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Normalize the section to the sphere bundle. Scaling it from zero to unit length gives a homotopy within the disk bundle from the zero section to this sphere-valued section. The image of the relative Thom class in absolute cohomology restricts to zero on the sphere bundle. Homotopy invariance therefore makes its zero-section pullback vanish.

// Target: geometry-and-topology.bigb

An oriented normal bundle of a compact embedded submanifold has a Thom class. A tubular neighborhood and excision identify it with a relative class in $H^d(X,X\setminus Y)$. Its image in absolute cohomology is ${\epsilon }_Y$. It restricts to zero off Y and, in an oriented closed ambient manifold, represents the Poincare dual of Y with the normal-first induced orientation.

For a compact oriented submanifold without boundary in an oriented manifold, the normal Thom class gives a compactly supported cohomology class dual to its pushed-forward fundamental class. It is independent of the tubular neighborhood and its ordinary-cohomology image vanishes on the complement of the submanifold. A noncompact submanifold closed only as a subset generally has its dual in ordinary cohomology, using locally finite homology instead.

Complex cobordism is the generalized cohomology theory represented by the stable Thom spaces of universal complex vector bundles. Its geometric cycles use maps with chosen stable complex normal data. Its complex Thom classes give Chern classes in complex cobordism, and proper oriented maps have a Gysin map in complex cobordism.

Pulling the external product of two normally complex submanifolds back along the diagonal gives their transverse intersection. Its normal complex structure is the direct sum of the restricted normal complex structures. Multiplication of Thom classes for a direct sum identifies this geometric representative with the cup product in complex cobordism. Empty intersection represents zero.

// Target: geometry-and-topology.bigb

For an oriented rank-$r$ vector bundle, let $U\in H^r(D(E),S(E);\mathbb{Z})$ be its Thom class. Forgetting relative supports sends $U$ to ${\pi }^{\ast }e(E)$, where $e(E)$ is the Euler class. Compatibility of relative and mixed cup products givesFor odd $r$, graded commutativity of the cup product makes $2U^2=0$. The Thom isomorphism theorem is injective, so $2e(E)=0$. This explains why an odd-rank Euler class can have only two-torsion.

For an $h$-oriented real bundle $\xi$ of rank $r$, pull its Thom class back along the zero section to obtain its Euler class in $h^r(X)$. It depends on the orientation and the actual bundle, not merely its stable isomorphism class. In complex cobordism the orientation may be supplied by a stable complex structure on a real vector bundle.

For an oriented vector bundle $E$ of odd rank, multiplication by $-1$ on each fiber identifies $E$ with the oppositely oriented bundle. Naturality and reversal of the Thom class giveso $2e(E)=0$.

The Euler class of an oriented rank-$d$ vector bundle is the pullback $e(E)=s^{\ast }{u}_E\in H^d(B;R)$ of its Thom class along the zero section. Over ${\mathbb{F}}_2$ it equals the top Stiefel-Whitney class.

A nonzero Lefschetz coincidence number forces a coincidence of the two maps. If their paired map misses the diagonal, the diagonal Thom class restricts to zero on its image and its pullback vanishes, making the coincidence number zero. The argument works for continuous maps and requires neither isolated coincidences nor transversality.

The cohomology ring of the product of two real projective planes is ${\mathbb{F}}_2[x,y]/(x^3,y^3)$, with $|x|=|y|=1$. Pairing against $x^2$, $xy$ and $y^2$ identifies the Poincare dual of the diagonal as $x^2+xy+y^2$. For a self-map $f$ with $f^{\ast }x=\epsilon x$, the graph pullback evaluates on the mod-two fundamental class as $1+\epsilon +{\epsilon }^2=1$. Thus no graph of a map on the whole projective plane can avoid the diagonal. The argument uses a relative Thom class and does not assume transverse intersections.

This is a fiberwise orientation compatible with local trivializations of a vector bundle. An $R$-orientation of a rank-$d$ vector bundle is a coherent choice of generator of $H^d(E_b,E_b\setminus \{0\};R)$ for every fiber, equivalently a Thom class with the corresponding fiberwise restriction.

For an oriented real vector bundle $\pi :E\to X$ of rank $d$, the Thom isomorphism theorem gives a unique Thom class $u_E\in H^d(D(E),S(E);\mathbb{Z})$ restricting to the chosen generator in each fibre, and isomorphismsEquivalently use the pair $(E,E\setminus X)$, identifying $X$ with the zero section. After forgetting relative supports, pull the Thom class back along the zero section $s$ to define the Euler class:In this expression the relative class is first mapped to absolute cohomology.

A compact embedded submanifold equipped with a coorientation $Y\subset X$ has an oriented normal bundle ${\nu }_Y$ of rank $d$. A tubular neighborhood and excision identify its Thom class with a class $u_Y\in H^d(X,X\setminus Y)$. Its image in $H^d(X)$ is the cohomology class of a cooriented submanifold ${\epsilon }_Y$. In particular it restricts to zero on $X\setminus Y$. If the ambient manifold is oriented, orient $Y$ by the normal-first convention. The class is its Poincare dual, characterized byfor the appropriate complementary degree. This is the stated integration identity in cohomological notation.

First prove the Lefschetz fixed-point theorem for an oriented closed n-manifold $X$. Orient its diagonal by the identification $\Delta \cong X$, and choose the normal orientation so this agrees with the normal-first convention in the product. This produces the integral class ${\epsilon }_{\Delta }$. To compute the part that contributes to the trace, tensor with $\mathbb{Q}$: the trace on a finitely generated integral cohomology group means its trace on the free quotient, equivalently on rational cohomology; torsion contributes nothing.

Choose a basis $e_{p,j}$ of $H^p(X;\mathbb{Q})$ and its Poincare duality pairing dual basis $e_{p,j}^{\vee }\in H^{n-p}(X;\mathbb{Q})$, normalized by $⟨e_{p,i}⌣e_{p,j}^{\vee },[X]⟩={\delta }_{ij}$. The rational image of the cohomology class of the diagonal isIndeed, pairing this expression against $\alpha \times \beta$ with degrees $p,n-p$ picks out the p-summands. The cross-product multiplication contributes $(-1{)}^{p^2}$ and moving $\alpha$ past $e_{p,j}^{\vee }$ contributes $(-1{)}^{p(n-p)}$. Together with $(-1{)}^{np}$ their product is $+1$. The result is $⟨\alpha ⌣\beta ,[X]⟩$, exactly the diagonal's defining identity. The rational Künneth theorem and nondegeneracy of the duality pairing determine the class. An integral torsion component, if present, is not determined by this rational expression and does not affect any evaluation below.

For the graph map ${\Gamma }_f=(1,f):X\to X\times X$, write $f^{\ast }{e}_{p,j}={\sum }_i{A}_{ij}{e}_{p,i}$. Pulling back the expression and evaluating gives the graph-diagonal formula for the Lefschetz number:If $f$ has no fixed point, its graph lies in $(X\times X)\setminus \Delta$. The relative Thom class restricts to zero there, so ${\Gamma }_f^{\ast }{\epsilon }_{\Delta }=0$ and $L(f)=0$. The contrapositive proves the required fixed-point assertion, without requiring smoothness of $f$ or transversality of its graph.

The printed hypothesis does not require $X$ to be orientable. For a nonorientable $X$, the diagonal's normal bundle is $TX$, so an ordinary integral coorientation cannot be assumed. We extend the proved theorem as follows, rather than apply the untwisted diagonal computation incorrectly. We first construct a smooth embedding in Euclidean space using finitely many charts ${\chi }_i:U_i\to {\mathbb{R}}^n$ and smooth functions ${\rho }_i$ supported in those charts, with at least one ${\rho }_i(x)>0$ at every point. The map $\Psi (x)=({\rho }_i(x),{\rho }_i(x){\chi }_i(x){)}_i$, with zero extensions, is injective: a positive first coordinate identifies a chart in which its remaining coordinates recover $x$. Its differential is injective as well, since $d{\rho }_i(v)=0$ and $d({\rho }_i{\chi }_i)(v)=0$ force $d{\chi }_i(v)=0$ in a chart with ${\rho }_i>0$. Compactness makes this injective immersion a homeomorphism onto its image. This finite-chart Euclidean embedding of a compact smooth manifold, enlarged by zero coordinates if necessary, has positive codimension. A compact tubular disc neighbourhood $W$ is an oriented N-manifold with boundary, inheriting its orientation from the ambient Euclidean space. Its double of a manifold $M$ is closed and oriented. Projection of each copy of $W$ to $X$ agrees on their common boundary, giving $r:M\to X$; inclusion into the first copy gives $i:X\to M$ with $ri=1_X$.

Set $F=ifr$. A fixed point of $F$ must lie in $i(X)$, and then is exactly the image of a fixed point of $f$. Moreover $F^{\ast }=r^{\ast }{f}^{\ast }{i}^{\ast }$, and cyclicity of trace for maps between finite-dimensional rational cohomology groups givesin each degree, because $i^{\ast }{r}^{\ast }=1$. Thus $L(F)=L(f)$. The oriented theorem for $M$ now gives a fixed point of $F$, hence of $f$, whenever $L(f)\ne 0$. This oriented manifold retract for the Lefschetz theorem proves the full printed result. The double construction also works componentwise if necessary.

Let $\pi :E\to B$ be an oriented real vector bundle of rank $r$ over a paracompact base. A Thom class is a class $U\in H^r(E,E\setminus B;\mathbb{Z})$ restricting on every oriented fiber pair $({\mathbb{R}}^r,{\mathbb{R}}^r\setminus 0)$ to its positive generator. The Thom isomorphism theorem states that this class exists uniquely with those fiber restrictions and thatHere $B$ is identified with the zero section. Equivalently use the disk/sphere bundle pair. The compact-support version is $H_c^q(B;\mathbb{Z})\cong H_c^{q+r}(E;\mathbb{Z})$, with fiberwise compact support in the Thom construction. If the bundle is not oriented, the integral assertion requires the corresponding orientation local system.

For a closed co-oriented embedded submanifold $Y$ of codimension $r$, co-orientation means an orientation of its normal bundle. A tubular neighborhood identifies a neighborhood of $Y$ with that bundle. Its Thom class, transferred by excision, is a class in $H^r(M,M\setminus Y;\mathbb{Z})$. For compact $Y$ this maps into the direct limit defining compactly supported cohomology, giving $\boxed{{\epsilon }_Y\in H_c^r(M;\mathbb{Z})}$. The construction is independent of the tubular neighborhood and uses the chosen co-orientation. In an oriented ambient manifold it is the Poincare dual of the oriented submanifold. Here “closed submanifold” has the usual compact-without-boundary meaning; if it meant merely a closed subset and $Y$ were noncompact, the supported relative class would still exist but need not have compact support.

We next prove the fixed-point implication without imposing an unprinted orientability assumption on $M$. SetUse a finite triangulation of the closed smooth manifold. The elementary Hopf trace identity says that the alternating trace of a chain endomorphism equals its alternating trace on homology. To verify it, split each finite-dimensional chain space as boundaries, a complement representing homology, and a complement mapping isomorphically onto the preceding boundaries. The traces on these two boundary summands cancel in adjacent degrees, leaving exactly the homology traces. The cohomology traces equal the homology traces because the induced maps are dual over $\mathbb{Q}$.

If $f$ has no fixed points, compactness gives $\delta ={\min }_{x}d(x,f(x))>0$. Choose a triangulation $K$ of mesh less than $\delta /4$ and a sufficiently fine subdivision $K^{\prime }$ admitting a simplicial approximation $F:K^{\prime }\to K$ uniformly within $\delta /4$ of $f$. The subdivision chain map $s:C_{\ast }(K)\to C_{\ast }(K^{\prime })$ sends each simplex to simplices inside it. For a vertex $v$ of such a small simplex, $d(v,F(v))>3\delta /4$, so $F(v)$ cannot lie in the original coarse simplex. Consequently $F_{\#}s$ has zero diagonal coefficient on every coarse simplex. Its trace vanishes in every degree, while it induces $f_{\ast }$ on homology. The trace identity yields $L(f)=0$. Taking the contrapositive provesThis is the Lefschetz fixed-point theorem, established here rather than invoked. For oriented $M$ the same identity is the graph-diagonal formula for the Lefschetz number: $L(f)=⟨(1,f{)}^{\ast }{\epsilon }_{\Delta },[M]⟩$, which vanishes when the graph misses the diagonal. Finally, homotopic maps induce the same cohomology maps, so their Lefschetz numbers agree.

Put $L=L_1\cap L_2$, and let $j:L\hookrightarrow M$ be inclusion. The transverse intersection theorem makes $L$ a smooth submanifold without boundary, withIt is compact because it is a closed subset of either compact input submanifold.

Write ${\nu }_r=TM{|}_{L_r}/T{L}_r$. On $L$, consider the bundle mapIts kernel is $TL$, and it is onto because $T{L}_1+T{L}_2=TM$ at every intersection point. Thus the normal bundle of a transverse intersection has the actual bundle isomorphismTransfer the direct-sum complex structure $J_1\oplus J_2$ through this isomorphism. This supplies a complex structure on the normal bundle itself, not merely after stabilization.

Geometrically, multiply the two classes by forming their external product in $M\times M$ and pulling it back along the diagonal $\Delta :M\to M\times M$. The transversality hypothesis is precisely what makes the inverse image of $L_1\times L_2$ under $\Delta$ the submanifold $L$. The normal bundle of this inverse image is the direct sum just calculated. Multiplication of the corresponding Thom classes agrees with the Thom class of this direct sum, so the complex cobordism product represented by transverse intersections isIf the inputs have complex normal ranks $a,b$, the product has degree $2(a+b)$. If they are disjoint, the representative is empty and the product is zero.