Thom classes and Euler classes

Written by GPT-6.1 Sol (OpenAI), at Ultra, October 2026. Self-checked by the writing AI, GPT-6.1 Sol, at Ultra. Independently authored text dedicated under CC0.

A fibre of a rank \(r\) vector bundle looks like \(\mathbb R^r\). Removing its origin leaves a relative cohomology class in degree \(r\). A Thom class joins these fibre classes into one global class. Multiplication by it transfers cohomology of the base into relative cohomology of the bundle. Pulling it back to the zero section gives the Euler class, which detects an obstruction to a nonzero section.

Learn first Vector bundles and their constructions and Grassmannians and classifying maps. The required singular-chain arguments are supplied here. We use elementary linear algebra and the definition of a free abelian group. The coefficient systems in this chapter are constant; local orientation coefficients will be introduced with the obstruction theory chapter. Hatcher's freely readable texts [AT] and [VB] give the singular-chain, Thom and Euler comparisons. The proof here works directly with the nonzero-vector pair, including the homological passage to arbitrary Hausdorff bases.

1. Chains, pairs and homotopies

Let \(C_m(X)\) be the free abelian group on the continuous maps \(\sigma:\Delta^m\to X\), with \(C_m=0\) for \(m<0\). Write \(\sigma[i_0,\ldots,i_k]\) for its restriction to the indicated ordered face. The differential is

Each twice-deleted face appears with opposite signs in \(\partial^2\sigma\), so \(\partial^2=0\). Homology is \(H_m=\ker\partial/\operatorname{im}\partial\). For \(A\subset X\), the relative complex is \(C_*(X,A)=C_*(X)/C_*(A)\). It is free on the simplices not entirely contained in \(A\). With an abelian coefficient group \(G\), use \(C_*(X,A)\otimes G\) for homology and \(\operatorname{Hom}(C_*(X,A),G)\) for cohomology; its coboundary is \(\delta a=a\partial\). Maps of pairs induce the evident chain maps and contravariant cochain maps.

Lemma 1.1 (homotopy invariance). Homotopic maps of pairs induce the same homology and cohomology maps, with every coefficient group.

Proof. For a homotopy \(H:X\times I\to Y\), triangulate \(\Delta^m\times I\) into the \(m+1\) ordered simplices

Apply \(H(\sigma(-),-)\) to their sum with coefficients \((-1)^i\), defining \(P\sigma\). Adjacent internal faces cancel. The bottom and top faces remain with signs minus and plus, and the side faces are \(-P\partial\sigma\). Hence \(\partial P+P\partial=H_{1*}-H_{0*}\). If \(H(A\times I)\subset B\), this formula descends to relative chains. Tensoring the homotopy, or precomposing cochains with it, proves the assertion. \(\square\)

The point complex has one generator in every nonnegative degree; its differential is zero in odd degrees and the identity in positive even degrees. Therefore a point has homology \(\mathbb Z\) in degree zero and zero elsewhere. By Lemma 1.1 the same is true of a contractible space. Moreover \(H_0(X)\) is free on the path components: the boundary of a path equates its two endpoints, and these are exactly the relations imposed by paths.

2. Making singular chains small

For an open cover \(\mathcal U\) of \(X\), let \(C_*^{\mathcal U}(X)\) be generated by simplices whose image lies in one cover member. Local arguments require the inclusion of these chains to be a chain homotopy equivalence, rather than just an intuitive subdivision picture.

Lemma 2.1 (small-chain lemma). There is a chain map \(\rho:C_*(X)\to C_*^{\mathcal U}(X)\) and a homotopy \(D\) with

Both preserve every subspace of \(X\), so the assertion also holds for relative complexes.

Proof. First construct subdivision \(S\) on the affine chains in each standard simplex. If \(b\) is its barycentre, define \(S\sigma=b*(S\partial\sigma)\) inductively, starting with vertices. Here the affine cone with apex first satisfies \(\partial(b*c)=c-b*(\partial c)\) for a positive-dimensional cycle, and the corresponding reduced formula for dimension zero. It follows inductively that \(\partial S=S\partial\). Composing these affine subdivisions with a singular simplex defines a natural chain map on singular chains.

Construct a subdivision homotopy \(T\) by induction too. On an \(m\)-simplex, the chain \(z=\sigma-S\sigma-T\partial\sigma\) is a cycle by the already established lower-dimensional identity. It has augmentation zero when relevant. Cone it to the barycentre and set \(T\sigma=b*z\). Then \(\partial T+T\partial=1-S\). All chains constructed remain inside the original simplex image. Thus subdivision and its homotopy preserve small chains and chains in any subspace.

A barycentric subsimplex has vertices that are barycentres of nested faces. If a face of size \(\ell\) lies in one of size \(k\), their barycentre difference has norm at most \((k-\ell)/k\) times the old diameter, at most \(m/(m+1)\) times it. The diameter of a Euclidean simplex is the maximum of its vertex distances. Iteration therefore makes the mesh tend to zero. Apply the Lebesgue-number argument of the preceding chapter to the inverse-image open cover of a standard simplex. Every singular simplex, and every finite chain, becomes small after sufficiently many subdivisions.

Put \(T_N=\sum_{j=0}^{N-1}S^jT\); then \(\partial T_N+T_N\partial=1-S^N\). Now construct \(\rho,D\) inductively on the basis of singular simplices. Given their values on the boundary of \(\sigma\), put \(z=\sigma-D\partial\sigma\), so \(\partial z=\rho\partial\sigma\) is small. Choose \(N\) such that \(S^Nz\) is small and define

The second term of \(\rho\sigma\) is small because \(T_N\) preserves smallness. The displayed identities give \(\partial\rho\sigma=\rho\partial\sigma\) and \(\sigma-\rho\sigma=\partial D\sigma+D\partial\sigma\). If \(\sigma\) is already small, its faces have \(D=0\) by induction; choose \(N=0\), obtaining \(\rho\sigma=\sigma\), \(D\sigma=0\). Every operation remains in the original image, which proves the subspace assertion. \(\square\)

Corollary 2.2 (excision). If \(Z\subset A\subset X\) and \(\overline Z\subset\operatorname{int}_X A\), the inclusion \((X\setminus Z,A\setminus Z)\to(X,A)\) induces homology and cohomology isomorphisms with every coefficient group.

Proof. Use the open cover \(\operatorname{int}A, X\setminus\overline Z\) of \(X\). After quotienting small chains by those in \(A\), the first summand disappears and the quotient is \(C_*(X\setminus\overline Z,A\setminus\overline Z)\). The restricted cover of \(X\setminus Z\) gives the very same quotient: its first member is \((\operatorname{int}A)\setminus Z\), whose chains again disappear. Lemma 2.1 identifies both ordinary relative complexes with these quotients up to chain homotopy. Tensor and dualize the equivalences. \(\square\)

3. Exact sequences and local gluing

A degreewise exact sequence of complexes \(0\to K\to L\to M\to0\) has a long exact sequence in homology. Its connecting map sends a cycle \(m\) to the class of \(\partial\ell\) in \(K\), where \(\ell\) lifts \(m\). Changing the lift changes this by a boundary; changing the cycle by a boundary does likewise. Exactness follows directly: a cycle of \(L\) maps to a boundary of \(M\) exactly when, after subtracting the boundary of a lift, it lies in \(K\); a cycle of \(M\) has zero connecting class exactly when its lift can be corrected by a chain in \(K\) to a cycle of \(L\); a cycle of \(K\) becomes a boundary in \(L\) exactly when it is the connecting image of the projected bounding chain. These three arguments prove exactness at all the repeating positions. The same lifting argument, with \(\delta\), proves the cohomological version and its naturality.

The pair sequence comes from \(0\to C_*(A)\to C_*(X)\to C_*(X,A)\to0\). It is degreewise split on the simplex bases, so it stays exact after tensoring or dualizing.

For an open cover \(X=U\cup V\), the sequence

is degreewise split exact: each simplex belongs to \(U\), to \(V\), or to both. Lemma 2.1 replaces the last complex by \(C_*(X)\). We obtain the Mayer–Vietoris sequences. For cohomology the middle-to-intersection map is the difference of restrictions. Dividing throughout by the corresponding chains in a subspace \(A\) gives the relative sequences for \((X,A)\), with pieces \((U,U\cap A)\), \((V,V\cap A)\) and \((U\cap V,U\cap V\cap A)\). Splitting is again obtained on the simplices outside \(A\).

We will use the following precise form of the five lemma. In a commutative diagram of exact five-term sequences, if maps at positions one, two, four and five are isomorphisms, the middle map is an isomorphism. For injectivity, an element with zero image maps to zero at position four, so comes from position two; lift its image-zero condition through position one and use the isomorphisms there to correct its preimage to zero. For surjectivity, map a target middle element to position four; lift there, note its image at position five vanishes, and lift back to a source middle element. Their difference lies in the image of position two, whose isomorphism provides the required correction. This proves the lemma and applies equally to homology and cohomology segments.

For example, cover \(S^m\), \(m\geq1\), by slightly enlarged hemispheres. Each is contractible, and their intersection retracts to \(S^{m-1}\). The reduced Mayer–Vietoris sequence gives \(\widetilde H_j(S^m;G)\cong\widetilde H_{j-1}(S^{m-1};G)\). Starting with two points, this yields \(\widetilde H_m(S^m;G)=G\) and zero in other degrees; the same argument gives reduced cohomology. The generator obtained by orienting the two hemispheres compatibly is the fundamental class of the oriented sphere. Concretely triangulate it as the boundary of an \(m+1\)-simplex, radially projected from an interior point. Each ray meets the boundary once, and its intersection distance is continuous face by face, so this is a homeomorphism. The alternating sum of oriented faces is a cycle because shared faces cancel. Its image at a point inside one face is the positive local disk generator by excision. Since the sphere homology group in degree \(m\) is infinite cyclic, that value one proves the cycle is a generator. To verify the same local value at any point, choose a different centre or a rotated simplex whose face interior contains that point; the radial projections are related through orientation-preserving rotations, which are homotopic to the identity by the plane-rotation argument of Section 6. This gives the fundamental class \([S^m]\), with local value one in every oriented chart.

4. Products, signs and evaluation on fibres

For cochains of degrees \(p,q\) with values in a commutative ring \(R\), define the Alexander–Whitney cup product by

Expanding boundaries, terms deleting vertices strictly before the joining vertex give \(\delta a\smile b\), those after it give \((-1)^p a\smile\delta b\), and the two joining terms cancel. Thus

The formula is natural, is associative (the three face intervals are identical in either parenthesization), and has the constant zero-cochain \(1\) as unit. It descends to cohomology. An absolute class can multiply a relative class for every pair; cochains on the subspace vanish in the relative factor.

We supply the chain comparison that explains external products and graded commutativity. On the total tensor complex use \(\partial(c\otimes d)=\partial c\otimes d+(-1)^{\deg c}c\otimes\partial d\). Define

Conversely define \(S(c\otimes d)\) by triangulating \(\Delta^p\times\Delta^q\) along all lattice paths from \((0,0)\) to \((p,q)\). Give a path the sign of the permutation of its horizontal and vertical steps relative to all horizontal steps followed by all vertical steps. Compose its affine simplex with \((c,d)\). Internal faces from paths interchanging adjacent unlike steps cancel; boundary faces give precisely \(\partial c\otimes d+(-1)^p c\otimes\partial d\). For \(A\), deleting a nonjoining vertex gives the matching tensor boundary and joining-vertex terms telescope. Hence both maps are chain maps, preserving augmentation.

Lemma 4.1 (product chain equivalence). These two maps are inverse up to natural chain homotopies. The homotopies preserve products of subspaces.

Proof. Here is an explicit inductive construction of the needed homotopies. Chains in a convex set have a contraction to any fixed vertex: cone each singular simplex to that vertex by linear interpolation. The cone boundary identity gives \(\partial K+K\partial=1-i\epsilon\), where \(\epsilon\) is augmentation and \(i\) inserts the vertex. Tensoring two such contractions gives a contraction of their tensor complex to \(\mathbb Z\): the homotopy is \(K\otimes1+(i\epsilon)\otimes L\), with the tensor sign \((-1)^{\deg c}\) in the second term. The first identity and then the second verify this formula; its boundary sum is \(1-(i\epsilon)\otimes(j\epsilon)\).

For \(1-SA\), assume a homotopy \(h\) is already defined on lower-dimensional simplices. On the universal diagonal simplex \(\Delta^m\to\Delta^m\times\Delta^m\), form \(z=(1-SA)\sigma-h\partial\sigma\). It is an augmented cycle. Cone \(z\) in the convex product to fill it. For an arbitrary \(\sigma\), push this filling forward by \(\sigma_X\times\sigma_Y\). This defines \(h\) naturally and gives \(\partial h+h\partial=1-SA\).

For \(1-AS\), repeat on a universal tensor generator \(1_{\Delta^p}\otimes1_{\Delta^q}\). The residual cycle is filled by the tensor contraction just constructed, then pushed forward by the two singular simplices. Induction on \(p+q\) gives the other homotopy. If either simplex lies in a subspace, every filling still maps into the corresponding product subspace. \(\square\)

The same induction compares \(A\) with the map obtained by swapping the two coordinates and then applying \(A\) and the signed tensor swap \(c_p\otimes d_q\mapsto(-1)^{pq}d_q\otimes c_p\). They agree on augmentation, and the tensor contraction fills the residual cycles exactly as in Lemma 4.1. Evaluate \(a\otimes b\) on this homotopy after the diagonal \(X\to X\times X\). For cocycles the resulting difference is a coboundary. Therefore

No assertion of cochain commutativity was used.

The equivalence also descends to \(C_*(X,A)\otimes C_*(Y,B)\) and the product complex modulo \(C_*(A\times Y)+C_*(X\times B)\). When \(A,B\) are open, small chains in their union identify this quotient up to chain homotopy with \(C_*(X\times Y,A\times Y\cup X\times B)\). If one subspace is empty the assertion holds for the other subspace without openness. For classes, define the external product by \(a\times b=\operatorname{pr}_1^*a\smile\operatorname{pr}_2^*b\); its cochain formula is evaluation of \(a\otimes b\) on \(A\). This defines external products on all pairs used here. Cup products of two relative classes likewise take values relative to their union when its two subspaces are open, by the same small-chain comparison.

For an integral relative cocycle \(u\) of degree \(r\), define a right cap operator on an \(m\)-simplex, \(m\geq r\), by

and set it to zero when \(m<r\). Boundary expansion and \(\delta u=0\) give \(\partial R_u=R_u\partial\); the faces at the cut cancel by the cocycle equation on the last \(r+1\) vertices. Thus it is a degree-\(-r\) map commuting with the unshifted differentials. If \(u\) vanishes on a subspace, this map descends from relative chains to absolute chains. For a cocycle \(a\) of degree \(m-r\), its defining identity is

We use this convention throughout. Reindexing chain groups here keeps their differential unchanged; it is not the alternative signed convention for suspended chain complexes. Replacing \(u\) by \(u+\delta v\) changes the cap map by a homotopy: on \(C_m\) set \(h=(-1)^{m-r+1}R_v\). The cup-product differential identity gives \(\partial h+h\partial=R_{\delta v}\). Products and this cap identity also make sense with an abelian coefficient group in the first factor and an integral cochain in the second, using integer multiplication on the group.

5. Coefficient groups without a compactness shortcut

To pass from local integral fibre generators to a global class, we need one coefficient theorem. We prove it rather than assuming that integral cohomology commutes with inverse limits of compact subsets; in general it does not.

Lemma 5.1. Every subgroup of a free abelian group is free.

Proof. Well-order a basis of \(F\). Let \(F_\alpha\) be generated by the basis elements preceding \(\alpha\), and put \(J_\alpha=J\cap F_\alpha\) for the subgroup \(J\). At a successor stage, \(J_{\alpha+1}/J_\alpha\) embeds in \(F_{\alpha+1}/F_\alpha=\mathbb Z\). It is either zero or generated by a least positive integer; in the latter case choose a lift in \(J_{\alpha+1}\). At a limit stage take the union. These chosen lifts form a basis: subtracting a multiple of the lift at the last nonzero coordinate reduces that last coordinate, and induction in the well-order gives spanning; the last nonzero coordinate also proves independence. All elements have finite support, and no infinite strictly decreasing sequence of ordinals occurs. \(\square\)

For an abelian group \(M\), choose a free presentation \(0\to J\to F\to M\to0\) and define \(\operatorname{Ext}(M,G)=\operatorname{coker}(\operatorname{Hom}(F,G)\to\operatorname{Hom}(J,G))\). This group does not depend on the presentation. Indeed a map between the quotient groups lifts to a map of the free groups by choosing preimages of their basis elements, and carries kernels to kernels. Two lifts differ by a map into the target kernel; precomposition by that difference becomes a restriction from the target free group and hence vanishes in the cokernel. Lifts of identity maps between two presentations therefore induce inverse cokernel maps. This also proves functoriality.

Theorem 5.2 (universal coefficient exact sequence). For a free abelian chain complex \(K\) there is a natural exact sequence

The last map evaluates a cocycle on cycles. In particular it is an isomorphism if \(H_{m-1}(K)=0\).

Proof. Put \(Z_m=\ker\partial\) and \(B_m=\operatorname{im}\partial\). These groups are free by Lemma 5.1. The surjection \(K_m\to B_{m-1}\) splits because \(B_{m-1}\) is free, so \(K_m=Z_m\oplus L_m\). A homomorphism \(H_m=Z_m/B_m\to G\) extends to \(K_m\) by zero on \(L_m\) and is a cocycle. This proves surjectivity of the evaluation map.

A cocycle in its kernel vanishes on \(Z_m\), so it factors through \(K_m/Z_m=B_{m-1}\). Two such cochains differ by a coboundary exactly when their maps on \(B_{m-1}\) differ by the restriction of a map on \(K_{m-1}\), equivalently of a map on \(Z_{m-1}\), since that summand splits too. Their quotient is the cokernel for the free presentation \(0\to B_{m-1}\to Z_{m-1}\to H_{m-1}\to0\). This identifies the kernel with the stated Ext group. Evaluation, factorization and the presentation comparison above commute with chain maps, so the sequence is natural; the chosen splittings are used only to prove surjectivity. \(\square\)

Over a field \(k\), the corresponding statement reduces to \(H^m(\operatorname{Hom}_k(K,k))=\operatorname{Hom}_k(H_m(K),k)\). The same splitting proof applies to vector spaces, and every linear map from a subspace extends by choosing a basis, so there is no Ext term. These statements apply to the singular relative complexes defined in Section 1.

6. The relative generator in a vector space

The pair \((\mathbb R,\mathbb R\setminus0)\) has homology \(\mathbb Z\) in degree one and zero elsewhere. Indeed the two components of the punctured line and the contractibility of the line, together with the pair sequence, identify \(H_1\) with the kernel of \(\mathbb Z\oplus\mathbb Z\to\mathbb Z\), \((a,b)\mapsto a+b\). The positively directed interval from \(-1\) to \(1\) represents the generator whose boundary is the positive point minus the negative point.

A free chain complex with this homology is chain homotopy equivalent to \(\mathbb Z\) concentrated in degree one. Here is the required algebra. Split \(K_m\to B_{m-1}\) by freeness, as in Theorem 5.2, and split \(Z_m\to H_m\), since these homology groups are free. Then \(K_m=B_m\oplus H_m\oplus L_m\), with \(\partial:L_m\to B_{m-1}\) an isomorphism. Projection to \(H_*\) and its inclusion are inverse up to the homotopy sending \(B_m\) to its inverse image in \(L_{m+1}\) and vanishing on the other summands. This proves the assertion, including a cocycle \(u_1\) evaluating to one on the chosen interval.

Apply Lemma 4.1 repeatedly to the product of \(r\) punctured-line pairs. Their exceptional union is exactly \(\mathbb R^r\setminus0\); all of its factor subspaces are open. The relative complex is therefore chain homotopy equivalent to \(\mathbb Z\) concentrated in degree \(r\). Its positive generator is the product of the \(r\) positive intervals, with coordinate axes in their listed order. Write \(u_r\) for its dual integral class. In particular the relative cohomology is \(G\) in degree \(r\) and zero otherwise for every abelian group \(G\).

An invertible real linear map acts on this generator by the sign of its determinant. To verify the sign, Gram–Schmidt writes a positive-determinant matrix as \(QR\), with \(Q\in SO(r)\) and \(R\) triangular with positive diagonal. Interpolating \(R\) to the identity stays invertible. A rotation in the plane spanned by a unit vector and the first coordinate vector takes that vector to the coordinate vector; for its antipode use a half-turn in a coordinate plane. This reduces an element of \(SO(r)\) to an element of \(SO(r-1)\), and induction gives a path to the identity. The cases \(r=1\) and \(SO(1)=\{1\}\) start the argument. A reflection in one coordinate exchanges the two components in the punctured-line calculation and acts by \(-1\). Every negative-determinant matrix is the product of a positive-determinant one and this reflection. Homotopy invariance proves the claim.

Thus an orientation determines an integral generator \(u_V\) for each fibre vector space \(V\). Modulo two the determinant sign disappears, giving a canonical generator without an orientation.

Proposition 6.1 (trivial Thom isomorphism). For every space \(B\), the class pulled back from \(u_r\) gives isomorphisms

The corresponding homology map \(\pi_*R_{u_r}\) is an isomorphism too. The same assertions hold with the mod-two fibre class and \(\mathbf F_2\) coefficients.

Proof. The relative product chain equivalence identifies the source relative complex with \(C_*(B)\otimes C_*(\mathbb R^r,\mathbb R^r\setminus0)\). Replace the second factor by its chain homotopy equivalent \(\mathbb Z[r]\). Tensoring the equivalence and its homotopy is legitimate for these free complexes; the graded tensor differential verifies the homotopy formula. On the original relative complex the projection to this shifted base complex is \((1\otimes u_r)A\). In degree \(j+r\) it evaluates \(u_r\) on the last \(r\) vertices and retains the first \(j+1\) vertices projected to \(B\). It is exactly \(\pi_*R_{u_r}\). Dualizing with values in \(G\) gives the displayed cup-product formula. This proves both isomorphisms and also proves that the fibre value one characterizes the trivial-bundle Thom class uniquely. \(\square\)

7. A global Thom class, including integral coefficients

Let \(\pi:E\to B\) be a real rank-\(r\) bundle, \(r>0\), over a Hausdorff base. Write \(E_0=E\setminus\text{zero section}\), an open subset. We use either integral coefficients with a chosen orientation, or \(\mathbf F_2\) without one. These alternatives will be called the chosen coefficients. In the integral case an orientation means continuous choices of oriented local frames; the preceding section explains the corresponding fibre generators.

Theorem 7.1 (Thom isomorphism). There is a unique class \(u_E\in H^r(E,E_0)\) whose restriction to every fibre pair is its chosen generator. The groups \(H^i(E,E_0)\) vanish for \(i<r\). Multiplication gives an isomorphism

The homology map \(\pi_*R_{u_E}:H_{j+r}(E,E_0)\to H_j(B)\) is also an isomorphism. In the oriented integral case the cohomology isomorphism holds with any abelian coefficient group \(G\), multiplying by the integral class \(u_E\).

Proof: finite covers. A bundle trivial on one open set satisfies all assertions by Proposition 6.1. Suppose they hold over \(U,V,U\cap V\). The two degree-\(r\) Thom classes agree on the intersection, because they have the same fibre values. Relative Mayer–Vietoris gives a common class on \(U\cup V\), uniquely since \(H^{r-1}(E_{U\cap V},(E_{U\cap V})_0)=0\). It still has the required fibre values.

Right multiplication by this class maps the base cohomological Mayer–Vietoris sequence to the relative total-space sequence. It commutes with coboundaries and restrictions by the cup differential formula, hence with the connecting maps from Section 3. The five lemma proves the cohomological isomorphism in every degree, including vanishing below \(r\). For integral orientation, the same diagram works with every group \(G\), multiplying on the right by the integral class.

On homology use small chains for the two inverse-image open sets. The operator \(\pi_*R_u\) takes a simplex supported above \(U\) to a simplex in \(U\), and likewise for \(V\) and their intersection. It commutes with boundaries, so it defines a map of their split exact chain sequences and their connecting maps. The homological five lemma gives the asserted isomorphism. Its induced map is independent of the cocycle representing \(u\), by the cap homotopy of Section 4.

Induction proves these statements for any finite trivializing cover. At the inductive step, the intersection of the last chart with the union of the earlier charts has a cover by fewer trivializing sets, so its assertion is already available. No infinite-cover induction has been assumed.

Proof: passage to arbitrary bases. For each compact subset \(C\subset B\), the restricted bundle has a finite trivializing cover. Denote its Thom class by \(u_C\). If \(C\subset D\), uniqueness shows that \(u_D\) restricts to \(u_C\).

Every singular chain in \(E\) projects to a compact subset of \(B\). The same holds for a finite collection of chains, including chains that witness a boundary. Consequently

For surjectivity a representing cycle has compact projected support; for injectivity a bounding chain has compact projected support too. Finite unions of compact subsets are compact, so these are filtered limits. This argument is valid for integral homology and for field homology.

The compatible maps \(\pi_*R_{u_C}\) therefore give isomorphisms \(T_j:H_{j+r}(E,E_0)\to H_j(B)\). In particular \(H_{r-1}(E,E_0)=0\), and \(T_0\) identifies \(H_r(E,E_0)\) with \(H_0(B)\). Positive fibre generators map to their component generators under \(T_0\). In the integral case compose \(T_0\) with the homomorphism \(H_0(B)\to\mathbb Z\) sending every component generator to one. Theorem 5.2 identifies this functional with a unique class \(u_E\in H^r(E,E_0;\mathbb Z)\). In the mod-two case use the field version of that theorem and the analogous functional with values in \(\mathbf F_2\).

On every compact \(C\), this class restricts to \(u_C\): their evaluation functionals agree, and the degree-\(r\) evaluation maps there are isomorphisms. Thus \(u_E\) has exactly the desired fibre values. Those values also characterize its evaluation functional, since the fibre generators generate \(H_r(E,E_0)\) by its identification with \(H_0(B)\). This proves global uniqueness.

The global cap map is now \(T_j\), because its restriction to each compact projected support is the already constructed cap map. It is therefore an isomorphism. Apply the natural universal coefficient exact sequences to this map of free complexes, reindexed by \(r\). It gives isomorphisms on the Hom and Ext terms because it is an isomorphism on homology. Exactness, or the five lemma with zero end terms, then gives an isomorphism on cohomology with every group \(G\) in the oriented case. By the cap evaluation identity this dual map is precisely \(a\mapsto\pi^*a\smile u_E\). In the mod-two case field duality gives the same conclusion with \(\mathbf F_2\). This also proves the stated low-degree vanishing. \(\square\)

For rank zero, \(E=B\), \(E_0=\varnothing\), and \(u_E=1\); the theorem is the identity map. No paracompactness or CW assumption was needed in Theorem 7.1. Those hypotheses will be useful for metrics and obstruction theory, but the compact-support chain argument already proves this theorem on every Hausdorff base.

The zero section \(z:B\to E\) and \(\pi\) are homotopy inverses: \((b,v,t)\mapsto(b,tv)\) contracts each fibre. Thus the theorem can equivalently be stated as the isomorphism \(H^j(E)\to H^{j+r}(E,E_0)\), given by multiplication by \(u_E\).

8. Euler classes and their product rule

Let \(j:H^r(E,E_0)\to H^r(E)\) forget the relative condition. Define

With integral coefficients this is the Euler class of an oriented bundle. With \(\mathbf F_2\) coefficients it is defined for every real bundle. Naturality of the Thom class is immediate from its uniqueness, with the following precise assumptions: a fibrewise linear isomorphism over a base map pulls the Thom class back to the Thom class, preserving the chosen orientation in the integral case. It induces a map of the nonzero-vector pairs. Consequently Euler classes are natural for pullbacks and oriented bundle isomorphisms.

The homotopy inverses above imply \(j(u_E)=\pi^*e(E)\). Therefore

In this formula the first relative factor can be forgotten before multiplying: both versions use the same cocycle on the same front face, so they agree.

Reversing the orientation replaces every fibre generator by its negative, hence replaces both the Thom class and the Euler class by their negatives. The automorphism \(v\mapsto-v\) has determinant \((-1)^r\) and fixes the zero section. For odd \(r\), pulling back by it yields \(e(E)=-e(E)\), so \(2e(E)=0\). This proves only the two-torsion assertion; it does not allow dropping torsion in an arbitrary base cohomology group.

The same conclusion can be seen in the multiplication law. For odd \(r\), graded commutativity gives \(u_E\smile u_E=-u_E\smile u_E\). Hence twice this square is zero. Since the square is the Thom image of \(e(E)\), injectivity of the Thom isomorphism again gives \(2e(E)=0\). This identifies why an odd-rank Euler class can survive as torsion even though it disappears with rational coefficients.

Proposition 8.1. If a positive-rank oriented bundle has a nowhere-zero section, its Euler class is zero. The same conclusion holds for the mod-two Euler class without orientation.

Proof. A section \(s:B\to E_0\) has zero pullback of \(j(u_E)\), because that class restricts to zero on \(E_0\). The sections \(s\) and \(z\) are homotopic as maps into \(E\) by scalar multiplication along each fibre. Lemma 1.1 gives \(e(E)=z^*j(u_E)=s^*j(u_E)=0\). \(\square\)

Proposition 8.2 (Whitney product). If \(E,F\) are oriented bundles on one base, orient \(E\oplus F\) by an oriented basis of \(E\) followed by one of \(F\). Then

The same formula holds modulo two without orientations.

Proof. First use the external product bundle over \(B\times C\). Its nonzero set is \(E_0\times F\cup E\times F_0\), a union of open subspaces. The external product \(u_E\times u_F\) is defined by Section 4 and restricts to the positive generator on each ordered product fibre, by the product calculation of Section 6. Uniqueness identifies it with \(u_{E\times F}\). Pull back to the product zero section to obtain \(e(E\times F)=e(E)\times e(F)\). For a common base, pull back along its diagonal. The pullback bundle is \(E\oplus F\), and diagonal pullback of the external product is the cup product by the defining Alexander–Whitney formula. \(\square\)

Reduction of coefficients \(\mathbb Z\to\mathbf F_2\) carries the integral Thom class to the unique mod-two Thom class, since it has the required fibre values. It therefore carries the integral Euler class to the mod-two Euler class. Write \(w_r(E)=e_{\mathbf F_2}(E)\) for this top mod-two characteristic class. Section 4 of Steenrod squares and Stiefel–Whitney classes constructs the full sequence: its top-square identity \(\operatorname{Sq}^r(u)=u\smile u\), together with the displayed Thom identity, proves that its top class is this same \(w_r\). That construction is not used in the preceding proofs.

9. Zeros on spheres and the real-line normalization

Here is the local calculation that turns an Euler class into a number. Let a rank-\(r\) bundle on an oriented sphere \(S^r\) have a section with finitely many isolated zeros. For mod-two coefficients no bundle orientation is needed. The section is a map of pairs \((S^r,S^r\setminus Z)\to(E,E_0)\). Choose disjoint small disks around the zeros. Excision identifies the degree-\(r\) relative group with the direct sum of their local groups. Its pulled-back Thom class has an integer local value in the oriented case, and a mod-two value in the other case. Forgetting the relative condition gives the ordinary Euler class, since the section is homotopic to zero. The fundamental cycle maps to the positive local disk generators, so

This equality follows from the relative evaluation pairing and excision, and has only been asserted here for the spheres whose fundamental cycles were proved in Section 3.

If a smooth zero is nondegenerate, write the section in an oriented chart and bundle frame as \(f(y)=Ay+o(\|y\|)\), with \(A\) invertible. On a sufficiently small disk the error has norm less than half the least singular value of \(A\) times \(\|y\|\). The straight homotopy from \(f\) to \(Ay\) stays nonzero off the origin, so it is a homotopy of the local pairs. The determinant calculation of Section 6 gives \(\operatorname{ind}_x(s)=\operatorname{sign}\det A\), or one modulo two.

For the unit sphere in \(\mathbb R^{r+1}\), the tangent section \(s(x)=e_{r+1}-x_{r+1}x\) has zeros exactly at the north and south poles. Its tangent derivatives there are \(-I\) and \(+I\), respectively. Therefore

For \(S^2\), this says \(e(TS^2)=2a\), where \(a\) evaluates to one on the outward oriented sphere. A nowhere-zero tangent field would contradict Proposition 8.1. For odd spheres the value is zero; an explicit field is \(Jx\), where \(J\) is the standard complex structure on the even-dimensional ambient space. It is perpendicular to \(x\) and has length one. The value \(1+(-1)^r\) also equals the Euler characteristic computed from the sphere homology in Section 3.

For a real line over a circle, the trivial bundle has a nowhere-zero section and mod-two Euler class zero. Represent the Möbius line by \((t+1,v)\sim(t,-v)\). The function \(v=\cos\pi t\) defines a section because its values change sign on translating \(t\) by one. It has exactly one nondegenerate zero per circle, at \(t=1/2\), so the local formula shows that its mod-two Euler class is the nonzero element of \(H^1(S^1;\mathbf F_2)\).

For any real line on a Hausdorff base, pull it back along each loop. The two circle calculations show that evaluation of its mod-two Euler class on that loop is exactly its orientation monodromy. Theorem 6.1 of the classifying-maps chapter identifies degree-one cohomology with all those loop evaluations, component by component. Hence

with the transport definition already proved there. That cohomology-class definition works on any base; complete isomorphism classification was proved for CW bases. This closes the normalization without assuming the cohomology ring of a Grassmannian or any Steenrod operation.

10. Exercises with solutions

Exercise 10.1 (easy). Compute the Euler class of an oriented trivial bundle of positive rank, and explain the rank-zero exception.

Solution. Choose a fixed nonzero vector as section; Proposition 8.1 gives Euler class zero. Alternatively pull back the bundle from a point, whose positive-degree cohomology is zero. A rank-zero bundle has Thom class and Euler class \(1\in H^0(B)\); it has no nonzero vector in a fibre, so the nonzero-section argument does not apply.

Exercise 10.2 (medium). Suppose an oriented rank-\(r\) bundle splits into two oriented odd-rank bundles. Prove that its Euler class has order at most two. What follows if \(H^r(B;\mathbb Z)\) has no two-torsion?

Solution. Write the summands as \(E,F\). By the orientation-reversing automorphism calculation, \(2e(E)=0\). The product rule gives \(2e(E\oplus F)=(2e(E))e(F)=0\). In a cohomology group without two-torsion the class is therefore zero. Both summands must be oriented to apply the integral product formula as stated.

Exercise 10.3 (medium). Let \(G\) be an abelian coefficient group. Show that an orientation-reversing bundle isomorphism changes the Thom isomorphism with coefficients \(G\) by multiplication by \(-1\). What happens when every element of \(G\) has order at most two?

Solution. The integral fibre generator, and hence the integral Thom class, changes sign. The map with coefficients \(G\) is multiplication by that integral class, so it changes by integer multiplication by \(-1\) on \(G\). If \(2G=0\), this multiplication is the identity. This explains the disappearance of orientation signs; it does not make an integral orientation-reversing map orientation preserving.

Exercise 10.4 (hard). Prove the mod-two Euler class of a real line over a CW complex is a complete isomorphism invariant, and determine the line with prescribed evaluations \(a,b\in\mathbf F_2\) on the two coordinate circles of a torus.

Solution. Section 9 identifies this Euler class with the first class defined by orientation monodromy. Theorem 9.2 of the classifying-maps chapter proves its complete classification on every CW complex, including infinite ones, by supplying the cover construction and an oriented positive frame. On the torus the line is represented by a square with fibre signs \((-1)^a\), \((-1)^b\) on the two pairs of opposite sides. Its orientation transport on the coordinate loops has the stipulated evaluations; uniqueness follows from that theorem. It is trivial exactly when both values vanish.

Exercise 10.5 (hard). In the proof of Theorem 7.1, justify the equality of the global class restricted to a compact \(C\) with \(u_C\). Explain why this step does not assume an integral inverse-limit theorem for all cohomological degrees.

Solution. Both classes have the same evaluation functional on \(H_r(E_C,(E_C)_0;\mathbb Z)\): the cap map sends a positive fibre generator to its component generator in \(H_0(C)\), and the functional sends each such generator to one. The finite-cover homology isomorphism shows that these fibre generators generate the degree-\(r\) relative group and that its degree-\(r-1\) group is zero. Theorem 5.2 therefore identifies degree-\(r\) cohomology with these functionals, proving equality. Only homology used a filtered compact-support limit. The global class was produced by the degree-\(r\) universal coefficient isomorphism; all other cohomology degrees were obtained by dualizing the global cap equivalence through the natural universal coefficient sequences.

References

[AT] Allen Hatcher, Algebraic Topology, author's text, Section 2.1 (prism homotopy, small chains and excision), Section 3.1 (universal coefficients), and Section 4.D, Theorems 4D.8–4D.10 (relative Leray–Hirsch, Thom isomorphism and existence).

[VB] Allen Hatcher, Vector Bundles and K-Theory, version 2.2, November 2017, author's text, Section 3.2, especially Proposition 3.13 (Euler naturality, products, odd-rank torsion and the section obstruction) and Proposition 3.14 (the even-sphere calculation).

[R] David Michael Roberts, Algebraic Topology, lecture notes, 2019, source. See the preceding chapter for its marked CC BY 4.0 interval-cover adaptation. No source text is reproduced in this chapter.