Let us group the vertices as $U=\{u_1,u_2,\dots,u_n\}$ and $V=\{v_1,v_2,\dots, v_n\}$ where $f(u_i)=v_i$. Let $L_0$ be the laplacian of the graph with vertex set $U$ and edges as restricted from $G$, let $L _1 = \operatorname{diag}\left( \sum _{j=1}^n \mu(u _i, v _j)\right)$ and $L=L _0+L _1$, also let $A$ be the symmetric matrix whose $a _{ij}$ is $-\mu(u _i,v _j)$. Clearly the Laplacian of $G$ is $M=\left( \begin {array} {cc} L & A \\ A & L \end {array} \right)$. Let $M^*$ stand for the matrix $M$ with deleted first row and column. We have $$\kappa(G)=\det \left( \begin {array} {cc} L & A \\ A & L \end {array} \right) ^ *=\det \left( \begin {array} {cc} B & C \\ D & (L+A)^ * \end {array} \right)$$ for some block matrices $B,C,D$ of size $n\times n,n\times (n-1),$ and $(n-1)\times n$ where this second matrix was obtained by adding the $i$'th i$th row of$M^*$M^{*}$ to it's $n+i$'th n+i$th row for$1\le i\le n-1$and then adding the first (or last)$n-1$columns to the$n$th column. So$D$is the matrix$(L+A)*$together with a last column of zeros, making$(L+A)^{*-1}D$with integer entries. Next we factor it using one of these identities $$\kappa(G)=\det(L+A)^ * \det(B-C(L+A)^{*-1}D)$$ and observe that$L+A$is the Laplacian of$G/f$so$\det(L+A)^ *=\kappa(G/f)$, and since the second factor is an integer we get the desired divisibility. 3 added 101 characters in body Let us group the vertices as$U=\{u_1,u_2,\dots,u_n\}$and$V=\{v_1,v_2,\dots, v_n\}$where$f(u_i)=v_i$. Let$L$L_0$ be the laplacian of the graph with vertex set $U$ and edges as restricted from $G$, let $L _1 = \operatorname{diag}\left( \sum _{j=1}^n \mu(u _i, v _j)\right)$ and $L=L _0+L _1$, also let $A$ be the symmetric matrix whose $a_{ij}$ a _{ij}$is$-\mu(u_i,v_j)$. -\mu(u _i,v _j)$. Clearly the Laplacian of $G$ is $M=\left( \begin {array} {cc} L & A \\ A & L \end {array} \right)$. Let $M^*$ stand for the matrix $M$ with deleted first row and column. We have $$\kappa(G)=\det \left( \begin {array} {cc} L & A \\ A & L \end {array} \right) ^ *=\det \left( \begin {array} {cc} B & C \\ D & (L+A)^ * \end {array} \right)$$ for some block matrices $B,C,D$ of size $n\times n,n\times (n-1),$ and $(n-1)\times n$ where this second matrix was obtained by adding the $i$'th row of $M^*$ to it's $n+i$'th row for $1\le i\le n-1$. Next we factor it using one of these identities $$\kappa(G)=\det(L+A)^ * \det(B-C(L+A)^{*-1}D)$$ and observe that $L+A$ is the Laplacian of $G/f$ so $\det(L+A)^ *=\kappa(G/f)$, and since the second factor is an integer we get the desired divisibility.
Let us group the vertices as $U=\{u_1,u_2,\dots,u_n\}$ and $V=\{v_1,v_2,\dots, v_n\}$ where $f(u_i)=v_i$. Let $L$ be the laplacian of the graph with vertex set $U$ and edges as restricted from $G$, also let $A$ be the symmetric matrix whose $a_{ij}$ is $\mu(u_i,v_j)$. -\mu(u_i,v_j)$. Clearly the Laplacian of$G$is$M=\left( \begin {array} {cc} L & A \\ A & L \end {array} \right)$. Let$M^*$stand for the matrix$M$with deleted first row and column. We have $$\kappa(G)=\det \left( \begin {array} {cc} L & A \\ A & L \end {array} \right) ^ *=\det \left( \begin {array} {cc} B & C \\ D & (L+A)^ * \end {array} \right)$$ for some block matrices$B,C,D$of size$n\times n,n\times (n-1),$and$(n-1)\times n$where this second matrix was obtained by adding the$i$'th row of$M^*$to it's$n+i$'th row for$1\le i\le n-1$. Next we factor it using one of these identities $$\kappa(G)=\det(L+A)^ * \det(B-C(L+A)^{*-1}D)$$ and observe that$L+A$is the Laplacian of$G/f$so$\det(L+A)^ *=\kappa(G/f)\$, and since the second factor is an integer we get the desired divisibility.