# $\forall r>0$ small enough $f(B(x_0,r))$ doesn't contain a ball around $f(x_0)$

Let $k<m$ and $f:\mathbb{R}^k \to \mathbb{R}^m$ be continuously differentiable around $x_0$. with $\operatorname{rank}Df(x_0) = k$.

Then $\forall r>0$ small enough $f(B(x_0,r))$ doesn't contain a ball around $f(x_0)$.

My attempt is the following:

Since $f$ is onto its image $\operatorname{Img}(f) \subset \mathbb{R}^m$ of degree at most $k$. Since $\operatorname{rank}Df(x_0) = k$, around $x_0$ the image of $f$ is a subset of $\mathbb{R}^m$ which is of dimension $k$. That is, around $x_0$ $f$ is injective (by the inverse function theorem).

The inverse function theorem applies for functions from $\mathbb{R}^k$ to itself, is this a legitimate use of this theorem, thinking of the image as a $k$ dimensional subset of $\mathbb{R}^m$? If so how do we show this?

Letting $U$ be the environment in which $f$ is injective, we take $r>0$ so that $B(x_0,r) \subset U$. If $\exists \delta > 0$ s.t $B(f(x_0),\delta) \subset f(B(x_0,r))$ then since $\dim(B(f(x_0),\delta)) = m$ we get $m \leq k$, in contradiction.

I'm not sure the use of the inverse function theorem is correct (due to the fact the image is in $\mathbb{R}^m$). And also the second part, couldn't I have concluded that without $f$ being injective in $U$?

• If you write $k<m$ then why write $\mathbb {R^k}$ and $\mathbb{R^m}$ instead of $\mathbb R^k$ and $\mathbb R^m$? That doesn't make sense so I changed it along with some other corrections. – Michael Hardy Jan 23 '17 at 15:09
• @MichaelHardy thanks! – Mariah Jan 23 '17 at 15:11
• @MichaelHardy any ideas on the question itself? – Mariah Jan 23 '17 at 15:45

HINT: Suppose $df(x_0)=\begin{bmatrix} I_k \\ 0 \end{bmatrix}$. Consider $g\colon\Bbb R^k\times \Bbb R^{m-k}\to\Bbb R^m$ given by $g(x,y) = f(x)+(0,y)$.
• Which part of my question is your hint addressed for? For my attempt to $f$ is injective around $x_0$? Also, do you mean to think of $\Bbb R^k\times \Bbb R^{m-k}$ as $\mathbb{R}^m$ and use the inverse function theorem? That may achieve $g$ as injective on some subset of $\mathbb{R}^m$, but why would it imply $f$ is too? – Mariah Jan 23 '17 at 22:48
• Ok, with that information, I'd like to understand if the last part of the proof is correct (my ball dimension argument) (where I claim this cannot happen since $f$ being injective implies the ball is of dimension $k$ and can't contain a subset of dimension $m$). I also don't understand how $B(f(x_0),\delta) \subset f(B(x_0,r))$ in any case since $k < m$. Do you have an idea for this? – Mariah Jan 23 '17 at 23:08
• Intuitively, the image of $f$ will (at least locally) be a $k$-dimensional submanifold in $\Bbb R^m$. But you now have $g^{-1}$ (on a small ball) to use. Could $f$ cover a whole ball in $\Bbb R^m$? – Ted Shifrin Jan 23 '17 at 23:16
• I still don't understand why not; If I differentiate around $(x_0,0)$, for example, I get a non singular Jacobian for $g$ and then $g$ is injective and onto from and on some open sets. Then $f$ is also injective on the restriction to $\mathbb{R}^k$. If we assume $f$ covers a ball, then the image of $g(x,0)$ covers that ball as well - I don't see a way to continue from here yet. – Mariah Jan 25 '17 at 12:18
Fix an $$x$$ where derivative is nonsingular. Notice that $$Df(x)(\mathbb{R}^k)$$ is a linear $$k$$-dimensional subspace of $$\mathbb{R}^m$$. Near the point $$x$$, $$f$$ is very close (little $$o$$ - close) to the derivative map. Thus, projection onto the affine plane $$f(x) + Df(x) (\mathbb{R}^k)$$, which is a copy of $$\mathbb{R}^k$$, we end up with a map between same dimensional Euclideans. Derivative agrees with the original derivative, hence is nonsingular. Now, we are in a position to impose IFT. If projection of $$f$$ is injective then we deduce that image of $$f$$ is (locally) $$k$$-dimensional.