Is $ S = \{(x,y) \in \mathbb{R}^2 \colon x^2 = y^3\} $ an immersed submanifold of $\mathbb{R}^2$? I know that it is not an embedded submanifold because it is not locally Euclidean with the subspace topology, and I feel that it is not an immersed submanifold, because its tangent space at $(0,0)$ has dimension $2$ which would mean that it has to be a smooth manifold of dimension $2$. But if that happens, then since we have that the inclusion map is a smooth immersion, it ends up being both a smooth immersion and a smooth submersion (since the dimensions of $S$ and $\mathbb{R}^2$ are equal) which means that it's a local diffeomorphism. But, it is also bijective onto its image meaning that it is a diffeomorphism onto $S$ with the subspace topology, which is a contradiction. Kindly let me know if my reasoning and conclusion are correct or not.
 A: (This is basically the same argument as the one in Example 5.45 of J.Lee's book Introduction to Smooth Manifolds)
The answer is no. Suppose that somehow we could make $S:=\{(x,y)\mid x^2=y^3\}$ into an immersed submanifold of $\mathbb{R}^2$. Then $S\setminus\{(0,0)\}$ can be considered as a smooth manifold in two ways- first, as an open submanifold of $S$, and second, as an embedded submanifold of $\mathbb{R}^2$ (for it is a disjoint union of graphs of smooth functions). Now it is a fact that if a subset of a smooth manifold is an embedded submanifold with respect to some smooth manifold structure, then there is no other smooth manifold structure making the same set into an immersed submanifold. Therefore, the two smooth manifold structure for $S\setminus\{(0,0)\}$ coincide. Since $S\setminus\{(0,0)\}$ is 1-dimensional, it follows that $S$ has to be $1$-dimensional, too.
Now let $\gamma:(-\varepsilon ,\varepsilon )\to M$ be any smooth curve starting at the origin. Write $\gamma(t)=(x(t),y(t))$. Because $y(t)$ attains its global minimum at $t=0$, we have $\ y'(0)=0$. We also have $x(t)^2=y(t)^3$ for all $t$; differentiating the both sides and evaluating at $t=0$ gives $x'(0)=0$. Thus $\gamma'(0)=0$. Since $\gamma$ was arbitrary, it follows that the tangent space $T_{(0,0)}M$ of $M$ at the origin is $0$-dimensional, a contradiction.
A: To find out the dimension of $S$ you have to consider the tangent space at a smooth point. 
Since  $S$ is the zero set of the function $f(x,y)=x^2-y^3$, a point $(x,y)$ is smooth (i.e. admits a neighbourhood diffeomorphic to euclidean space) iff $Df(x,y)\neq 0$.
This shows that $0$ is not a smooth point and it also shows (by using the implicit function theorem) that $S\backslash \{0\}$ is indeed a smooth 1-dimensional submanifold of $\mathbb{R}^2$.
On the other hand you can see that the map $\gamma:\mathbb{R}\rightarrow \mathbb{R}^2, \gamma(t) = ( t^3,t^2)$ is smooth map onto $S$. This is a topological embedding but not an immersion (since $\gamma'(0)=0$).
Edit: $S$ is not an immersed submanifold. Assume otherwise, then there is an immersion $I=(I_x,I_y):\mathbb{R}\rightarrow  \mathbb{R}^2$ onto $S$. We may assume that $I(0)=(0,0)$. Again by the implicit function theorem we find a neighbourhood $(-\epsilon,\epsilon)$ of $0$, such that $I((-\epsilon,\epsilon))$ is an embedded submanifold of $\mathbb{R}^2$. $I$ has to be injective, because otherwise its image $S$ would contain a loop, which is impossible since we already know that $S$ is homeomorphic to $\mathbb{R}$. Injectivity of $I$ implies that $I_x$ is strictly monotic (assume increasing), thus $I((-\epsilon,\epsilon))=S\cap\{I_x(-\epsilon)<x<I(\epsilon)\}$. (We have just assured that the singularity at $(0,0)$ does not come from a self intersection nor from approaching itself) But this implies that $S\cap\{I_x(-\epsilon)<x<I(\epsilon)\}$ has a one dimensional tangent space at $(0,0)$ which is not true as you've already found out.
