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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.

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  • $\begingroup$ 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. $\endgroup$ – copper.hat Mar 27 at 5:03
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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 $x$.

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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.

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