Given a graph $G=(V,E)$, here is a Linear Relaxation of the edge cover polytope:
(1) For each $v \in V, \sum_{e \in \delta(v)} x_e \geq 1.$
(2) For each $e \in E$, $0 \leq x_e \leq 1.$
Here $\delta(S)$ is the set of edges with exactly one endpoint in $S$.
It is known that the above polyhedron has integral extreme points when G is a bipartite graph. It is also known that, in a general graph, any extreme point of the polyhedron is half-integral - that is, it's coordinates are either integers or multiples of 1/2.
I am trying to prove the half-integrality on general graphs by using a transformation to a new bipartite graph $\tilde G$, obtained from $G$ by placing a new vertex 'in the middle' of each edge. Note that this makes $\tilde G$ a bipartite graph.
However, I'm struggling with the details. Given an extreme point of the polytope for $G$, I need to somehow create a feasible point for $\tilde G$. Then I can apply the integrality and transform it back somehow.
Can anyone help me out with how to proceed?
I am not sure how to make your $\tilde{G}$ work, but there is a typical auxiliary bipartite graph for this situation: let $G'$ be the graph whose vertices are two copies $U$ and $V$ of $V(G)$ where for $u \in U$ and $v \in V$, $uv$ is an edge in $G'$ if and only if the vertices corresponding to $u$ and $v$ in $G$ are adjacent in $G$.
In this way, $G'$ is like a bipartite doubling of $G$, where every edge turns into two, which makes the correspondence between edge covers of $G$ and edge covers of $G'$ easy. Given an edge cover of $G$, we can form an edge cover of $G'$ by giving the weight of edge $e \in E(G)$ to each of its two copies in $G'$. Given an edge cover of $G'$, we can form an edge cover of $G$ by assigning to $e \in E(G)$ the average weight of its two copies in $G'$.
All that remains is to check that these operations do indeed form edge covers and that an integral edge cover of $G'$ corresponds to a half-integral edge cover of $G$.