where $G_i$ are connected components. A subgraph $H \subset G$ is acyclic if and only if the intersection $H_i
{\stackrel{\mbox{\scriptsize def}}{=}}H \cap G_i$ is acyclic for all $i$. Hence $\alpha(H) = \alpha(H_1)
\dots \alpha(H_m)$, and therefore $\SumSub(\alpha,G) = \SumSub(\alpha, G_1)
\dots \SumSub(\alpha, G_m)$. By the induction hypothesis $\SumSub(\alpha,
G_i) = (-1)^{k_i}\sigma(G_i)$ where $k_i$ is the number of edges of $G_i$. So $$\SumSub(\alpha,G) = \SumSub(\alpha, G_1) \dots \SumSub(\alpha, G_m) =
(-1)^{k_1 {+ \dots +} k_m} \sigma(G_1) \dots \sigma(G_m) = (-1)^k \sigma(G).$$ Now it will suffice to prove Theorem \[Th:Direct\] for connected graphs $G$.

### $G$ is connected and not strongly connected. {#SSec:NotSSC}

In this case $G$ contains an edge $e$ which is not contained in any directed cycle. For such $e$ if $H \subset G$ is acyclic and $e \notin H$ then $H \cup \{e\}$ is acyclic, too. The converse is true for any $e$: if an acyclic $H \subset G$ contains $e$ then $H \setminus \{e\}$ is acyclic. Therefore $$\SumSub(\alpha,G) = \sum_{\substack{H \subset G \setminus \{e\},\\ H \text{
is acyclic}}} (-1)^{\#\text{edges of $H$}} + (-1)^{\#\text{edges of $H \cup
\{e\}$}} = 0 = \sigma(G).$$ So it will suffice to prove Theorem \[Th:Direct\] for strongly connected graphs $G$.

### $G$ is strongly 