$y^2 - x^3$ not an embedded submanifold How can I show that the cuspidal cubic $y^2 = x^3$ is not an embedded submanifold of $\Bbb{R}^2$? By embedded submanifold I mean a topological manifold in the subspace topology equipped with a smooth structure such that the inclusion of the curve into $\Bbb{R}^2$ is a smooth embedding. I don't even know where to start please help me. All the usual tricks I know of removing a point from a curve and see what happens don't work. How can I extract out information about the cusp to conclude it is not? Also can I put a smooth structure on it so it is an immersed submanifold? THankz.

Suppose $C=\{(x,y):y^2=x^3\}$ is an embedded submanifold of $\mathbb R^2$. Consider a smooth curve $\gamma:(-\epsilon,\epsilon)\to C$ such that $\gamma(0)=(0,0)$. Since the inclusion is smooth $\iota:C\to\mathbb R^2$, we have a smooth curve $\iota\circ\gamma:(-\epsilon,\epsilon)\to\mathbb R^2, t\mapsto (\gamma_1(t),\gamma_2(t))$. Note the relation $\gamma_2(t)^2=\gamma_1(t)^3$, which yields $\gamma_1(t)=\gamma_2(t)^{2/3}$. The function $x\mapsto x^{2/3}$ is smooth for $x>0$. Since $\gamma$ is smooth, its derivative is smooth too, and hence we have
$$
\dot\gamma_1(0)=\lim_{t\to 0}\dot\gamma_1(t)=\lim_{t\to 0}\frac 23\gamma_2(t)^{-1/3}\dot\gamma_2(t).
$$
Since $\lim_{t\to 0}\gamma_2(t)^{-1/3}=\infty$, the above limit can only be finite if $\lim_{t\to 0}\dot\gamma_2(t)=0$, and hence $\dot\gamma_2(0)=0$. This means that $\gamma(t)=(\gamma_1(t),0)$. Since $\gamma_1(t)^3=0$, we have in fact $\gamma\equiv 0$. We've shown that the tangent space of $C$ at $(0,0)$ is zero-dimensional, which contradicts that $C$ is a curve.
