Given a submersion $\pi:M\to N$ and a curve $\alpha:(-\varepsilon,\varepsilon)\to N$, find $\gamma:(-\varepsilon,\varepsilon)\to M$ I came across the following exercise:

Let $\pi:M\to N$ a submersion and $\alpha$ a smooth curve on $N$ with $\alpha(0)=q$. Show that given $p\in\pi^{-1}(q)$ and $v\in T_p M$ such that $(d\pi)_p v=\alpha'(0)$, there exists a smooth curve $\gamma:(-\varepsilon,\varepsilon)\to M$ with $\alpha=\pi\circ\gamma$, $\gamma(0)=p$ and $\gamma'(0)=v$.

The most natural way to approach this is using that a submersion has local sections. 
Indeed, for $q=\pi(p)$ there exists an open neighborhood $U$ and $\sigma:U\to M$ smooth such that $\pi\circ\sigma=\mathrm{Id}|_U$ and $\sigma(q)=p$. 
Then we define $\gamma:(-\varepsilon,\varepsilon)\to M$ given by $\gamma=\sigma\circ\alpha$ (where $\alpha(-\varepsilon,\varepsilon)\subset U$). This $\gamma$ satisfies $\alpha=\pi\circ\gamma$, $\gamma(0)=p$.
But $\gamma'(0)=(d\sigma)_q (d\pi)_p v$ and I don't know how to show that $$(d\sigma)_q (d\pi)_p v=v$$ (Which seems weird because it is the another order $(d\pi)_p (d\sigma)_q=\mathrm{id}_{T_q U}$, which is the identity).
Any help will be greatly appreciated!
 A: So I think you should be able to find a $\sigma$ with the added property that
$(d\sigma)_q(d\pi)_p v = \gamma'(0) = v$. Since submersions have local normal forms, just pretend that you're working with a submersion from $\mathbb{R}^M$ to $\mathbb{R}^N$ (with $M \geq N$) where $\pi(x_1, ..., x_M) = (x_1, ..., x_N)$. (Normal forms are nice in that way.) If you can prove what you want for this case, then you can build back up to your manifolds using charts and such.
For simplicity, you can even just consider the example
$\pi:\mathbb{R}^3\rightarrow\mathbb{R}^2$ defined by $\pi(x, y, z) = (x, y)$ and $\alpha(t) = (t, 2t)$ as a map from $(-\epsilon, \epsilon)$ into $\mathbb{R}^2$. We have $\alpha(0) = (0, 0) = q$. For any point $p = (0, 0, z)\in\pi^{-1}(q)$ and velocity vector $v = (0, 0, a)$ in $T_pM$ with $d\pi_p(v) = \alpha'(0)$, we would end up with the curve $\gamma(t) = (t, 2t, at + z)$ which would have $\gamma'(0) = (0, 0, a)$ and $\gamma(0) = 0, 0, z)$ as required.
I think you can generalize the ideas from this example to get your result. Hopefully that's helpful and gives some more intuition.
