I am studying a basic course in differentiable manifolds.I am confused in the following: Consider $f: \mathbb R \to \mathbb R^2$ given by $f(t)=(t^2,t^3)$.
I have shown that it is a smooth topological embedding onto its image but I do not understand
why is this not differentiable submanifold of $\mathbb R^2$
can we make it as an immersed submanifold in $\mathbb R^2$
any help would be appericiated...
Let $N$ denote the image of $f$. Suppose $N$ were an immersed submanifold of $\mathbb{R}^2$ with the standard differentiable structure. Suppose $g:M\to \mathbb{R}^2$ is an immersion with $g(M) = N$. Consider the map $p:\mathbb{R}^2\to \mathbb{R}$ given in standard coordinates by $p(x,y) = x^3-y^2$. Of course $p$ is a smooth map, and $p\circ g = 0$, so by the chain rule, $\mbox{im}\, dg \subset \ker dp$, and since $\ker dp$ has dimension $1$ away from $(0,0)$, we have $\mbox{im}\, dg$ has dimension at most $1$, and clearly $\dim M \ge 1$, so $\dim M = 1$. Let $a\in M$ be such that $g(a) = (0,0)$. Since $g$ is an immersion, $g$ is locally injective at $a$. Let $U$ be an open subset of $M$ containing $a$ on which $g$ is injective with a coordinate chart (diffeomorphism) $\phi:U \to B_\delta(0)\subset\mathbb{R}$, the open ball of radius $\delta>0$ around $0$ in $\mathbb{R}$, with $\phi(a) = 0$. Let $h = g\circ \phi^{-1}$ and let $t$ be the standard coordinate on $B_\delta(0)\subset\mathbb{R}$. Note that $h$ is also an injective immersion. Write $h(t) = (h_1(t),h_2(t))$. We have $h_1(t) = h_2(t)^{2/3}$, so $$h_1'(t) = \frac{2}{3}h_2(t)^{-1/3}h_2'(t)$$ for $t\ne 0$. Since $h_2'$ is continuous, we must have $h_2'(0) = 0$, otherwise $h_1'(t)$ would not exist at $0$. So since $dh_0$ is nonzero, we must have $h_1'(0) \ne 0$. Without loss of generality, we assume $h_1'(0) > 0$. Thus there is some $\delta\ge\epsilon>0$ so that $h_1$ is increasing in $B_\epsilon(0)$. But then for some $-\epsilon<t_0<0$ we must have $h_1(t_0)<h_1(0)=0$, and so $h(t_0)\notin N$, a contradiction.