An unoriented real bundle has no integral Thom class

Suppose, toward a contradiction, that $u\in H^1(D(\mu ),S(\mu );\mathbb{Z})$ restricts to a generator on every fiber. Pulling the disk/sphere pair back along the quotient parameter gives the product pair $P=([0,1]\times [-1,1],[0,1]\times \{-1,1\})$ and the quotient pair map $Q:P\to (D(\mu ),S(\mu ))$, $Q(t,v)=[t,v]$.

Write $j_t:([-1,1],\{-1,1\})\to P$ for inclusion at $t$. The maps $j_0$ and $j_1$ are homotopic through maps of pairs. The prism of [F3] preserves the boundary subcomplex, descends to relative chains, and after cochain precomposition shows $j_0^{\ast }{Q}^{\ast }u=j_1^{\ast }{Q}^{\ast }u$.

Put $g=j_0^{\ast }{Q}^{\ast }u$. The clutching relation gives $Q{j}_1(v)=[1,v]=[0,-v]=Q{j}_{0}r(v)$, so step 2.1 says $g=r^{\ast }g$ for the reflection $r(v)=-v$. Reflection swaps the endpoint class $(0,1)$ with $(1,0)=-(0,1)$ modulo diagonal constants, so the ordered connector calculation in [F4] gives $r^{\ast }g=-g$ in $H^1([-1,1],\{-1,1\};\mathbb{Z})\cong \mathbb{Z}$. Thus $2g=0$, impossible for the generator required in step 1.1. Hence no integral fiberwise-generating class exists.

Reducing the same clutching action modulo two makes $-1=1$, so [F1] gives the canonical mod-two orientation. The Möbius bundle is numerable over the CW complex $S^1$, and [F5] therefore supplies its normalized mod-two class. Thus the example isolates the nontrivial integral orientation system rather than a failure of the disk-pair construction.

The circle base and interval fibers are nonempty and the coefficient rings are fixed and nonzero. The rank-one fiber, zero vector, both interval endpoints, both base-loop endpoints, identity transport before clutching, sign reversal at clutching, degree-one generator, zero class and mod-two sign degeneration all occur in steps 1.1–4.1. The integral nonexistence proof is finite and choice-free; AC is used only through [A1] for the positive mod-two existence statement. No converse beyond this explicit witness is asserted.