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Let $f: X \to Y$ be a differentiable map of manifolds where $dim \;X = dim\;Y = n$. If $df(p)$ is nonsingular for all $p \in X$, show $f$ is an open map.


So here is what I was thinking:

As $df(p)$ is nonsingular for all $p \in X$, that is, it is an isomorphism between tangent spaces (bijective homomorphism), then for an open subset, $U \subset \mathcal{R}^n$, s.t. $U \overset{\text{$f$ is injective & continuous}}{\longrightarrow} \mathcal{R}^n$, then the image of $U$ ($f(U) \subset \mathcal{R}^n$) is open, as well as $f$ is a homeomorphism betweeen $U$ and $f(U)$ (homeomorphism meaning bijective and continuous $f$ and $f^{-1}$ [via the claim $df(p)$ is nonsingular]).

Is this, and quoting Brouwer's fixed point theorem enough to reasonably answer this question?

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up vote 4 down vote accepted

I don't think you need Brouwer's fixed point theorem. Careful reading of the statement of the Inverse Function theorem (for maps in $\mathbb R^n$) shows that the map is open. Indeed, the theorem gives you a neighborhood of $f(x)$ that is diffeomorphic to a neighborhood of $x$ via $f$. In particular, this neighborhood of $f$ is in the range of $f$.

The reduction of the manifold case to the Euclidean case is immediate: compose $f$ with charts on both sides, and unwind the definition of $df$ to see that the composition has invertible derivative matrix.

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Oh yeah, I guess I overshot the inverse function theorem a little... Thank you, @LVK ! –  nate Sep 25 '12 at 19:53

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