# Chromatic self-maps on finite complete linear hypergraphs

Let $$X\neq\varnothing$$ be a set and let $${\cal E}\subseteq {\cal P}(X)\setminus\{\varnothing\}$$ be a collection of non-empty subset. We say that a map $$f: {\cal E}\to X$$ is a chromatic self-map if

1. $$f(e) \in e$$ for all $$e\in {\cal E}$$, and

2. if $$e_1\neq e_2 \in {\cal E}$$ and $$e_1\cap e_2 \neq \varnothing$$, then $$f(e_1)\neq f(e_2)$$.

Consider for $$n\in \mathbb{N}$$, $$n>1$$ the set $$[n] = \{1,\ldots,n\}$$. We say that a collection $${\cal C}\subseteq {\cal P}([n])$$ is complete linear if

1. $$|{\cal C}|=n$$,
2. $$c\neq d \in {\cal C} \implies |c\cap d| = 1$$, and
3. $$|c| > 1$$ for all $$c\in{\cal C}$$.

Given $$n>1$$, does every complete linear collection on $$[n]$$ have a chromatic self-map?

By domotorp's answer https://mathoverflow.net/q/362229 to a previous question, De Bruijn-Erdos theorem characterizes complete linear collections as near-pencils or finite projective planes.

The image of a chromatic self-map is a system of distinct representatives. This is exactly the setting of Hall's marriage theorem (https://en.wikipedia.org/wiki/Hall%27s_marriage_theorem). It is a folklore exercise that Hall's condition is satisfied for finite projective planes. And it is straightforward to define a chromatic self-map for the near-pencil.