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Show that at a zero $x$, the derivative $d(\vec v_x ) :T_x (X)→R^N$ actually carries $T_x (X)$ into itself.

I know that we have the vector field $\vec v:X\to R^n$. If $X=R^n \times \{0\}$, then the claim is true obviously..

The book tells me to find a local parametrization, so let $\phi:U\to R^n$ where $U$ is the open neighborhood in $X$. But I can't see how this gonna help me

XiaoXiao Zhen
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    I think that's not true without additional assumptions. Take, for example $X = \mathbb R.$ Then $TX = \mathbb R \times \mathbb R$, and $X$ embeds into $TX$ via $X \simeq \mathbb R \times {0} \subseteq TX.$ Let's denote this embedded copy of $X$ in $TX$ by $\widetilde X \subseteq TX$. This implies the following. Since $\dim TX = 2,$ we have $T_{(0,0)}TX = \mathbb R \times \mathbb R,$ and by construction $T_0\widetilde X = \mathbb R \times {0} \subseteq T_{(0,0)}TX.$ Now, consider the vector field $v:X \rightarrow TX,p \mapsto (p,p).$ $v$ has a zero at $p = 0$. – jflipp Nov 29 '14 at 18:23
  • [continued] Moreover, consider the curve $\gamma:\mathbb R \rightarrow X, t \mapsto t.$ We have $\gamma(0) = 0$ and $\gamma'(0) = 1 \in T_0X.$ If the claim from your question were true, we would have $(v\circ\gamma)'(0) \in T_0\widetilde X.$ But we calculate $(v\circ\gamma)'(0) = (1,1) \notin T_0\widetilde X.$ – jflipp Nov 29 '14 at 18:26
  • I have the same feeling and that gave me a headache for 2 days. I copied the claim exactly word by word from the book, so I don't think I left out some thing important. – XiaoXiao Zhen Nov 29 '14 at 18:32

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