Constant coefficients miss monodromy

Give $S^1$ one vertex and one oriented edge. If $g$ denotes the positive loop, a lift of the edge has boundary $g\tilde{v}-\tilde{v}$. Under the published right-chain convention this is $\tilde{v}\cdot (g^{-1}-1)$, while the corresponding left fiber action of $g^{-1}$ is the specified monodromy $T$. Thus [F1] gives the two-term complex $0\to \mathbb{Z}\stackrel{T-1}{\to }\mathbb{Z}\to 0$. For $T=-1$, its differential is $-2$, with zero kernel and cokernel $\mathbb{Z}\mathrm{/}2$. For $T=1$, its differential is zero, so both degree-one and degree-zero groups are $\mathbb{Z}$.

The two systems have isomorphic stalk $\mathbb{Z}$ at every point but different loop transport and different homology. Hence stalk data without monodromy cannot replace a local system. The sign of the differential could be $+2$ after reversing the chosen cell, with the same kernel and cokernel. No other degrees occur and no AC is used.