# Injective edge choice functions in linear hypergraphs

A linear hypergraph is a hypergraph $$H=(V,E)$$ such that

1. for $$e\in E$$ we have $$|e|\geq 2$$, and
2. if $$e\neq e_1\in E$$, then $$|e\cap e_1| \leq 1$$.

An injective edge choice function of a linear hypergraph is an injective map $$f:E\to V$$ such that:

for all $$e\in E$$ we have $$f(e)\in e$$.

Obviously, if $$|E|>|V|$$ there cannot be such a function.

Question. If $$H=(V,E)$$ is a linear hypergraph with $$V$$ finite and $$|e|\geq 3$$ for all $$e\in E$$, does $$H$$ necessarily have an injective edge choice function?

• Maybe I misread something, but does 2. hold if $e\cap e_1=\emptyset$ anyway? So the question is if $H$ has a set of distinct representatives.Surely it does not, if $|E|>|V|$. – Péter Komjáth May 15 at 15:26
• You are right @PéterKomjáth! I should formulate this in a more elegant way. - An interesting subquestion is whether $|e|>2$ for all $e\in E$ implies $|E|\leq |V|$? – Dominic van der Zypen May 15 at 17:52
• No, there's an affine plane on 9 points with 12 lines, and every line has length 3. – Bullet51 May 16 at 2:51
• Thanks - can you post this as an answer such that we can close this thread? – Dominic van der Zypen May 16 at 7:34

Even $$|E|\leq|V|$$ does not hold: An affine plane contains more lines than points.