one vertex less) than $G$, and $\beta_1(G/e) = \beta_1(G)$, so $\sigma(G/e) =
\sigma(G)$. The graph $H/e$ contains one edge less than $H$, so $\alpha(H/e) = -\alpha(H)$. Now by the induction hypothesis $$\SumSub(\alpha,R_e^+) = -\SumSub(\alpha,G/e) = -(-1)^{k-1} \sigma(G/e) =
(-1)^k \sigma(G),$$ and then implies $$\SumSub(\alpha,G) = \SumSub(\alpha,R_e^-) + \SumSub(\alpha,R_e^+) = 0 + (-1)^k
\sigma(G) = (-1)^k \sigma(G).$$

### $G$ is strongly connected and contains no crucial edges.

Let $e = [ab]\in G$ be an edge and not a loop: $b \ne a$. Recall that $G_e^{\vee}$ will denote a graph obtained from $G$ by reversal of the edge $e$. Since $e$ is not a crucial edge, $G \setminus \{e\} = G_e^{\vee}
\setminus \{e\}$ is strongly connected. So $G_e^{\vee}$ is strongly connected, too, implying $\sigma(G_e^{\vee}) = \sigma(G)$.

\[Lm:Reversal\] If the graph $G$ is strongly connected and $e = [ab] \in G$ is not a crucial edge then $\SumSub(\alpha,G) = \sigma(G)$ if and only if $\SumSub(\alpha,G_e^{\vee}) = \sigma(G_e^{\vee}) = \sigma(G)$.

Acyclic subgraphs $H \subset G$ are split into five classes:

1.  \[It:NoEAtoB\] $e \notin H$, but $a \succeq b$ in $H$ (that is, $H$ contains a directed path joining $a$ with $b$).

2.  \[It:NoEBtoA\] $e \notin H$, but $b \succeq a$ in $H$.

3.  