I have casually almost (i.e. up to details that shoud work) proved the following discrete version of Brouwer's fixed point theorem. I should have obtained this result as a corollary of quite complicated things and I do not understand if the result is trivial and can be easily proved directly or it deserves to be stressed. I would like to hear your opinion about that.


Let $n\geq1$ be a fixed integer and denote by $X=[-n,n]^2\subseteq\mathbb Z^2$. Given $(x,y)\in X$ I denote by $A(x,y)$ the set formed by the following *at most* five points: $(x-1,y),(x,y),(x+1,y),(x,y-1),(x,y+1)$. At most means that if one of those points does not belong to $X$, I will not consider it.


The result would be: let $f:X\rightarrow X$ such that for all $(x,y)\in X$ one has $f(A(x,y))\subseteq A(f(x,y))$. Then $f$ has a fixed point.


Is that trivial?


Thank you in advance,


Valerio



A Brouwer fixed point theorem on finite sets
I believe the following is a counter-example:


$f: \lbrace -1,0,1\rbrace^2 \to \lbrace -1,0,1\rbrace^2$


$\forall x:$


$ f(-1,x) = (1,x)$


$f(0,x)=(1,x)$


$f(1,x)=(0,x)$


