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Let $f\colon \mathbb R^2 \to \mathbb R^2$ be a continuous function such that $|f(p)-f(q)|\geq a|p-q|,\quad \forall p,q\in\mathbb R^2$ and $a>0$. Show that $f$ is injective and surjective (therefore has inverse) and that its inverse is continuous.

This is a problem from a metric space topology test that I did. The most important contents of the test were uniform convergence, equicontinuity, Arzelà-Ascoli Theorem, iterated functions, Stone-Weierstrass Theorem etc. The exercise is very simple and I believe it is possible to solve it by more elementary concepts.

I already showed that $f$ is injective so my problem is surjectivity. For surjectivity i've tried something like this: since $f$ is injective it follows that $\exists g \colon \mathbb R^2 \to \mathbb R^2$ such that $\left(g\circ f\right)(x)=x$. On the other hand $f$ is a right inverse for $g$ which implies that $g$ is surjective. I was trying to show that $g$ is injective but i did not get anything.

I have also shown that $f^{-1}$ is continuous (assuming $ f $ is surjective).

Thank you very much!

nom
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    The argument for $f$ surjective is going to have to rely on some property of $\mathbb{R}^2$ which is not true of general metric spaces. For instance, $f(x,y) = (2x, 2y)$ on ${ (x, y) \in \mathbb{R}^2 \mid x^2 + y^2 \ge 1 }$ is not surjective. – Daniel Schepler Jun 19 '18 at 17:55
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    I am uncertain about the duplicate target. The candidate is about a 1-dimensional case. I think having two dimensions complicates matters significantly. To wit, the answer to the dupe candidate relies on our ability to order the elements of $\Bbb{R}$, and we don't have a counterpart of that in two dimensions. At least I don't see it. – Jyrki Lahtonen Jun 20 '18 at 13:09

1 Answers1

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For first think you can observe that the map is closed because if you consider $C\in\mathbb{R}^2$ closed you have that if $y\in\mathbb{R}^2$ is in the clousure of $f(C)$ than there exists $\{x_n\}_n\subset C$ for which $f(x_n)\to y$. So $\{f(x_n)\}_n$ is a Cauchy sequence and by your property of $f$ you have that also $\{x_n\}_n$ is a Cauchy sequence and so it is convergent to an $x\in C$ because $C$ is closed. But $f$ is continuos and so $f(x_n)\to (f(x)=y)$. So $y\in f(C)$ and than $f(C) $ is closed.

Now my idea is that if you prove that $f$ is also an open map than it is surjective because each open-closed continuos map from a topological space to a connected topological space is surjective.

Another idea is to prove that the image of the function is dense in $\mathbb{R}^2$ because in this case $\mathbb{R}^2 =cl(f(\mathbb{R}^2))=f( \mathbb{R}^2)$

Federico Fallucca
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