Leray–Hirsch, the Thom Isomorphism, and Gysin Sequences — Examples

Define $e_i=1\times b_i={\mathrm{pr}}_B^{\ast }1⌣{\mathrm{pr}}_F^{\ast }{b}_i$. On the fiber $\{b\}\times F$, the first factor restricts to $1$ and the second to $b_i$, so $e_i{\mathrm{\mid }}_{F_b}=b_i$. The supplied list is therefore a basis on every fiber.

Apply [F2]. Its map sends $(a_i)$ to ${\sum }_i{p}^{\ast }{a}_i⌣e_i={\sum }_i{a}_i\times b_i$. This is exactly the cross-product map in [F1], and the basis identifies its source with $H^{\ast }(B;R){\otimes }_R{H}^{\ast }(F;R)$. Thus both isomorphisms agree, not merely their abstract modules.

If $m=0$, the fiber cohomology and both sides are zero; $m=1$ gives one shifted copy. A point fiber has basis $1$, and a point base recovers $H^{\ast }(F)$. Empty $F$, degree zero, zero classes and all finite direct-sum endpoints are included in the formulas. The zero ring is outside the stated PID convention. No basis is chosen: it is supplied. AC is used exactly through [A1] in Künneth and cohomological Serre/Leray–Hirsch.

Put $n=1$ in [F1]. The pair is $(B\times [-1,1],B\times \{-1,1\})$, its quotient is $\mathrm{\Sigma }{B}_{+}$, and [F2]'s fiber generator is the connector of $(0,1)$ on the two boundary endpoints. Thus its pullback to the product is the suspended unit and cup by it is the one-fold relative suspension isomorphism.

Put $n=2$. The quotient is ${\mathrm{\Sigma }}^2{B}_{+}$, the ordered generator is the second connector applied to the rank-one generator, and [F2] gives the twice-iterated suspension isomorphism. This fixes the orientation sign rather than choosing an unspecified generator.

For empty $B$ the spaces are the one-point based space and cohomology maps are zero; for a point base the spaces are $S^1$ and $S^2$. The zero ring, zero/unit classes, both interval endpoints, the two coordinate orders and both suspension endpoints are explicit in steps 1.1–1.2. All connectors are finite and formulaic, so no AC is used.

The generator of $H^1([-1,1],\{-1,1\};\mathbb{Z})$ is the pair connector of the endpoint class $(0,1)$ modulo diagonal constants. The Möbius clutching map is reflection $t\mapsto -t$, which swaps the endpoint coordinates. It sends $(0,1)$ to $(1,0)$, and $(1,0)=-(0,1)$ modulo the diagonal $(1,1)$. Thus its integral orientation monodromy is $-1$.

A global integral generator would have to return to itself after the base loop, while step 1.1 returns its negative. Since a generator of the free module $\mathbb{Z}$ is nonzero, this is impossible. Modulo two, $-1=1$ and the same transition fixes the unique nonzero generator, giving the canonical ${\mathbb{F}}_2$ orientation of [F1].

Apply [F2] with rank one and $R={\mathbb{F}}_2$. It produces a unique normalized degree-one Thom class and the displayed shift for every $k$. The two fiber endpoints, the base-loop start/end, the zero vector, degree-zero unit and zero classes are explicit above. The base and fibers are nonempty, and the coefficient rings are fixed, so empty and zero-ring cases are inapplicable. The monodromy calculation is finite and choice-free; AC is used exactly through [A1] in [F2].

By definition, ${\mathbb{C}\mathbb{P}}^{\mathrm{\infty }}$ is the weak colimit of complex lines in ${\mathbb{C}}^{N+1}$, so [F1] identifies it with ${\mathrm{Gr}}_1({\mathbb{C}}^{\mathrm{\infty }})$ and $\lambda$ with the pairs $(\mathrm{\ell },v)$, $v\in \mathrm{\ell }$. Its unit vectors therefore form the standard principal $S^1$-bundle. A principal chart with local unit section $s_i(\mathrm{\ell })$ gives the linear chart $(\mathrm{\ell },c{s}_i(\mathrm{\ell }))\mapsto (\mathrm{\ell },c)$ of $\lambda$; hence the support-subordinate numeration in [F2] is also a numeration of $\lambda$. The base is the stated weak CW complex.

A nonzero vector $v$ in a complex fiber orders its underlying real plane by $(v,iv)$. Replacing $v$ by $zv$ for $z\in {\mathbb{C}}^{\times }$ changes this ordered basis by the real matrix of multiplication by $z$, whose determinant is $\mathrm{\mid }z{\mathrm{\mid }}^2>0$. Thus all complex-linear transition functions preserve the corresponding generator in [F3], and these generators define the complex orientation of the underlying real rank-two bundle.

Apply [F4] with $R=\mathbb{Z}$, $n=2$, the numeration from step 1.1, and the orientation from step 1.2. It gives the unique normalized $u_{\lambda }$ and exactly the displayed isomorphism for every $k$; no Euler or characteristic-class identification is used.

The base is nonempty and the coefficients are the nonzero ring $\mathbb{Z}$, so empty-base and zero-ring cases are outside this example. The single complex line, its zero vector and a coordinate-line point ${\mathbb{C}\mathbb{P}}^0\subset {\mathbb{C}\mathbb{P}}^{\mathrm{\infty }}$ are included in steps 1.1–1.2. At the input-degree endpoint $k=0$, the unit maps to $u_{\lambda }$; negative-degree and zero inputs map between zero groups or to zero as covered by [F4]. AC is used only through [A1] in [F4]; the explicit orientation and the supplied numeration require no additional choice.

Write $K=(S^1\times [0,1])\mathrm{/}(z,1)\sim (r(z),0)$. Product charts away from the seam and charts changing fiber coordinate by $r=r^{-1}$ across the seam make $K\to S^1$ a fiber bundle. Positive-loop transport is $r$, so [F2] and [F4] give $T_1=-1$ on $H_1(S^1;\mathbb{Z})=\mathbb{Z}$ and $T_0=1$ on $H_0(S^1;\mathbb{Z})=\mathbb{Z}$.

Cohomological transport on $H^1(S^1;\mathbb{Z})$ is likewise multiplication by $-1$, since evaluation on the homology generator changes by the degree in step 1.1. It fixes no generator. The contrapositive of [F1] therefore says that no global class on $K$ can restrict to an integral basis of $H^{\ast }(S^1;\mathbb{Z})$.

The degree-one part of [F3], using step 1.1, gives $0\to \mathbb{Z}\mathrm{/}2\to H_1(K;\mathbb{Z})\to \mathbb{Z}\to 0$. The fixed point $1\in S^1$ defines the section $[t]\mapsto [(1,t)]$, so the projection onto the last $\mathbb{Z}$ splits and $H_1(K;\mathbb{Z})\cong \mathbb{Z}\oplus \mathbb{Z}\mathrm{/}2$. The same explicit mapping-torus model is path connected, hence $H_0(K;\mathbb{Z})=\mathbb{Z}$.

Apply [F5] in degree one. Since $H_0(K)=\mathbb{Z}$ is free, its Ext term is zero, and every homomorphism $\mathbb{Z}\mathrm{/}2\to \mathbb{Z}$ is zero. Consequently $H^1(K;\mathbb{Z})\cong \mathrm{Hom}(\mathbb{Z}\oplus \mathbb{Z}\mathrm{/}2,\mathbb{Z})\cong \mathbb{Z}$.

By [F4] and [F5], both base and fiber have one copy of $\mathbb{Z}$ in cohomological degrees zero and one. Falsely declaring the fiber basis $(1,u)$ constant would make the degree-one Leray–Hirsch source $H^1(S^1)\otimes ⟨1⟩\oplus H^0(S^1)\otimes ⟨u⟩\cong {\mathbb{Z}}^2$, whereas step 3.1 gives only $\mathbb{Z}$. The failed conclusion and its missing global-basis hypothesis are therefore witnessed explicitly.

The base, fiber and total space are nonempty, and the coefficient ring is fixed as nonzero $\mathbb{Z}$. Steps 1.1–4.1 include the one base loop, its two seam endpoints, the degree-zero unit, the zero kernel of multiplication by two, identity action on $H_0$, reflection action on $H_1$, and the degenerate false identity-monodromy comparison. AC is used only through [A1] in [F1] and [F5]; the mapping-torus and Wang calculations are choice-free. No converse claim is made.

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.