I try to solve the following problem:For each $a\in \mathbb{R}$, let $M_a$ be the subset of $\mathbb{R}^2$ defined by $$ M_a=\{(x,y):y^2=x(x-1)(x-a)\}.$$ For which values of $a$ is $M_a$ an embedded submanifold of $\mathbb{R}^2$? For which values can $M_a$ be given a topology and smooth structure making it into an immersed submanifold?
My attempt: Let $F(x,y)=y^2-x(x-1)(x-a)$ so that $M_a=F^{-1}(0)$.\ Then $DF(x,y)=[-(3x^2-(2a+2)x+a)~~ 2y]$. Therefore, $0$ is a regular value of $F$ unless there is a point $(x,y)$ such that $$y=0,3x^2-(2a+2)x+a=0 ~\text{and}~ F(x,y)=-x(x-1)(x-a)=0.$$ In this case, $-x(x-1)(x-a)=0$ implies that $x=0$ or $x=1$ or $x=a$.\ When $x=0$, $3x^2-(2a+2)x+a=0$ implies that $a=0$. When $x=1$, $3x^2-(2a+2)x+a=0$ implies that $a=1$. The case $x=a$ gives the above values. Thus we have the following cases to consider: Case 1: $a=0, (x,y)=(0,0)$. When $a=0$, the point $(0,0)$ is local minimum of $F$, so $(0,0)$ is an isolated point of $M_0$, and hence $M_0$ can not be an embedded or immersed submanifold. my difficulty is in the following case: Case 2: $a=1, (x,y)=(1,0)$. when $a=1$, the point $(1,0)$ is saddle point of $F$, and thus the curve $M_1$ is self-intersecting at $(1,0)$. In such cases I saw examples saying it can not be an embedded submanifold but by giving an appropriate topology and smooth structure we can make $M_1$ an immersed submanifold of $\mathbb{R}^2$. My difficulty is to justify how it cannot be embedded submanifold, and how we define the topology and smooth structure to make it immersed submanifold. Clearly for $a\neq 0$ and $a\neq 1$, $M_a$ is an embedded submanifold.
Concernig $M_1$ the problem is not the self-intersection in $(1,0)$.
It is clear that $M_1$ is the image of a map $m : \mathbb{R} \to \mathbb{R}^2$. On $[0,\infty)$ it is defined by $m(x) = \sqrt{x(x-1)^2}$ and produces the upper half $M_1^+ = \lbrace (x,y) \in M_1 \mid y \ge 0 \rbrace$ and on $(-\infty,0]$ the lower half $M_1^-$ via $m(x) = -\sqrt{\lvert x \rvert (\lvert x \rvert-1)^2}$. $M_1^\pm \backslash \lbrace (0,0) \rbrace$ are submanifolds of $\mathbb{R}^2$.
Let us consider the self-intersection. If you understand an immersed submanifold as an immersion $f : N \to \mathbb{R}^2$, then there is no problem at all since only the local behavior is relevant. If you want to find an injective immersion (which means to retopologize $M_1$), then the self-intersection can be resolved by cutting $M_1$ into two pieces which form the components of a new space $N_1$. The two pieces are $N_1' = m((-1,\infty))$ and $N_1'' = m((-\infty,-1))$ but in $N_1$ they are disjoint open sets.