Thom isomorphism for a trivial oriented bundle

Projection to the second factor is a map of pairs, so $u={\mathrm{pr}}_2^{\ast }{v}_n$ is defined. Its restriction to every fiber is literally $v_n$; hence it is normalized by [F5].

For one interval, the pair sequence for $(B\times I,B\times \mathrm{\partial }I)$ has restriction $H^q(B;R)\to H^q(B;R)\oplus H^q(B;R)$ equal to the diagonal, by the two endpoint deformation retractions and finite additivity in [F3]. Exactness identifies the next relative group with the cokernel of that diagonal, and $(x,y)\mapsto y-x$ identifies the cokernel with $H^q(B;R)$; the following diagonal is injective. Thus the connecting map is an isomorphism $H^q(B;R)\cong H^{q+1}(B\times I,B\times \mathrm{\partial }I;R)$.

Normalize the interval generator as the connector of the boundary class $(0,1)$. Applying [F4]'s second connector formula to ${\pi }^{\ast }a$ and $(0,1)$ says that the isomorphism of step 1.2 sends $a$ to $(-1{)}^{\mathrm{\mid }a\mathrm{\mid }}{\pi }^{\ast }a⌣v_1$. Multiplication by this degree-dependent sign is itself an automorphism, so cup product with $v_1$ is an isomorphism in every degree.

We first record the relative form needed for iteration. If $(X,A)$ is a pair with an explicit collar of $A$, the one-interval connector gives H^q(X,A;R)\xrightarrow{\cong}H^{q+1}\bigl(X\times I,\,A\times I\cup X\times\partial I;R\bigr). \tag{1} To verify this rather than assume it, use the collar to subdivide chains until every simplex in the union lies in one of its two members. The small-chain equivalence in [F3] then identifies the target relative cochain complex with the kernel in the termwise split restriction sequence for the pairs $(X\times I,X\times \mathrm{\partial }I)$ and $(A\times I,A\times \mathrm{\partial }I)$. The resulting cohomology sequence forms a natural exact ladder over the pair sequence of $(X,A)$. The absolute vertical maps for $X$ and $A$ are the one-interval isomorphisms of step 2.1, so [F3]'s five lemma makes (1) an isomorphism. Connector naturality in [F4] identifies (1), up to the same invertible degree sign, with relative cup by the interval generator.

Write the $n$-cube as an ordered product of intervals. Each pair $(B\times I^j,B\times \mathrm{\partial }{I}^j)$ has an explicit radial collar in the cube coordinate, so apply (1) successively for $j=0,\dots ,n-1$. The generator $v_n$ in [F2] is the iterated ordered connector generator. At the cochain level, the iterated relative products are represented by repeated Alexander--Whitney cup products; associativity in [F4] identifies either parenthesization with the single product ${\pi }^{\ast }a⌣(v_1⌣\cdots ⌣v_1)$. Connector naturality identifies the parenthesized fiber product with $v_n$, up to the product of the displayed invertible signs. Thus the composite is $a\mapsto {\pi }^{\ast }a⌣v_n$ up to a unit sign and is an isomorphism. Radial identification of cube and disk pairs, and [F1], give the stated target.

For $n=0$, $v_0=1_R$ and the map is the identity of $H^k(B;R)$. Empty $B$, the zero ring, $n=1$, negative/zero $k$, both interval endpoints, a constant class, and degenerate singular simplices all remain inside the exact pair calculation. Every sign is a unit and hence affects neither bijectivity nor normalization. The construction uses finitely many fixed connectors and no arbitrary family, so it is choice-free.