$\ell$ such that $G_\ell$ is strongly connected. Since $\ell < m$, the graph $G_{\ell+1}$ exists and is not strongly connected, and therefore $G_\ell
\setminus \{e_{\ell+1}\} = G_{\ell+1} \setminus \{e_{\ell+1}\}$ is not strongly connected either. So, the edge $e_{\ell+1}$ is crucial for the graph $G_\ell$, and by Clause \[SSec:SSCCrucial\] one has $\SumSub(\alpha,G_\ell) = (-1)^k \SumSub(G_\ell) = (-1)^k \sigma(G)$. The graphs $G_0=G {, \dots ,} G_\ell$ are strongly connected, so for any $i = 0 {, \dots ,}
\ell-1$ the edge $e_{i+1}$ is not crucial for the graph $G_i$. Lemma \[Lm:Reversal\] implies now $$\begin{aligned}
\SumSub(\alpha,G_{\ell-1}) = (-1)^k \sigma(G_{\ell-1})
&\Longrightarrow \SumSub(\alpha,G_{\ell-2}) = (-1)^k \sigma(G_{\ell-2})\\
&{\Longrightarrow \dots \Longrightarrow} \SumSub(\alpha,G) = (-1)^k \sigma(G).
 \end{aligned}$$ Theorem \[Th:Direct\] is proved.

Theorem \[Th:Diag\] follows from Theorem \[Th:Direct\]’ {#SSec:DiagFromDirect}
-------------------------------------------------------

Note first that the operation $B_i$, and hence $\Lapl$, preserves the sinks of the graph: if $\Lapl H = \sum_G x_G G$ and $x_G \ne 0$ then $G$ has the same sinks as $H$. Therefore if $I = \{i_1 {, \dots ,} i_s\}$ then $\Lapl(\det_{n,k}^I) = \sum_G x_G G$ where all the graphs $G$ in the right-hand side have the sinks $i_1 {, \dots ,} i_s$ and have no loops.

Let $G$ be a graph with sinks $i_1 {, 