Prove that a mapping $f:[-1,1]^2\to\mathbb R^2$ with certain properties has the value $(0,0)$. 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


*

*for each $x\in[-1,1]$ it holds $f_2(x,1)>0$, 

*for each $y\in[-1,1]$ it holds $f_1(-1,y)<0$, 

*for each $x\in[-1,1]$ it holds $f_2(x,-1)<0$ and

*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.
 A: 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. 
