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Consider a vector-valued function $f: [0,1]^n\rightarrow[0,1]^n$. Write $f(x)=\{f_1(x), ..., f_n(x)\}$ with $x\in[0,1]^n$, where the $f_i: [0,1]^n\rightarrow[0,1]$ are continuous functions with the following properties:

1) $f_i(\{x_1, x_2, ..., x_n\})=0$ if $x_i=0$.

2) $f_i(\{x_1, x_2, ..., x_n\})=1$ if $x_i=1$.

Fix $p\in[0,1]^n$. I want to prove that there exists a solution $x\in[0,1]^n$, such that $f(x)=p$.

If feel that this should be true, since by the intermediate value theorem, we know that for fixed $x_2$, ..., $x_n$ we can find $x_1$ such that $f_1(\{x_1, x_2, ..., x_n\})=p_1$ (and similarly for $f_2$ we can find an $x_2$ if we fix the other $x_i$'s).

Any help is appreciated.

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This has its own Wikipedia entry Poincaré–Miranda theorem and is similar to the Brouwer fixed-point theorem.

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  • $\begingroup$ Exactly what I needed $\endgroup$ – MthQ Nov 10 '15 at 16:19
  • $\begingroup$ It surprises me that it seems to have taken 57 years between the conjecture and the proof. Is there something wrong with the following elementary argument? Take the boundary of the domain and continuously contract it onto the origin O. Let h be the signed distance from p to the nearest point on the image of the contracting surface. The sign indicates whether p is inside (-) or outside (+). Once the surface is fully contracted to the origin, h is either zero, in which case O is the solution, or positive. If the latter, then by the intermediate value theorem h had a zero somewhere. $\endgroup$ – Ben Crowell Nov 10 '15 at 16:21
  • $\begingroup$ The problem is in making the sign well-defined. $\endgroup$ – user78588 Nov 10 '15 at 16:27
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    $\begingroup$ Continuous maps can be very complicated! So it is not at all clear that your suggestion defines a continuous function. More or less this is the en.wikipedia.org/wiki/… Jordan–Brouwer separation theorem, which is hard! $\endgroup$ – user78588 Nov 10 '15 at 16:37
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    $\begingroup$ The reference on the wikipedia page says Poincare did prove the Poincare-Miranda theorem. (Although I don't know if his proof without homology/degree theory would be accepted today.) $\endgroup$ – user78588 Nov 10 '15 at 16:43
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This is a straightforward application of degree theory.

First of all, by induction you can reduce to the case when $p\in(0,1)^n$.

Then, note that the a linear homotopy between $f$ and the identity map such that the image of the boundary of the unit box remains in the boundary. Thus

$$\deg(f,(0,1)^n,p)=\deg(I,(0,1)^n,p)=1$$.

It follows that $f(x)=p$ has at least one solution.

Reference: Ambrosetti and Machiodi Nonlinear Analysis and Semilinear Elliptic Equations

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