# Set of points such that derivative is injective is an open set

Suppose that $$A$$ and $$B$$ are finite dimensional vector spaces. Let $$U \subseteq A$$ be open and $$f:U \to B$$ be $$C^{\infty}$$. Show that $$\{a \in U : (Df)_a \text{ is injective}\}$$ is open.

I tried showing that the rank of the $$(Df)_x$$ is constant for x in a neighborhood of $$a$$ but I don't think I have enough tools for that. I suppose this somehow follows from the inverse function theorem but I just don't see it. My guess is that it being injective implies that it is nonzero, and so it must be nonzero on a neighborhood, but I don't know if that's correct.

• The rank may increase locally just as the rank of the $1\times 1$ matrix $x$ is zero at $x=0$ but is one for $x$ nearby. – copper.hat Mar 27 at 5:03

Note that $$Df(x)$$ is injective iff $$Df(x)^T Df(x)$$ is invertible and the map $$x \mapsto Df(x)^T Df(x)$$ is continuous.
Since $$\det$$ is continuous and $$\det (Df(x)^T Df(x)) > 0$$, we see that it is strictly positive (we only care about being non zero) in a neighbourhood of $$x$$ and so $$Df(y)$$ is injective for $$y$$ in some neighbourhood of $$x$$.
Let $$M(n,m)$$ be the space of matrices of size $$n\times m$$. The set $$U$$ of matrices with maximal rank is clearly open, since its complement is the zero-locus of all the maximal minors (in particular, closed).
Now if you pick a basis for $$A$$ and $$B$$, the map $$Df$$ can be considered as $$Df:A\to M(\dim B,\dim A)$$ and the points where $$Df$$ is injective is exactly $$Df^{-1}(U)$$, which is open.
As stated, you should consider the cases where $$\dim A> \dim B$$ and $$\dim B<\dim A$$ separately, but I am sure you can work that out.