Is it true that the edge weight of a bridge equals to its least-squares potential difference within a graph? Consider a (not necessarily connected) directed multigraph $G=(V,E,t,h,w,u)$, where V is the vertex set, E is the edge set,
$t: E \rightarrow V$ is the function of edge tails, $h: E \rightarrow V$ is the function of edge heads, while $w: E \rightarrow \mathbb{R}^+$ and $u: E \rightarrow \mathbb{R}$ are two (fixed) weight functions.
Suppose that $\pi: V \rightarrow \mathbb{R}$ is a potential function minimizing the following (least-squares) objective function:
$\mathit{\Omega}( \mathit{\Pi})=\sum\limits_{e \in E}{w_e{[u_e-\mathit{\Pi}(h_e)+\mathit{\Pi}(t_e)]}^2}$.
Is it true that if $e^* \in E$ is a bridge (i.e., an edge whose deletion increases the number of components within G - multiple edges are not bridges here), then $\pi(h_{e^*})-\pi(t_{e^*})=u_{e^*}$? If so, how to prove this conjecture?

Suppose that $C$ is the connected component containing $h_{e^*}$ in $G - e^*$. Then adding the same constant to $\pi(v)$ for all vertices $v$ in $C$ changes $w_{e^*}[u_{e^*}  - \pi(h_{e^*}) + \pi(t_{e^*})]^2$ and no other term of $\Omega(\pi)$.
In particular, by adding the right constant, we can make $u_{e^*}  - \pi(h_{e^*}) + \pi(t_{e^*}) = 0$ and set this term to $0$, and since $\pi$ is supposed to be a minimizer, it must the the case that $u_{e^*}  - \pi(h_{e^*}) + \pi(t_{e^*})$ is already $0$.
