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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?

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Just to be clear, you don't want to show that $g(N)$ is just contained in a path component of $N'$ but actually is a path component. Correct? – Michael Albanese Jan 8 '14 at 11:05
@MichaelAlbanese Well, yeah... It's already connected; hence contained in a path component since $N$ is assumed connected. I wanna show it's a component of $N'$. – Mia Jan 8 '14 at 11:07
That's why I asked. – Michael Albanese Jan 8 '14 at 11:08
@MichaelAlbanese Do you suppose we could show that $g$ itself is onto? – Mia Jan 8 '14 at 13:11
up vote 3 down vote accepted

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.

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