Intersection Pairings Self Intersection and Euler Classes

First assume both transversality conditions. By [F1], $S=F^{-1}(B)$ is a neat one-manifold with $\mathrm{\partial }S=(F{\mathrm{\mid }}_{\mathrm{\partial }W}{)}^{-1}(B)$: its dimension is $(a+1)-(n-b)=1$. It is compact because $B$ is closed and $W$ is compact. Its boundary consists of the finite transverse intersections with $A_1$ and $A_0$.

For arbitrary $F$, choose a collar $c(x,t)$ by [F4]. Choose a smooth function $h:[0,1)\to [0,1)$ with $h=0$ near zero and $h(t)=t$ outside a smaller collar. Replace $F(c(x,t))$ there by $F(c(x,h(t)))$. Interpolation between $t$ and $h(t)$ gives a homotopy fixed on the boundary, and the new map $F^{\mathrm{\prime }}$ is constant in the normal coordinate near the boundary. Consequently using $F^{\mathrm{\prime }}$ on both halves extends smoothly to a map $H:DW\to M$. If $\mathrm{\partial }W$ is empty, use the disjoint double without this modification.

Orient $Q=TM{\mathrm{\mid }}_B\mathrm{/}TB$ by normal-first order, so a positive quotient determinant followed by a positive determinant of $TB$ is positive in $TM$. Orient $S$ by $\det TW=\det TS\otimes \det F^{\ast }Q$. At an endpoint choose an outward vector $r\in TS$; neatness makes it outward also in $W$, and it is transverse to $\mathrm{\partial }W$, not tangent to it. A positive boundary determinant $u$ then makes $(r,u)$ positive in $TW$. The quotient image of $u$ has sign equal to the local intersection sign of the boundary map with $B$, by the normal-first definition of $Q$. Thus the boundary point sign of $S$ is that local sign. On $A_1$ this is ${\epsilon }_{A_1}$, and on the oppositely oriented $A_0$ it is $-{\epsilon }_{A_0}$. This determinant-ray argument includes zero-dimensional boundary factors.

By [F2] the signed boundary sum is zero; step 2.1 identifies it with $I(A_1,B)-I(A_0,B)$. Reducing the same finite sum modulo two gives the parity equality.

A submersive parameter family for $H$ from [F4] remains submersive in its parameter directions after restriction to the seam $\mathrm{\partial }W$. Parametric transversality on $DW$ and on $\mathrm{\partial }W$ therefore excludes only two null sets of parameters. Their union is null; choose a good parameter arbitrarily near zero (in parameter dimension zero the bad sets are empty). Its restriction to $W$ and to $\mathrm{\partial }W$ is transverse, and its boundary maps are homotopic to the original inclusions along the parameter segment. Apply step 3.1 to the perturbed maps and then [F3] to recover the original numbers. If the original boundary map was transverse, $H$ is transverse on a seam neighbourhood because it is constant in the collar direction; relative transversality in [F4] instead fixes that neighbourhood. Countable Choice is inherited from [F2] and [F4]; the finite sign calculations add none.

Choose a metric by [F1]. In the derivative calculation and final quotient-coordinate transport of The tubular neighbourhood theorem in a smooth ambient manifold, the constructed derivative sends a tangent vector and a normal lift to their sum, hence induces the identity on the quotient at every zero vector. Compactness of $Z$ allows a uniform small metric disk inside its domain: cover $Z$ by finitely many smaller trivializing patches with compact closures, and take the minimum of their positive allowable radii. This also makes the closed disk compact, since on each such patch its fibre coordinates lie in a bounded closed ball. Rescale the metric so this disk is the unit disk. Radial retraction of $D(\nu )\setminus Z$ onto $S(\nu )$ and the pair sequence [F6] give an isomorphism $H^r(D(\nu ),D(\nu )\setminus Z;R)\to H^r(D(\nu ),S(\nu );R)$. Lift $u_{\nu }$ uniquely along it and use tubular excision to define $\alpha$. For $r=0$, the punctured bundle and sphere are both empty, so this comparison is the identity.

Let $U$ be the open tube and $K$ a smaller closed disk bundle inside it. The inclusion of the outer annulus $U\setminus K$ into the punctured tube is a fibrewise homotopy equivalence, by radial movement to a radius strictly between the inner and outer radii. Pair sequences therefore identify the Thom lift with a class $u_K\in H^r(U,U\setminus K;R)$. It defines $u_c\in H_c^r(U;R)$. Excision extends $u_K$ to $H^r(M,M\setminus K;R)$, and its absolute image is $\bar{\alpha }$. The support compatibility in [F3] gives $\bar{\alpha }\cap [M]=j_{\ast }{D}_U(u_c)$ for $j:U\hookrightarrow M$: represent $[M]$ and its restriction $[M{]}_K$ by the same chain, and cap with the cocycle vanishing outside $K$.

The projection $\pi :U\to Z$ and zero section $\zeta :Z\to U$ are homotopy inverses by fibrewise contraction. Put $b={\pi }_{\ast }{D}_U(u_c)\in H_z(Z;R)$. To compute its restriction at $x\in Z$, restrict to a trivializing product of a tangent ball $T$ and a normal disk $N$. This localization is legitimate at chain level: shrink a tangent ball about $x$, represent the Thom class with support inside a smaller normal disk, and subdivide the finitely many chains until small for the product neighbourhood and its complement. In the quotient modulo $Z\setminus \{x\}$, pieces projected outside the tangent ball vanish. The relative cap and small-chain comparison of [F6] therefore reduce the restriction of $b$ to the cap on this disk product.

Let $d$ and $a$ be the positive tangent and normal relative orientation cycles in this product, with degrees $z$ and $r$. Its ambient orientation cycle is the shuffle $S_{T,N}(d\otimes a)$: its restrictions have the prescribed tangent-first local orientation, so [F4] identifies it with that relative orientation class. The local Thom cocycle is pulled back from a normal cocycle $\eta$ with $\eta (a)=1$, by [F2]. For a product chain $c$, the cap definition gives ${\mathrm{pr}}_{T\mathrm{\#}}({\mathrm{pr}}_N^{\ast }\eta \cap c)=(\eta \otimes \mathrm{i}\mathrm{d}){\mathrm{AW}}_{N,T}({\tau }_{\mathrm{\#}}c),$ where $\tau :T\times N\to N\times T$ swaps the factors and only normal degree $r$ is contracted. In each path of [F5], exchanging the $z$ tangent and $r$ normal steps reverses the order of each of the $zr$ unlike pairs, so ${\tau }_{\mathrm{\#}}{S}_{T,N}(d\otimes a)=(-1{)}^{rz}{S}_{N,T}(a\otimes d)$; and ${\mathrm{AW}}_{N,T}{S}_{N,T}\simeq \mathrm{i}\mathrm{d}$. These identities remain valid on the product relative complexes: the model homotopies preserve the coordinate subspaces, and [F6] supplies the small-chain comparison for their union. Contracting a chain homotopy with the closed $\eta$ gives a boundary (with the cap boundary sign), so the resulting local homology class is $(-1{)}^{rz}\eta (a)d=(-1{)}^{rz}d$. Thus $b$ restricts at every $x$ to $(-1{)}^{rz}$ times the local orientation of $Z$.

By [F4], $b=(-1{)}^{rz}[Z]$, including disconnected $Z$. Fibrewise contraction and homotopy invariance give $D_U(u_c)={\zeta }_{\ast }b$, so step 2.1 yields $\bar{\alpha }\cap [M]=(-1{)}^{rz}{j}_{\ast }{\zeta }_{\ast }[Z]=(-1{)}^{rz}(i_Z{)}_{\ast }[Z]$. Empty $Z$ gives zero throughout. For rank zero the normal Thom class is the supplied orientation unit, and the same determinant comparison gives the oriented component classes; no assumption that that unit is always $+1$ is made. In characteristic two the sign is $1$. AC is used through [F1]–[F3], and no extra orientation selection is made.

The zero section $s_0:M\to E$ is a smooth embedding [F1], so by [F2] its normal bundle is the quotient ${\nu }_{s_0}=TE{\mathrm{\mid }}_{s_0(M)}\mathrm{/}T(s_0(M))$. The canonical splitting $TE{\mathrm{\mid }}_{s_0(M)}\cong TM\oplus E$ along the zero section, whose vertical summand is the fibre direction, identifies this quotient with $E$ as smooth bundles over $M$ by [F3] and [F4]; moreover $s_0(M)$ is closed in $E$ because in every bundle chart its complement is the open set of nonzero vectors.

The section $s$ is transverse to $s_0(M)$ in the sense of [F6], so [F5] makes $Z=s^{-1}(s_0(M))$ an embedded submanifold of $M$ of codimension $r$; it is closed because $s_0(M)$ is closed and $s$ is continuous, so $\dim Z=n-r$ when nonempty. The quotient map $D^{v}s:TM{\mathrm{\mid }}_Z\to E{\mathrm{\mid }}_Z$ has kernel $TZ$ by [F5] and is surjective by transversality. In bundle charts it is the derivative of the local section components at their zeros, hence smooth; it induces a smooth fibrewise isomorphism $TM{\mathrm{\mid }}_Z\mathrm{/}TZ\to E{\mathrm{\mid }}_Z$, whose inverse is smooth by the inverse-matrix formula. Here $E{\mathrm{\mid }}_Z$ is the pullback along the inclusion $Z\hookrightarrow M$, not along the section $s:M\to E$. Combining with step 1.1 gives the canonical isomorphism ${\nu }_Z\cong E{\mathrm{\mid }}_Z$ of smooth bundles over $Z$.

Orientation clause. The isomorphism of step 1.1 carries the orientation of $E$ to the normal orientation of $s_0(M)$ induced by the ambient $E$ and the zero-section orientation, by the third-orientation rule [F9] applied with the total-space orientation in which a positive tangent basis of $s_0(M)$ followed by a positive fibre basis is positive (the tangent-first convention of this pair). Pullback along $s$ preserves this ordered determinant-line comparison because $ds$ maps the normal directions of the transverse preimage isomorphically onto the normal directions of $s_0(M)$ by [F5] and [F8], and the induced orientation of $Z$ in $M$ is the one for which a positive tangent basis of $Z$ followed by a positive normal basis is positive in $M$ [F9]. Hence the orientation of ${\nu }_Z$ induced from $M$ and $Z$ corresponds to the supplied fibre orientation of $E{\mathrm{\mid }}_Z$; over ${\mathbb{F}}_2$ both sides carry their unique nonzero generator.

Pair comparison and bounding. Choose a smooth metric by Every smooth vector bundle admits a smooth bundle metric. The radial retraction of the punctured disk bundle to the sphere bundle, together with Long exact sequence of a pair in singular cohomology, makes the restriction $H^r(D(E),D(E)\setminus s_0(M);R)\to H^r(D(E),S(E);R)$ an isomorphism. Let ${\tilde{u}}_E$ be the inverse image of $u_E$; in rank zero both removed subspaces are empty. Replace $s$ by $\rho \circ s$ with $\rho (v)=v\mathrm{/}\sqrt{1+\mathrm{\mid }v{\mathrm{\mid }}^2}$; this is smooth and lands in the open unit disc bundle, has the same zero locus $Z$ because ${\rho }^{-1}(0)=\{0\}$, and $d{\rho }_0=\mathrm{i}\mathrm{d}$, so transversality of $s$ to the zero section at every point of $Z$ is unchanged. Thus $s$ is a map of pairs from $(M,M\setminus Z)$ to $(D(E),D(E)\setminus s_0(M))$, and below $s^{\ast }(u_E)$ abbreviates the precisely typed $s^{\ast }({\tilde{u}}_E)$. Take a tubular chart $\mathrm{\Phi }:\mathrm{\Omega }\subseteq {\nu }_Z\to U$ from [F1]. Its vertical differential followed by the quotient $q_z:T_{z}M\to {\nu }_{Z,z}$ gives $B_z=q_z\circ d{\mathrm{\Phi }}_{0_z}{\mathrm{\mid }}_{{\nu }_{Z,z}}$. Since $\mathrm{\Phi }$ fixes $Z$, its tangent differential is the identity on $T_{z}Z$; the invertibility of $d{\mathrm{\Phi }}_{0_z}$ therefore makes $B_z$ invertible. In local bundle coordinates its matrix consists of smooth first derivatives, and the inverse matrix is smooth by the cofactor formula. Thus $B$ is a smooth bundle automorphism. Replace $\mathrm{\Phi }$ by $\mathrm{\Psi }=\mathrm{\Phi }\circ B^{-1}$ on $B(\mathrm{\Omega })$; this is a tubular chart and $q_z\circ d{\mathrm{\Psi }}_{0_z}{\mathrm{\mid }}_{{\nu }_{Z,z}}=\mathrm{i}\mathrm{d}$.

Local normalization. On a trivializing chart $U$ write $s(x)=(x,f(x))$ with $f:U\to {\mathbb{R}}^r$, $f(0)=0$ and $d{f}_0$ surjective; $Z\cap U=f^{-1}(0)$ and $T_{0}Z=\ker d{f}_0$. The normalized Thom class restricts on $(D^r,S^{r-1})$ to the orientation generator [F2], so the local fibre component represents the fibre orientation class in the punctured-fibre pair; the invertible normal derivative permits computing its pullback on a small normal disk by a local diffeomorphism. Under the normal identification ${\nu }_Z\cong E{\mathrm{\mid }}_Z$ of [F6], the composite ${\nu }_{Z,0}\stackrel{d{f}_0}{\to }{\mathbb{R}}^r$ is an orientation-preserving isomorphism, the vertical part of $d{f}_0$ being exactly the comparison defining the induced orientation; so the local generator is the normal generator. This uses a tubular chart with normal differential the identity; an arbitrary orientation-reversing fibre reparametrization would reverse the integral generator.

Global identification. The tube of [F1] identifies a neighbourhood of $Z$ with a neighbourhood of the zero section of ${\nu }_Z$. On each sufficiently small normal disk at $z\in Z$, the fibre component of $s$ has its only zero at $z$ and has invertible derivative there by [F6]. Choice-free smooth inverse function theorem in Euclidean space gives a local diffeomorphism. Its differential preserves the supplied $R$-orientation by [F6] and the normalization in step 1.1, so its pullback carries the fibre orientation generator to the prescribed normal generator; over ${\mathbb{F}}_2$ the local degree is $1$ regardless of sign. Thus step 1.2 says precisely that the relative pullback restricts to the prescribed generator on every normal fibre pair. Thom uniqueness [F3] identifies it on the tube with $u_{{\nu }_Z}$. Excision [F4] identifies $H^r(M,M\setminus Z;R)$ with the relative cohomology of that tube and its complement of $Z$; choose a small metric disk subbundle inside the chart domain (possible uniformly because $Z$ is compact), and use excision to restrict to its interior. The punctured-disk comparison from step 1.1 identifies the resulting group with $H^r(D({\nu }_Z),S({\nu }_Z);R)$ after rescaling the metric. Hence $s^{\ast }(u_E)$ is the image of $u_{{\nu }_Z}$ under the tubular identification.

Euler class. Let $s_0:M\to D(E)$ be the zero section and $u_E^{\mathrm{a}\mathrm{b}\mathrm{s}}=j^{\ast }{u}_E\in H^r(D(E);R)$ its absolute restriction. The forget-support map $H^r(M,M\setminus Z;R)\to H^r(M;R)$ factors the pullback, so the image of $s^{\ast }(u_E)$ is $s^{\ast }(u_E^{\mathrm{a}\mathrm{b}\mathrm{s}})$. The affine homotopy $(x,t)\mapsto (1-t)s(x)$ stays inside the disc bundle and is a homotopy from $s$ to $s_0$ through sections, so Homotopic maps induce equal maps in singular cohomology gives $s^{\ast }(u_E^{\mathrm{a}\mathrm{b}\mathrm{s}})=s_0^{\ast }(u_E^{\mathrm{a}\mathrm{b}\mathrm{s}})$. By [F5] this is $e(E)$. Empty $Z$ and rank zero follow from the same formulas; AC is inherited from the Thom and tubular suppliers.

By [F1], $Z$ has codimension $r$ and normal bundle $E{\mathrm{\mid }}_Z$ with the prescribed orientation. By [F5] every bundle and base used here meets the Thom hypotheses. Let $\alpha \in H^r(M,M\setminus Z;R)$ be its normal Thom extension and $\bar{\alpha }$ its absolute image. By [F3], $\bar{\alpha }=e(E)$.

Apply [F2] with $z=n-r$ to obtain $e(E)\cap [M]=(-1{)}^{r(n-r)}(i_Z{)}_{\ast }[Z]$. If $Z$ is empty its relative group is zero and [F3] gives $e(E)=0$, including $r>n$, when transversality forces the zero locus to be empty. Cap duality [F4] gives the stated equivalence of nonzero classes.

If $r=n$, each zero is nondegenerate, and its point orientation is the sign of $D^{v}s:T_{x}M\to E_x$ in the supplied orientations. The cap formula of [F4] followed by augmentation therefore gives $⟨e(E),[M]⟩={\sum }_{x\in Z}\mathrm{sgn}(D^v{s}_x)$. If $r=0$, $Z=M$ has dimension $n$; its induced orientation is the ambient orientation multiplied by the supplied rank-zero orientation unit $o$ (over $\mathbb{Z}$, $o^{-1}=o$). Thus $[Z]=o[M]$ and the formula reads $e(0_M,o)\cap [M]=o[M]$, as [F5] requires. For the zero section the vertical derivative is zero at every base point, so it is surjective exactly when $r=0$ if the base is nonempty; there are no points to check when the base is empty. Empty manifolds and the canonical mod-two orientations satisfy the same formulas. AC is inherited from the stated suppliers.

By [F1] choose a smooth $f:A\to M$, homotopic to $i_A$, transverse to $B$. Then $I(A,B)=I(f,B)$ and $P=f^{-1}(B)$ is finite by [F5]. Homotopy invariance of homology gives $f_{\ast }[A{]}_A=[A]$. By [F2] and $\mathrm{P}\mathrm{D}[A]\cap [M]=[A]$, $⟨\mathrm{P}\mathrm{D}[A]⌣\mathrm{P}\mathrm{D}[B],[M]⟩=⟨f^{\ast }\mathrm{P}\mathrm{D}[B],[A{]}_A⟩\mathrm{.}$ This uses the map $f$ on the fixed oriented source $A$, and does not replace $A$ by its image.

Pull back ${\alpha }_B$ along the map of pairs $(A,A\setminus P)\to (M,M\setminus B)$ to get $\beta$. At $p\in P$, the quotient derivative $q_p:T_{p}A\to {\nu }_{B,f(p)}$ is an isomorphism. A normalized tubular chart for $B$ and a local trivialization of its normal bundle give a normal-component map with derivative $q_p$. The inverse function theorem makes it a local diffeomorphism. In the induced normal coordinates the fibre Thom generator pulls back to ${\sigma }_p$ times the source orientation generator, where ${\sigma }_p$ is the orientation-ray sign of $q_p$: using the local diffeomorphism as a source chart proves this directly. Excision [F4] and the finite direct sum of the point-supported relative complexes then give $⟨\bar{\beta },[A{]}_A⟩={\sum }_{p\in P}{\sigma }_p\mathrm{.}$ In dimension zero the same statement is multiplication of the supplied point and normal orientation units, without an inverse-function argument.

Let $u$ be a positive tangent determinant of $B$ and $v$ a positive normal determinant. By tangent-first normal orientation, $(u,v)$ is positive in $M$. If $d$ is a positive determinant of $T_{p}A$, then $(u,d{f}_p(d))$ has sign ${\sigma }_p$ because quotienting its second block gives $q_p(d)$. Swapping the blocks of dimensions $b,a$ shows that $(d{f}_p(d),u)$ has sign $(-1{)}^{ab}{\sigma }_p$. Thus [F5] gives ${\epsilon }_p=(-1{)}^{ab}{\sigma }_p$, including the point-ray case. By [F3], $f^{\ast }\mathrm{P}\mathrm{D}[B]=(-1{)}^{ab}\bar{\beta }$. Combining with steps 1.1–2.1 yields the displayed formula. Empty $P$ gives zero on both sides. Over ${\mathbb{F}}_2$ the same finite local evaluation applies with every orientation sign equal to one. AC is inherited through transverse representatives, Thom existence and duality.

The sets. Because $s$ is small, $s(A)$ lies in the interior of the disc bundle, and $\phi$ is injective on the disc bundle. For $p,q\in A$, $\phi (p)=\phi (s(q))$ holds iff $p=s(q)$ as points of the total space (with $p$ viewed as the zero vector over $p$), which forces $p=q$ and $s(p)=0$. Hence $A\cap A_s=\phi (Z(s))$; transversality of $s$ to the zero section is exactly the statement that the vertical differential is surjective at each zero [F1], so the graph $A_s$ meets $A$ transversely there [F2] and by [F4] the tube is a slice chart in which the two submanifolds are the zero section and the graph of the local expression of $s$.

Local model. At $x\in Z(s)$ choose a slice chart of the tube with domain coordinates $(u,v)\in {\mathbb{R}}^a\oplus {\mathbb{R}}^a$ in which $A=\{v=0\}$, $\phi$ is the identity chart, and the orientation of $M$ is the ordered product of the tangent orientation of $A$ and the normal orientation of ${\nu }_A$; this is the tangent-first convention of this pair (An oriented transverse normal bundle orients an embedded submanifold). In these coordinates $A_s$ is the graph of $Ju+O(\mathrm{\mid }u{\mathrm{\mid }}^2)$, where $J={\mathrm{\partial }}_{\nu }{s}_x$ is the vertical derivative. The tangent space of $A_s$ at $x$ is spanned by the vectors $e_j+{\sum }_k{J}_{kj}{e}_{a+k}$; the map $T_{x}A\oplus T_x{A}_s\to T_{x}M$ written in the ordered ambient basis has matrix $\left(\begin{array}{cc}{I}_a& I_a\\ 0& J\end{array}\right)$, whose determinant is $\det J$ (the matrix is square because $2a=n$). By [F3] the sign ${\epsilon }_{(A,A_s)}(\phi (x))$ is the sign of this determinant, proving the displayed identity, including the factor order (first $A$, then $A_s$).

For $a=0$, each point of $A_s=A$ carries the transported point sign ${ϵ}_A$. The local ordered intersection sign is ${ϵ}_A^2{ϵ}_M={ϵ}_M$. The tangent-first normal orientation is ${ϵ}_{\nu }={ϵ}_M{ϵ}_A$, so the vertical map has ray sign ${ϵ}_A{ϵ}_{\nu }={ϵ}_M$ as well. The set and transversality arguments in step 1.1 need no orientations.

Global statement. Adding the finitely many points of $Z(s)$ (finite because $A$ is compact and the zeros of a transverse section are isolated) gives $I(A,A_s)={\sum }_{x\in Z(s)}\mathrm{sign}\det {\mathrm{\partial }}_{\nu }{s}_x$, which is the self-intersection number of The self-intersection number of a complementary-dimensional oriented submanifold. The definition of $A\cdot A$ uses only transverse push-off sections, so no perturbation of a non-transverse section is needed here.