A [hypergraph](https://en.wikipedia.org/wiki/Hypergraph) $H=(V,E)$ is said to be *complete and linear* if


1. whenever $e\_1\neq e\_2\in E$ then $|e\_1\cap e\_2|=1$, and
2. for $v,w\in V$ there is $e\in E$ such that $\{v,w\}\subseteq e$.


Assuming that $V$ is infinite, is there an injective function $f:E\to V$ such that $f(e)\in e$ for all $e\in E$? (There is a Hall-marriage-theorem argument that the [answer is positive for finite $V$](https://mathoverflow.net/questions/374766/injective-choice-function-for-finite-fano-planes).)



Injective choice function for infinite complete linear hypergraphs
After the discussion above, here is what I think is the cleanest proof and it has the property that $f$ is bijection (unless there is an edge of order 1).


If there is an edge of order 1, then we must have $E=\{\{v\}, V\}$ for some $v\in V$, in which case the desired injection is trivial. If there is an edge of order 2, then $H$ must be a near-pencil and $f$ can easily be found and is necessarily a bijection.


So suppose every edge has order at least 3 in which case $H$ is a non-degenerate projective plane where $\kappa:=|V|=|E|$, every edge has the same cardinality $\lambda$, and every vertex has degree $\lambda$. Now let $B$ be a bipartite graph with parts $V$ and $E$ such that $\{v,e\}\in E(B)$ if and only if $v\in e$. Note that $B$ is a $\lambda$-regular bipartite graph and thus has a perfect matching by [bof's answer to my question](https://mathoverflow.net/questions/375823/perfect-matchings-in-infinite-regular-bipartite-graphs). This perfect matching is the function $f$ you are looking for and $f$ is in fact a bijection.


