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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!

user110320
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1 Answers1

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$\gamma$ 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 $1$.

A. Goodier
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  • Thanks!! So then, the fact that the only value that maps the derivative to $0$ is not equivalent to the condition that $\gamma '(t)$ is maximal rank? Or how are these conditions are related. It's just I always thought about finding injective maps by showing the only value that mapped to $0$ was the $0$ element in the domain. – user110320 Nov 03 '17 at 00:22
  • $\gamma'(0)$ is the zero map, which is not injective. A linear map is injective if and only if its matrix has maximal rank. So if there is a value of $t$ such that the rank of the Jacobian matrix is not maximal, for that value of $t$, $\gamma'(t)$ is not injective. – A. Goodier Nov 03 '17 at 07:30