Let $H=(V,E)$ be a [hypergraph](https://en.wikipedia.org/wiki/Hypergraph) that is a finite Fano plane, that is, $V$ is a finite set and $E$ has the following properties:


1. for $e\_1\neq e\_2\in E$ we have $|e\_1|=|e\_2|$, as well as $|e\_1\cap e\_2|=1$, and
2. for $v\neq w\in V$ there is a (unique) $e\in E$ with $\{v,w\}\in e$.


Is there always an injective map $f:E \to V$ with $f(e)\in e$ for all $e\in E$?



Injective choice function for finite Fano planes
Yes, there is always such a map. Let $k$ be the number of vertices in each edge of $H=(V,E)$. Consider an arbitrary vertex $v \in V$ and choose $e \in E$ such that $v \notin e$. For each $w \in e$ there is a unique edge $f\_w$ such that $\{v,w\} \subseteq f\_w$. Moreover, for distinct $w,w' \in e$, $f\_w \neq f\_{w'}$. Since every edge containing $v$ must also intersect $e$, we conclude that there are exactly $|e|=k$ edges which contain $v$. Now, let $G$ be the bipartite graph with bipartition $(E,V)$, where $e \in E$ is adjacent to $v \in V$ if and only if $v \in e$. Since $G$ is $k$-regular, $G$ has a perfect matching by Hall's theorem, which gives the required injective map (it is bijective actually).


