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The mapping $f:[-1,1]^2\to\mathbb R^2$ is known to be continuous. Also the image of the upper edge of the rectangle is contained in the upper half-plane, the left edge's image is contained in the left half-plane, and so on for the bottom and right edge. Formally speaking this means:

If $f_1,f_2:[-1,1]^2\to\mathbb R$ are defined to be the components of $f$, i. e. $$f(x,y)=(f_1(x,y),f_2(x,y))\,,$$ then the following conditions hold

  1. for each $x\in[-1,1]$ it holds $f_2(x,1)>0$,
  2. for each $y\in[-1,1]$ it holds $f_1(-1,y)<0$,
  3. for each $x\in[-1,1]$ it holds $f_2(x,-1)<0$ and
  4. for each $y\in[-1,1]$ it holds $f_1(1,y)>0$.

Under these assumptions prove that there exist $x_0,y_0\in[-1,1]$ such that $f(x_0,y_0)=(0,0)$. This appears to be obvious intuitively, but I'm looking for a formal and rigorous proof.

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  • $\begingroup$ If $f$ has two components, shouldn't it map to $\mathbb{R}^2$? $\endgroup$ – Roland Feb 5 '14 at 13:02
  • $\begingroup$ @Roland Yes, I'll correct it. Thank you. $\endgroup$ – Ralph Tandetzky Feb 5 '14 at 13:08
  • $\begingroup$ I am not so sure this is true. For example, $f$ could map continuously the unit square onto a "square annulus", which does not include the origin. I don't see any reason why such a mapping should not exist. $\endgroup$ – Giuseppe Negro Feb 5 '14 at 13:24
  • $\begingroup$ @GiuseppeNegro: It's probably impossible because of the mean value theorem or something similar. $\endgroup$ – Najib Idrissi Feb 5 '14 at 13:26
  • $\begingroup$ @GiuseppeNegro Mind the boundary conditions. How would you map onto a square annulus and still retain them? $\endgroup$ – Ralph Tandetzky Feb 5 '14 at 13:29
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Here's a really high powered proof by contradiction:

Let $g:\mathbb{R}^2 \setminus (0,0) \to [-1,1]^2$ be the mapping $$ g(x,y) = \begin{cases} \frac{1}{|x|}(x,y) & |x| > |y| \\ \frac{1}{|y|} (x,y) & |x| \leq |y| \end{cases} $$ Observe that the image of $g$ is the boundary of $[-1,1]^2$, and $g$ is continuous.

Assume $f$ is as given, such that $f^{-1}((0,0)) = \emptyset$. Let $h(x,y) = g(-f(x,y))$.

By definition,

  • $h$ is continuous, as $g$ is continuous, and $-f$ is a continuous function whose image is contained in $g$'s domain.
  • $h(x,y) \neq (x,y)$ if $\max(|x|,|y|) < 1$. This is because $(x,y)$ is in the interior of the square, while $h(x,y)$ is on the boundary.
  • $h(x,y) \neq (x,y)$ if $\max(|x|,|y|) = 1$: this is because if $(x,y)$ is on the upper half plane and on the boundary, $h(x,y)$ is in the lower half plane; similarly for the other boundaries.

Thus $h:[-1,1]^2 \to [-1,1]^2$ is a continuous self-mapping of a convex, compact region in $\mathbb{R}^2$ that has no fixed points; this contradicts Brouwer's fixed point theorem.

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    $\begingroup$ Awesome! Just for clarity it would be nice to make clear that the proof is indirect at the beginning by saying something like "Assume that $f$ has no zero". $\endgroup$ – Ralph Tandetzky Feb 5 '14 at 16:22

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