Checking that $f(t) = (t^2-1,t^3- t)$ is an immersion.

Question

I am reading Tu's book "Introduction to Manifolds". In Example 11.8, it says that $f(t) = (t^2 - 1,t^3-t)$ is an immersion however it says that the equation $f'(t) = (2t,3t^2 -1) = (0,0)$ has no solution in $t$ and this means the map $f$ is an immersion. This confuses me since I thought the definition of an immersion was that the differential of the map $f$ is injective at every point and so since we cannot find a point $t_0$ such that $f(t_0) = (0,0)$, then wouldn't that imply that $f'$ is not injective? Or why is it that the equation $f'(t) = (2t,3t^2 -1) = (0,0)$ having no solution in $t$ implies this map $f$ is injective. I'm sorry if this question is really easy, it's just I'm still trying to grasp the concepts of immersion and submersion and I am having a hard time showing maps are either immersions or submersions.

Thanks for your help!

Answer 1

$f:\mathbb{R}^n\to\mathbb{R}^m$ is an immersion if the rank of the matrix of the linear map $d_xf:T_x\mathbb{R}^n\to T_{f(x)}\mathbb{R}^m$ is $n$ for every $x\in\mathbb{R}^n$. This is equivalent to $d_xf$ being injective for every $x\in\mathbb{R}^n$.

My notation for the differential (your $f'(t)$) is $d_tf$.

For your map $f:\mathbb{R}\to\mathbb{R}^2$, the fact that $d_tf=(0,0)$ has no solution means that the rank of $d_tf$ must be $1$. Therefore, $f$ is an immersion.

If $d_tf=(0,0)$ did have solutions, it would mean that at these values of $t$, the rank of $d_tf$ is $0$. Also note that since $d_tf$ is a $1\times2$ matrix, the rank can only be either $0$ or $1$.

Answer 2

You have $f : \mathbb{R} \rightarrow \mathbb{R}^2 $ defined as $f(t) =(f^1(t), f^2(t)) =(t^2 - 1,t^3-t)$. The differential at $t_0$ is a map $d_{t}f : T_{t} \mathbb{R} \rightarrow T_{f(t)}\mathbb{R}^2$. At any point $t \in \mathbb{R}$ , basis of the tangent space at that point $T_t \mathbb{R}$ is $\frac{d}{dt}$. So for any $t$, $$ d_tf : c\frac{d}{dt} \mapsto d_tf \bigg(c\frac{d}{dt}\bigg) \in T_{f(t)}\mathbb{R}^2 ,\quad \forall c \in \mathbb{R} $$ In coordinate representation $d_tf$ is the Jacobian matrix, which map the component of vectors in $T_t \mathbb{R}$ to components of vectors in $T_{f(t)}\mathbb{R}^2$, that is $$ c \mapsto \begin{pmatrix} \frac{df^1}{dt} (t) \\ \frac{df^2}{dt} (t) \end{pmatrix} \cdot c $$ Now to check that this map is an immersion we have to check $\text{Ker}(d_tf) = \{0\}$ or not. Suppose that $d_tf$ map any vector to zero vector in $T_{f(t)}\mathbb{R}^2$, that is $$ \begin{pmatrix} \frac{df^1}{dt} (t) \\ \frac{df^2}{dt} (t) \end{pmatrix} \cdot c = \begin{pmatrix} 0 \\ 0 \end{pmatrix} $$ But you have $f'(t) = (\frac{df^1}{dt} , \frac{df^2}{dt}) =(2t,3t^2 -1) $ So we have equations $$ \frac{df^1}{dt} = 2t =0 \implies t =0 $$ and $$ \frac{df^2}{dt} = 3t^2 -1 = 0 \implies t = \pm 1/\sqrt{3} $$ So there is no point $t \in \mathbb{R}$ that satisfy this equations. Which is means that the only way it image is zero vector is when $c=0$ which is zero vector in the domain $T_t\mathbb{R}$. Hence $Ker(d_tf) = \{0\}$. I hope this help.