TITLE: Intermediate value for a vector-valued function
QUESTION [7 upvotes]: 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.

REPLY [7 votes]: This has its own Wikipedia entry Poincaré–Miranda theorem and is similar to the Brouwer fixed-point theorem.

REPLY [5 votes]: 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