A smooth topological embedding that is not a smooth embedding

Question

I am reading Introduction to smooth manifolds by Lee. In example 4.18 it says that the map $\gamma: \mathbb{R} \rightarrow \mathbb{R}^2$ given by $\gamma(t) = (t^3,0)$ is not a smooth embedding, since $\gamma '(0)= 0$. I am confused why the fact that $\gamma '(0)= 0$ implies that $\gamma$ is not a smooth immersion. I know that $\gamma ' = (3t^2,0)$ and so the only value that gives $0$ is $t = 0$, doesn't this say that $\gamma '$ is injective? Can anyone clarify this for me? Thanks!

Answer 1

$\gamma$ is an immersion if for every $t\in\mathbb{R}$, $\gamma'(t)$ is injective. Equivalently, the rank of the $\gamma'(t)$ must be maximal for every $t$. If $\gamma'(0)=0$, then the derivative has rank $0$ at $t=0$, so $\gamma$ is not an immersion. For $t\neq 0$, the rank of $\gamma'(t)$ is $1$.