1
$\begingroup$

Let $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 $U$?

  • 1
    If you write $k$\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. – 2017-01-23
  • 0
    @MichaelHardy thanks! – 2017-01-23
  • 0
    @MichaelHardy any ideas on the question itself? – 2017-01-23

2 Answers 2

2

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 $g(x,y) = f(x)+(0,y)$.

  • 0
    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? – 2017-01-23
  • 0
    Yes to everything. Write out the definition of injective :) – 2017-01-23
  • 0
    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? – 2017-01-23
  • 0
    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 $\Bbb R^m$? – 2017-01-23
  • 0
    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. – 2017-01-25
  • 0
    If $g(x,0)$ covers a ball, how is $g$ a bijection to that ball? – 2017-01-25
0

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.

For detailed argument refer to https://mathoverflow.net/a/323429/121665 .