An untwisted E2 table misses mapping-torus monodromy

Lift one positive circuit of the base interval in the mapping-torus model $(F\times [0,1])\mathrm{/}(x,1)\sim (h(x),0)$. Its endpoint identification on the fiber is $h$, so [F1] and [F2] give monodromy $h_{\ast }$ on $H_q(F;R)$. Therefore the correct base groups retain this local system rather than replacing it by a constant copy of its stalk.

Let $F=S^1$ and let $h$ be a reflection. On $H_1(S^1;\mathbb{Z})=\mathbb{Z}$, $h_{\ast }=-1$. Give the base circle one vertex and one edge. Its lifted edge boundary evaluates through [F3] to $T-1:\mathbb{Z}\to \mathbb{Z}$. For the correct monodromy $T=-1$, this is multiplication by $-2$, so the $q=1$ row is $H_0(S^1;{\mathbb{Z}}_{-1})=\mathbb{Z}\mathrm{/}2$ and $H_1(S^1;{\mathbb{Z}}_{-1})=0$. If monodromy is discarded, $T=1$ makes the differential zero, giving $H_0=\mathbb{Z}$ and $H_1=\mathbb{Z}$ instead.

In the $q=0$ row the reflection acts trivially on $H_0(S^1;\mathbb{Z})$, so the untwisted and correct rows agree there; the discrepancy is specifically caused by monodromy, not by the fiber groups. This comparison computes only the proposed coefficient rows and does not invoke or assert convergence of a Serre spectral sequence. No AC is used.