Given any collection $\mathcal{C} = \{E\_1, E\_2, ..., E\_m\}$ of finite and nonempty discrete sets, is there a set $I$ such that
$$ \forall E\_k \in \mathcal{C}, \; E\_k \cap I \neq \emptyset,$$
and
$$ \forall i \in I, \; \exists E\_k \in \mathcal{C}, E\_{k} \cap I = \{i\} $$
? 


For example, for $E\_1 = \{1,2\}$, $E\_2 = \{2,3\}$ and $E\_3 = \{1,3\}$ with $m = 3$, a possible $I$ is $\{1,2\}$.


I've been trying to answer this question ever since stumbling upon it in my research, under a different form. I'm hardly a real mathematician so I submitted this problem to some nearby theoretical computer scientists hoping they'd tell me it was trivial, but no such luck.


All I've found until now are a few of rather trivial observations. To begin with, the second property to be verified by $I$ is equivalent to there being an injection $g$ from $I$ to $\{1, ..., m\}$ such that $E\_{g(i)} \cap I = \{i\}$ for all $i \in I$. 
Also, the problem can be formulated as a graph problem using a bipartite graph with $m$ vertices on one side representing the sets in the collection and vertices on the other side for each element in the union set $\bigcup\_{i=1}^m E\_k$, but graph theory isn't my forte either. Finally, notice that we only really need to consider the collections where $E\_1, ..., E\_m$ also verify the following properties:


* any one set $E\_j$ is not included in any other $E\_k$;
* there exists no element $e$ in exactly one set $E\_k$, in other words $e$ is either in none or in at least two sets of $\mathcal{C}$.



Hitting set problem variant
Let $I$ be a minimal set that intersects each $E\_j$, where minimal means that no point can be removed from $I$ without it no longer intersecting each $E\_j$. Take any $i\in I$. We know $i$ lies in some $E\_j$, otherwise $I$ was not minimal. If every $E\_j$ that contains $i$ also contains another element of $I$, then we can remove $i$ from $I$ and it still intersects each $E\_j$, so again $I$ would not be minimal. So some $E\_j$ that includes $i$ contains no other element of $I$.


