \mathbb R^2:t\mapsto (t^2,t^3-t)$$

Edit
The idea for the parametrization comes from algebraic geometry:
Consider the one parameter family of lines $L_t=\{y=t(x-1)\}$ through $(1,0)$ and look at the other point of intersection of that line $L_t$ with the curve.
You obtain the point $P_t=(t^2,t(t^2-1))$ and the immersion (=parametrization) is the map associating to the slope $t$ of the line $L_t$ the point $P_t$.