Odd circles and doubly covered edges

Question

Let $G= (V,E)$ be a simple, undirected graph. A set $C\subseteq V$ is said to be an edge cover if for all $e\in E$ we have $e\cap C \neq \emptyset$.

It is most satisfying to find an edge cover $C$ such that every edge is covered exactly once (that is, $|C\cap e| = 1$ for all $e\in E$) -- but unfortunately this is only possible if $G$ does not contain odd circles (i.e. $\chi(G) \leq 2$).

However: given any graph $G$, is it always possible to find an edge cover $C\subseteq V(G)$ such that the number of edges $e\in E$ with $e\subseteq C$ is at most the number of odd circles in $G$?

Answer 1

Consider a triangular prism with $6$ vertices and $9$ edges. You need to pick two vertices from each triangle and that also chooses both ends of another edge. That is a problem if you view it as two odd cycles (the triangles. ) I suppose there are 8 odd cycles if you include the $5$-cycles.

Of course with $n$ edges all on one vertex you can get a satisfying edge cover using $1$ vertex and another using $n$ vertices.

Similarly, there many be many different (size) sets $D$ of edges including at least one from each odd cycle. For each one there is an edge cover that uses both vertices of each edge in $D$ but only one vertex from every other edge (except those with both ends also on edges in $D$)

LATER

Actually it is somewhat more complicated than that. And the original question is still not answered. I reformulated the question with some variations here.