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 $X=[-n,n]^2$. 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$has a fixed point. Is that trivial? Thank you in advance, Valerio 2 misprint fixed: y+1 replaced to (x,y+1) I have casually almost (i.e. up to details that shoud work) proved the following discrete version of Brouwer's fixed point theorem. I should have obtained this result as a corollary of quite complicated things and I do not understand if the result is trivial and can be easily proved directly or it deserves to be stressed. I would like to hear your opinion about that. Let$n\geq1$be a fixed integer and denote by$X=[-n,n]^2$. Given$(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),(y+1)$. (x-1,y),(x,y),(x+1,y),(x,y-1),(x,y+1)$. 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 $f(A(x,y))\subseteq A(f(x,y))$. Then $f$ has a fixed point.

Is that trivial?

Valerio

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# A Brouwer fixed point theorem on finite sets

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 $X=[-n,n]^2$. Given $(x,y)\in X$ I denote by $A(x,y)$ the set formed by the following at most five points: $(x-1,y),(x,y),(x+1,y),(x,y-1),(y+1)$. 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 $f(A(x,y))\subseteq A(f(x,y))$. Then $f$ has a fixed point.

Is that trivial?