$$\alpha: \mathbb{R}\to\mathbb{R}^2\quad \alpha(t)=(t^3,t^2)$$ The condition for an immersion is that $\alpha'(t)=(3t^2,2t)$ is injective for all $t\in\mathbb{R}$.

The book seems to imply that there is a problem at $t=0$ but I don't understand why. At $t=0$ the tangent is undefined because if you take the limit from both sides you get opposite results. However that does not make $\alpha'$ non-injective.

What am I missing?

  • 1
    $\begingroup$ Think about what "$\alpha'(t)$ is injective" means. What injectivity are we really talking about? $\endgroup$
    – Arthur
    Apr 26, 2018 at 14:40
  • 1
    $\begingroup$ The diferential should be injection at any point. This implies that $\alpha'(t)\neq 0 $ for any t (?) $\endgroup$ Apr 26, 2018 at 14:48

3 Answers 3


The condition for immersion is not that the function $\alpha'$ is injective. It is that, for each $t$, the linear transformation $\alpha'(t):\mathbb R\to\mathbb R^2$, given by $$ (3t^2,2t)\,s=(3t^2s,2ts). $$ is injective. At $t=0$, the linear transformation is the zero map, which of course is not injective.

Part of the problem here is an awful convention that creeps up from calculus, and pre calculus, which is to call a function $f$ as $f(x)$. So, in this case, where $\alpha'$ is a function but $\alpha'(t)$ is a function for each $t$, the poor notation wreaks havoc with the inexperienced student.

  • $\begingroup$ Perhaps worth clarifying that what OP wrote as $\alpha'(t)$ is actually the vector $v$ such that $\alpha'(t)(s) = sv$. $\endgroup$ Apr 26, 2018 at 14:42
  • $\begingroup$ Thank, Najib, I clarified what the linear transformation is. $\endgroup$ Apr 26, 2018 at 14:43
  • $\begingroup$ @MartinArgerami OK, I understand now. Thanks $\endgroup$
    – Skortya
    Apr 26, 2018 at 14:45
  • $\begingroup$ This is incorrect. As Najib said, $\alpha'(t)$, as a derivative, is a map $\mathbb{R} \to \mathbb{R}^2$ given by $s \mapsto sv$. For instance, your maps cannot be an injection for any $t$, for dimensional reasons. $\endgroup$
    – Aloizio Macedo
    Apr 26, 2018 at 14:55
  • $\begingroup$ You are totally right. I guess contravariant maps are more natural to me. Edited. $\endgroup$ Apr 26, 2018 at 14:57

The map is a submersion if the differential is injective. However the differential is a linear map.

For instance at a point $t_0\in\mathbb{R}$ the differential at $t_0$ is $d\alpha_{t_0}(x)=(3t_0^3,2t_0)x$.

This linear map is injective except at $t_0=0$ which gives the null map.


Keep in mind that a map between manifolds $f:M\to N$ is called an immersion at the point $p\in M$ when the differential $f_{*p}:T_pM\to T_{f(p)}N$ is $1-1$, that is when $\ker(f_{*p})=\{0\}$. $f$ is called an immersion when it is an immersion on every point of the domain manifold $M$.

Let's see if we can apply this to your map $a:\mathbb{R}\to\mathbb{R^2}$ with $a(t)=(t^3,t^2)$. For a point $p$ of $\mathbb{R}$, it is $a_{*p}:T_p\mathbb{R}\to T_{(p^3,p^2)}\mathbb{R}^2$ with $\displaystyle{a_{*p}\big{(}\frac{d}{dt}\vert_p\big{)}=\frac{dt^3}{dt}\vert_p\frac{\partial}{\partial x}|_{(p^3,p^2)}+\frac{dt^2}{dt}\vert_{p}\frac{\partial}{\partial y}\vert_{(p^3,p^2)}=3p^2\frac{\partial}{\partial x}\vert_{(p^3,p^2)}+2p\frac{\partial}{\partial y}\vert_{(p^3,p^2)}}$,

so for any vector $\lambda\frac{d}{dt}\vert_p\in T_p\mathbb{R}$:

$\displaystyle{a_{*p}\big{(}\lambda\frac{d}{dt}\vert_p\big{)}=3\lambda p^2\frac{\partial}{\partial x}\vert_{(p^3,p^2)}+2\lambda p\frac{\partial}{\partial y}\vert_{(p^3,p^2)}}$ Therefore, at the point $p=0$ you have $a_{*0}(v)=0$ for any $v\in T_0\mathbb{R}$. Hence $\ker(a_{*0})=T_0\mathbb{R}$, and $a$ is not an immersion at point $0$. Now if you exclude 0 and consider the domain of $a$ to be the submanifold $(0,+\infty)$, $a$ is indeed an immersion.


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