Is $x^2 = y ^2$ an immersed submanifold of $\mathbb{R}^2$? So, I know that this isn't an embedded submanifold, since with the subspace topology inherited from $\mathbb{R}^2$, it is not locally Euclidean at $0$. Further, I also know that if I can parametrize it nicely, I will be able to say that it is the image of an injective smooth immersion and thus an immersed submanifold, but I can't seem to do that and don't know how to prove that it is not one.
 A: If "immersed submanifold" means "the image of an immersion, a map whose differential is injective everywhere", then the crossed lines are the image of two parallel lines under a simple map. 
More explicitly, let's define a map from $X=\{(x,y)\in\mathbb{R}^2|y^2=1\}$ to $Y=\{(x,y)\in\mathbb{R}^2|x^2=y^2\}$ via $f(x,y) = (x,\text{sign}(y)x).$ It maps two parallel lines to your two crossed lines. Since $\text{sign}(y)$ is constant on the connected components, its non-differentiability at $y=0$ doesn't come up. So $f$ is differentiable, and is an immersion.
However as Jack Lee points out in the comments, the more usual definition of "immersed submanifold" is a subset with its own topology and smooth structure (not necessarily subspace) such that the inclusion is an immersion. 
We can realize $Y$ as an immersed submanifold in this sense as well. Let's define $f,g,h\colon \mathbb{R}\to\mathbb{R}^2$ by $f(x)=(x,x)$, and $g(x) = (-e^x,e^x)$ and $h(x)=(e^x,-e^x).$ So $f$ parametrizes the line through the origin and quadrants III and I, $g$ parametrizes the ray through quadrant II (approaching but never reaching the origin), and $h$ parametrizes the ray through quadrant IV (also approaching but never reaching the origin). These functions define a topology on their image which is not the subspace topology. Consider for example that the space with its immersed topology has three connected components, while it has only one in the subspace topology. They give the set $Y=\{(x,y)\in\mathbb{R}^2|x^2=y^2\}$ the structure of an immersed submanifold of $\mathbb{R}^2$.
