I am provided with a smooth map $g : N \rightarrow N^\prime $ between differentiable manifolds.
$N$ is assumed to be compact and connected.
Moreover, the differential $dg_x: T_x N \rightarrow T_{g(x)} N^\prime $ is surjective for any $x \in N$.
I want to show that the image of $N$ by $g$ is a path component of $N^\prime$.
I know that the components of $N'$ are submanifolds of $N'$ which have the same dimension as $N'$; but that's about it.
Can anyone give me a hint as how to employ surjectivity of $dg_x$ to get my result?
For just the hint: Try to prove that $dg_x$ surjective implies it is open.
If you want the full proof:
First things first: since $N$ is compact and $g$ is continuous, $g(N)$ is compact, hence closed. Now we show it is also open, and so it's a disjoint union of path components. Since $N$ is connected, it's a single path component.
We actually show that $dg_x$ surjective implies it is open. Being open is a local property, so without loss of generality we can assume $g: U \rightarrow V$, with $U$ and $V$ open sets in $\mathbb{R}^{n+k}$ and $\mathbb{R}^{k}$ respectively. Now the constant rank theorem tells us that, up to reparametrization, and eventually restricting $U$ again, $g$ is a linear projection, hence an open map.