Prove the existence of such an immersion This question was asked in my quiz on smooth manifolds and I was unable to solve it. I tried it again at home and still not able to solve it. Question is:

Let $\sigma$ be an integral curve for a vector field X on a manifold M, with $\sigma(0)=p$. If $X_p\neq 0$, show that  there exists $\epsilon >0$ such that $\sigma : (-\epsilon , \epsilon) \to M $ is an immersion.

What I could prove is that: If $p\in M$ then there exists $\epsilon >0$ and an integral curve of X such that $\sigma : (-\epsilon ,\epsilon)\to M$ with $\sigma(0)=p$.
But How can I prove that a $\sigma$ also exists which is an immersion.
Can you please tell?
 A: First of all one has to be clear what does it means integral curve, so it would be useful to remember this concept:
An integral curve of starting point $p\in M$  of a vector field $X\colon M\to TM$ is a curve $ \sigma \colon (-\epsilon, \epsilon)\to M$ such that
$\sigma(0)=p$
$\sigma’(t)=X(\sigma(t))$
The first important theorem is the following one:
There exists always a unique integral curve of starting point $p$ related to a  vector field $X$
The proof of the theorem is quite easy, because in a coordinate chart $(U,\phi)$, the condition became
$(\sigma\circ \phi)^{-1}(0)=0$
$(\sigma\circ\phi^{-1})’(t)= \sum_i d\phi_{i,\sigma(t)}(X(\sigma(t))\frac{\partial}{\partial \phi_{i,\sigma(t)}}$
In other words
$(\sigma\circ \phi)^{-1}(0)=0$
$\frac{d}{dt}(\sigma\circ\phi^{-1})_i= d\phi_{i,\sigma(t)}(X(\sigma(t))$ for each $i=1,\dots n$
This is a Cauchy problem in $t,y_1=\sigma\circ\phi^{-1}_1,\cdots ,y_n= \sigma\circ \phi^{-1}_n$ variables and so by locally existence and uniqueness Cauchy-Lipschitz theorem (or equivalently Picard-Lindelöf theorem) there exists a unique solution $\gamma(t)\colon (-\epsilon, \epsilon)\to \phi(U)$ for an $\epsilon$ enough small.
Then our integral curve will be $\sigma:=\phi^{-1} \circ \gamma $ and we have done.
But now you have an important condition to your problem:
$X_p\neq 0$
So by continuity (X is a smooth map) of $X$ there exists an appropriate neighbourhood of $p$ such that $X_q\neq 0$.
If you intersect your integral curve $\sigma$ with that neighbourhood then you will get $d\sigma_t(d/dt)=\sigma’(t)=X(\sigma(t))\neq 0$
But $d/dt$ is a basis of $T_t(-\epsilon,\epsilon)$ and so the differential of $\sigma$ in $t$, $d\sigma_t$, has to be injective, that is the definition of a smooth map to be an immersion.
Thus we have done. If there is some mistake or error please tell me and we will discuss
