We investigate the infinite version of the $k$-switch problem of Greenwell and Lov\'asz. Given infinite cardinals ${ \kappa}$ and ${ \lambda}$, for functions $x,y\in { }^{ \lambda}\kappa $ we say that they are totally different if $x(i)\ne y(i)$ for each $i\in { \lambda}$. A function $F:{ }^{ \lambda}\kappa \longrightarrow { \kappa} $ is a proper coloring if $F(x)\ne F(y)$ whenever $x$ and $y$ are totally different elements of ${ }^\lambda{ \kappa} $. We say that $F$ is weakly uniform iff there are pairwise totally different functions $\{ r_{ \alpha}:{ \alpha}<{ \kappa}\}\subset { }^{ \lambda}{ \kappa}$ such that $F(r_{ \alpha})={ \alpha}$; $F$ is tight if there is no proper coloring $G:{ }^{ \lambda}\kappa \longrightarrow { \kappa}$ such that there is exactly one $x\in { }^{ \lambda}{ \kappa}$ with $G(x)\ne F(x)$. We show that given a proper coloring $F:{ }^{ \lambda}{ \kappa}\to { \kappa}$, the following statements are equivalent $F$ is weakly uniform, there is a ${ \kappa} ^{ +}$-complete ultrafilter $\mathscr{ U}$ on ${ \lambda}$ and there is a permutation ${ \pi}\in Symm({ \kappa})$ such that for each $x\in { }^{ \lambda}{ \kappa}$ we have $$F(x)={ \pi}({ \alpha})\ \Longleftrightarrow \ \{ i\in { \lambda}: x(i)={ \alpha}\} \in \mathscr{ U}.$$ We also show that there are tight proper colorings which cannot be obtained such a way.
Categories: math.CO