Let $G=(V,E)$ be an undirected graph. What I call a selection function of $G$ is a partial function $f:V \to E$ such that for every node $v$, if $f(v)$ is defined then it is one of the adjacent edges to $v$. In other words, every node can either select no edge or select an edge that is adjacent to it. The image of $f$ is simply $\mathrm{Im}(f)= \{e \in E \mid \exists v \in V \text{ s.t. } e=f(v)\}$ i.e., the set of edges that are selected.
I am interested in the following counting problem:
INPUT: a graph $G$
OUTPUT: $|\{\mathrm{Im}(f) \mid f \text{ is a selection function of }G\}|$
Question 1: does it have an established name? (It seems to be to be rather natural so maybe it has already been considered)
Question 2: is this problem #P-hard? (Ideally, I would like it to be hard on bipartite graphs)
Note that this problem is in #P, which is not entirely obvious at first glance. To show this, it is enough to show that for every $S \subseteq E$, we can determine in PTIME if $S$ is the image of some selection function. This can be done as follows. First, construct the bipartite graph $G_S$ with nodes in the left partition being the nodes of $G$ and nodes in the right partition being edges in $S$, and connect $v$ in the left partition to $e$ in $S$ if $v$ is an endpoint of $e$. Then, compute the size $m$ of a maximum cardinality matching of $G_S$. Accept if $m=|S|$, and reject otherwise (observe that $m$ is always $\leq |S|)$. It is easy to check that this accepts iff $S$ is the image of some selection function.