This confuses me.
One easy case of counting is when the decision problem is in $P$ and there are no solutions.
A lecture show that the problem of counting the number of perfect matchings in a bipartite graph (equivalently, counting the number of cycle covers in a directed graph) is $\#P$ -complete.
They give reduction from counting vertex covers of size $k$ to counting cycle covers in a digraph using gadgets.
Theorem 27.1 The number of good cycle covers in $H$ is $(k!)^2$ times the number of vertex covers of $G$ of size $k$.
Using gadget they leave only the "good" cycles.
My understanding of the lecture is that $G$ doesn't have vertex cover of size $k$ iff the transformed digraph $G'$ doesn't have cycle cover. Checking if $G'$ has cycle cover can be done in polynomial time, implying $P=NP$ since we can transform the decision problem to finding solution.
What am I misunderstanding?
The permanent of the adjacency matrix of digraph counts cycle covers and is $\#P$-complete.
The decision problem "Is the permanent of (0,1) matrix zero" is in P since finding cycle cover is in $P$.
$P \ne NP$ implies there is no reduction of counting $NP$-complete problems to counting $(0,1)$-permanent which maps $0 \mapsto 0$.
Edit Related MO question
Markus Bläser points out that bad cycle are still "there",
but the sum of their weights vanishes.
Appears to me the weight of bad cycle in a widget is zero.
From page 148 (11 of the pdf):
The full adjacency matrix B with submatrices A corresponding to these four-node widgets counts 1 for each good cycle cover in H and 0 for each bad cycle cover
Wouldn't maximum weight cycle cover contain only the good cycles, corresponding to a $k$ vertex cover in the original graph?
In CC every vertex must be in exactly one cycle.