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For a closed smooth embedding $i:N\hookrightarrow M$ of codimension $r$ whose normal bundle is oriented over a coefficient ring $R$, a tubular neighborhood, excision and the Thom isomorphism theorem identifyThe relative-to-absolute map defines $i_{!}:H^j(N;R)\to H^{j+r}(M;R)$. For closed oriented $M,N$, it is characterized bywith compatible orientations. It obeys $i_{!}(b⌣i^{\ast }c)=i_{!}b⌣c$. If $j:M\setminus N\hookrightarrow M$ and $\delta :H^q(M\setminus N;R)\to H^{q+1-r}(N;R)$ is the connecting map followed by the inverse Thom isomorphism, then, over ${\mathbb{F}}_2$,These identities follow from naturality of the relative cup product and show how the exact sequence determines multiplication in complements. Over ${\mathbb{F}}_2$ every real normal bundle has the required orientation.

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.

Remove tubular neighborhoods of diffeomorphic embedded oriented surfaces with opposite normal Euler numbers, and glue their boundary circle bundles by a fiber-reversing identification. The resulting smooth manifold is a fiber sum. The square-zero case uses trivial normal bundles. Additional symplectic hypotheses give a symplectic fiber sum along a square-zero surface.

For a closed smooth submanifold $M\subseteq N$ of codimension $c>0$, a tubular neighborhood and the homological Thom isomorphism theorem identifySubstitution in the long exact sequence in relative homology gives the displayed embedding Gysin sequence. It does not require orientability over ${\mathbb{F}}_2$.

The complement of a closed tubular neighborhood of a smooth plane conic has the displayed integral homology. The Excision theorem and Thom isomorphism theorem reduce the calculation to the relative long exact sequence: the ambient degree-four fundamental class restricts with coefficient one, while the degree-two map is intersection with the conic and has coefficient two. A collar neighborhood identifies the open and compact exterior homotopy equivalence types. The boundary instead has first homology $\mathbb{Z}/4$, from the Gysin sequence and the normal Euler class four.

The exterior of a link is the complement of the interior of a closed tubular neighborhood of the entire link. Its boundary consists of one torus per component; its first homology has the component meridians as a free basis.

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.

The symplectic neighborhood theorem says that a symplectomorphism between compact symplectic submanifolds, together with a symplectic vector bundle isomorphism of their symplectic normal bundles, extends to a symplectomorphism of neighborhoods. To outline the proof, use tubular neighborhoods to extend the bundle identification smoothly. The two pulled-back symplectic forms agree as bilinear forms along the zero section. The relative Poincaré lemma writes their difference as $d\alpha$, with $\alpha$ vanishing there. Their linear interpolation ${\omega }_t$ remains nondegenerate on a sufficiently small neighborhood. Solvingand integrating $V_t$ gives the relative Moser theorem; the flow fixes the submanifold and carries one form to the other.

For the symplectic fiber sum along a square-zero surface, the self-intersection is the Euler number of the oriented rank-two normal bundle. Thus each normal bundle is trivial. Rescale one ambient symplectic form if necessary to equalize the areas of the two copies of $C$. The Moser theorem for the surface then supplies a base identification preserving their area forms. The symplectic neighborhood theorem identifies each neighborhood with a product carryingRemove small disk neighborhoods and glue collars with the base identification andfor a suitable positive constant $a$. Both signs reverse, so the normal two-forms match; the base forms already match. They define a closed nondegenerate two-form across the neck, agreeing with the ambient forms elsewhere. This constructs the symplectic sum. Its construction uses the area normalization and the chosen normal-bundle gluing.

For a smooth counterexample, take $X=Y={\mathbb{CP}}^2$, and in a small four-ball in each choose an unknotted sphere bounding a three-ball. These spheres are null-homologous and have self-intersection zero. Their standard smooth fiber sum isIndeed, the complement of the standard $S^2\times D^2$ in $S^4$ is $S^1\times D^3$; doubling these complements gives $S^1\times S^3$, and the two ambient projective-plane summands remain as connected sums. This is the smooth fiber sum along unknotted null-homologous spheres.

In the induced orientation, both summands in $Z={\mathbb{CP}}^2\#({\mathbb{CP}}^2\#(S^1\times S^3))$ have positive positive index of the intersection form. The symplectic connected-sum obstruction therefore rules out a symplectic form: connected-sum vanishing of the Seiberg–Witten invariant of a four-manifold contradicts Taubes nonvanishing theorem. In the opposite orientation the intersection form is negative definite, also ruling out a symplectic form, whose class has positive square. Thus the smooth fiber sum need not be symplectic.

Choose a complex projective line in each Complex projective plane summand. Make the connected sum at small disks on these lines, so the lines themselves join through the neck to form a smoothly embedded sphere $S\cong S^2$. Its class is the sum $h_1+h_2$ of the two line classes. The intersection form is diagonal with entries one, henceThe oriented normal bundle of $S$ is a rank-two real bundle with Euler number two, because its Euler number equals the self-intersection number. Oriented plane bundles over $S^2$ are classified by that integer, so this normal bundle is isomorphic to $T{S}^2$. The boundary of a closed tubular neighborhood $N$ is consequently the unit tangent bundle of $S^2$.

A unit tangent pair $(x,v)$ determines the oriented orthonormal frame $(x,v,x\times v)$ in ${\mathbb{R}}^3$. Thus $UT{S}^2\cong SO(3)\cong {\mathbb{RP}}^3$, the latter identification coming from the double covering by unit quaternions. Therefore $\partial N$ is the required smoothly embedded ${\mathbb{RP}}^3$. Its complement is disconnected: the nonempty interior of $N$ and the nonempty exterior $X\setminus N$ are disjoint open subsets of $X\setminus \partial N$ that exhaust it. The neighborhood can be chosen small enough that its exterior is nonempty. This is the construction of separating projective three-space in a positive connected sum.

Write $M{U}^{\ast }={\Omega }_U^{\ast }$ for complex cobordism. A complex orientation of a smooth map is chosen stable complex normal data, not a complex structure on either manifold separately. More concretely, replace the continuous map by a homotopic smooth map $f_0$, choose an embeddingand give its normal bundle ${\nu }_j$ a stable complex structure on a real vector bundle. The auxiliary dimension can be chosen so the normal rank is even; further stabilization makes the normal bundle actually complex. Equivalence means stabilization by trivial complex summands and homotopy of the oriented embedding data. The stable identityidentifies this as orientation of the relative stable normal bundle. For odd relative codimension one uses an odd auxiliary suspension coordinate; the degree formula below is unchanged. This is the geometric factorization model for complex-oriented maps used in Algebraic and Geometric Topology 18, geometric cobordism discussion.

Let $d=\dim Y-\dim X$ and $r={\mathrm{rank}}_{\mathbb{R}}{\nu }_j=d+N$. The chosen normal orientation gives a Thom class $u_{{\nu }_j}$ of degree $r$ and the Thom isomorphism in complex cobordism. For compact manifolds the Pontryagin-Thom collapse of a tubular neighborhood to its Thom space isFor $a\in M{U}^q(X)$, define the Gysin map in complex cobordism by the compositeThus $f_{!}$ raises degree by the real relative codimension $d$. This definition depends on the specified orientation and is invariant under the stated equivalences of its normal data. The plus sign on a space denotes adjoining a disjoint basepoint.

For the embedding $i:L\hookrightarrow M$ with normal rank $2n$, let $c:M_{+}\to \mathrm{Th}(\nu )$ be its tubular collapse. The Thom construction gives $i_{!}(1)=c^{\ast }{u}_{\nu }$. Restricted to $L$, this collapse is exactly the zero section $z:L\to \mathrm{Th}(\nu )$. ThereforeThis is the self-intersection formula in complex cobordism. For an actual complex rank-$n$ normal bundle its top Chern class in complex cobordism is the zero-section pullback of its Thom class, so the right side is $c_n^{MU}(\nu )$. Equivalently, under the splitting principle for complex vector bundles both are the product of the first Chern classes of the line summands. The same argument gives $i^{\ast }{i}_{!}(a)=a{e}_{MU}(\nu )$ for every $a$.

There is a real distinction if the printed word “stable” is interpreted literally. The Euler class of a stably complex real bundle need not equal $c_n^{MU}$ of its stable complex class. For the diagonal embedding $S^4\hookrightarrow S^4\times S^4$, the normal bundle is $T{S}^4$. It has a stable complex structure withso its stable complex rank-two class is trivial and has second Chern class zero. However the ordinary integral image of $i^{\ast }{i}_{!}(1)$ is $e(T{S}^4)$, which evaluates to the Euler characteristic $2$. Thus the always-valid answer is the cobordism Euler class. The requested top-Chern identification is valid for an actual complex normal bundle, or if “top Chern class” here is explicitly being used as the Euler-class terminology for the chosen normal orientation.

We prove the Weinstein neighborhood theorem with the canonical sign convention ${\omega }_0=-d\alpha$. The essential first step is to match the two symplectic forms as bilinear forms on the entire tangent space along $X$, not only after pulling them back to $TX$.

Here is the needed symplectic splitting along a Lagrangian submanifold. Write $E=TM{|}_X$ and $L=TX$. Choose a smooth vector bundle complement $N$ of $L$, for example using a Riemannian metric. Since $L$ is a Lagrangian subspace in every fiber, the pairing mapis an isomorphism: its kernel is $N\cap L^{\omega }=N\cap L=0$, and both bundles have rank $n$. Let $s:L^{\ast }\to N$ be its inverse and put $b(\eta ,\zeta )=\omega (s\eta ,s\zeta )$. Define $a:L^{\ast }\to L$ byand set $\tilde{s}=s+a$. Nondegeneracy of the dual pairing gives a unique smooth $a$. Since $b$ is alternating,Thus $K=\tilde{s}(L^{\ast })$ is a smooth Lagrangian complement to $L$. The mapis a vector bundle isomorphism, is the identity on $TX$, and satisfiesThe last equality uses the canonical splitting of the tangent bundle of $T^{\ast }X$ along its zero section.

Choose a Riemannian metric near $X$ for which $K$ is orthogonal to $TX$. The tubular neighborhood construction using its Riemannian exponential map gives a diffeomorphismafter shrinking around the zero section. It restricts to the given inclusion of $X$ and has differential $A$ along it. Consequently $\beta ={\psi }^{\ast }\omega$ and ${\omega }_0$ agree pointwise as ambient bilinear forms at every point of the zero section.

We use the following precise relative Moser theorem. If two closed symplectic forms ${\gamma }_0,{\gamma }_1$ near a compact embedded submanifold $S$ agree as ambient bilinear forms along $S$, then, on smaller neighborhoods, there is a diffeomorphism $\kappa$ fixing $S$ pointwise with ${\kappa }^{\ast }{\gamma }_1={\gamma }_0$. One sufficient version allows a symplectic path ${\gamma }_t$ with ${\partial }_t{\gamma }_t=d{\eta }_t$ and ${\eta }_t=0$ as ambient covectors along $S$: solve ${\iota }_{Z_t}{\gamma }_t=-{\eta }_t$ and use its flow. The vector field vanishes on $S$, so the flow fixes $S$, and compactness permits a common smaller neighborhood for the full time interval.

For completeness, all hypotheses can be checked directly here. Put $\delta =\beta -{\omega }_0$ and ${\gamma }_t={\omega }_0+t\delta$. At the zero section, every ${\gamma }_t$ equals ${\omega }_0$. The nondegenerate property is open, and compactness of $X\times [0,1]$ allows a single smaller neighborhood on which all ${\gamma }_t$ are symplectic forms. Choose it fiberwise star-shaped. Write $H_s(q,p)=(q,sp)$ and let $E$ be the vertical radial vector field. The radial homotopy primitive near a zero sectionis smooth, is zero as an ambient covector along the zero section, and satisfiesThe radial homotopy operator gives this identity, using $d\delta =0$ and $H_0^{\ast }\delta ={\pi }^{\ast }{i}_0^{\ast }\delta =0$. Smoothness at $s=0$ follows because contraction with $E$ supplies a factor of $s$ after evaluation at $(q,sp)$; the vanishing of $\delta$ along $X$ supplies additional vanishing. Therefore solve ${\iota }_{Z_t}{\gamma }_t=-\eta$. By Cartan's magic formula, its flow maps ${\kappa }_t$ satisfyAfter shrinking the starting neighborhood, these flows exist for $0\le t\le 1$ and fix $X$. This is a relative local construction, not an assertion of completeness throughout the noncompact cotangent bundle.

Finally set $\phi =\psi \circ {\kappa }_1$, let $U_0$ be its sufficiently small domain, and let $U=\phi (U_0)$. ThenThus the map is a symplectomorphism of neighborhoods and agrees with the specified inclusion at every point of $X$.

A Heegaard splitting of a closed oriented three-manifold is a decomposition $Y=H_0{\cup }_{\Sigma }{H}_1$ into two handlebodies, with their common boundary the Heegaard surface $\Sigma$. With boundary present, the corresponding pieces are compression bodies, whose negative boundaries account for $\partial Y$.

For the given triangulation, take the barycentric subdivision. A regular neighbourhood $H_0$ of the original one-skeleton is a handlebody: thicken vertices to balls and edges to one-handles, then contract a spanning tree. The complementary region is a regular neighbourhood $H_1$ of the dual one-skeleton, whose vertices are tetrahedron centres and whose edges cross triangular faces. It too is a handlebody. Both graphs are connected, and their common boundary supplies the Heegaard splitting. If the original triangulation has $v$ vertices and $e$ edges, its Heegaard surface has genus $e-v+1$; the dual count gives the same number by $v-e+f-t=0$.

Choose an oriented meridian of a knot $\mu$ and the Seifert longitude $\lambda$ supplied by a Seifert surface for the null-homologous knot. For relatively prime integers $p,q$, rational Dehn surgery removes the interior of a tubular neighborhood and attaches a solid torus with its meridian of a solid torus on the unoriented slope $p\mu +q\lambda$. The choices $(p,q)$ and $(-p,-q)$ describe the same slope; $q=0$ is the original meridional filling. The boundary gluing reverses boundary orientation so that the oriented three-manifold extends across the filling.

For integral coefficients $n_i$, attach two-handles to $B^4$ along the components of the framed link, with their indicated Seifert framing shifts. The boundary operation removes $S^1\times D^2$ and inserts $D^2\times S^1$, with meridian $n_i{\mu }_i+{\lambda }_i$. Thus the compact oriented surgery trace satisfiesFor a finite rational coefficient, use a negative continued fractionReplace that component by an integrally framed chain of successive meridians with coefficients $a_0,\dots ,a_k$. Repeated slam-dunk moves give back $p/q$. Perform this replacement for every rationally framed component, leaving meridional fillings out. The resulting integral framed link has the same filled boundary, so its surgery trace proves that every such rational filling bounds a compact oriented four-manifold.

The drawn component $L_1$ is an unknot. Its exterior is a solid torus $V$, in which $L_2$ is a pattern of a satellite knot. A meridian of a solid torus of $V$ is ${\ell }_1$, whereas its longitudinal direction is $m_1$. Thus the prescribed gluing sends these directions to $m$ and $\ell$, respectively. This identifies $V$ with a tubular neighborhood of $K$ with its zero Seifert framing.

Removing the pattern before making this identification givesEquivalently, meridionally filling the remaining boundary first removes $L_2$ from the construction and leaves $E(K)\cup V=S^3$; its filling core is the required satellite knot $C(K)$.

In the original link diagram, the two parallel passages through a meridional disk of $V$ run in the same direction; closing them uses a single interchange. The winding number of a satellite pattern is therefore two. Untwisting the surrounding disk shows that $P(U)$ is the unknot (the pattern is the $(2,1)$ cable knot, up to the harmless sign of its single interchange). In particular ${\Delta }_{P(U)}=1$. The Satellite formula for the Alexander polynomial now gives the concise answerHere $\doteq$ allows multiplication by $\pm t^j$. Reversing the winding orientation replaces $t^2$ by $t^{-2}$, giving the same Alexander polynomial of a knot up to such a unit.

Euler number and self-intersection. Orient the rank-two normal bundle ${\nu }_S$ by the orientations of $S$ and $M$: the ordered tangent and normal spaces have the ambient orientation. Choose a smooth normal section $s$ transverse to the zero section. Its zeros are isolated, and their signed number is the evaluation $⟨e({\nu }_S),[S]⟩$ of the Euler class on the fundamental class. Scale the section sufficiently small to lie in a tubular neighborhood. Its graph is a push-off $S^{\prime }$ isotopic to $S$.

The intersections of $S$ with $S^{\prime }$ occur exactly at the zeros of $s$. In a positively oriented local splitting $TS\oplus {\nu }_S$, the basis formed from the tangent spaces to $S$ and to the graph has block matrixIts determinant is $\det Ds$, so the local intersection sign is precisely the local zero index of the section. Summing proves the Euler number equals self-intersection identity

The conic. The given map is the degree-two Veronese map $[u:v]\mapsto [u^2:uv:v^2]$ followed by the projective linear transformationThe Veronese map is injective with nonvanishing differential: on the chart $u\ne 0$ its coordinates are $[1:t:t^2]$, and on $v\ne 0$ they are $[s^2:s:1]$. It is therefore a smooth embedding, and $S$ is a smooth plane conic diffeomorphic to ${\mathbb{CP}}^1\cong S^2$. Its equation in the given coordinates is $z_0^2-z_2^2=4z_1^2$.

Let $L$ be a projective line, with its complex orientation. The cohomology ring of complex projective space gives $[L]\cdot [L]=1$ in ${\mathbb{CP}}^2$. The hyperplane $z_1=0$ meets $S$ at $[u:v]=[1:0]$ and $[0:1]$. The zeros of $uv$ are simple in the respective local coordinates, and complex intersections have positive signs. Hence $[S]\cdot [L]=2$, so $[S]=2[L]$. The self-intersection number and the normal Euler class arewhere $u\in H^2(S^2;\mathbb{Z})$ has $⟨u,[S^2]⟩=1$.

The boundary of the tubular neighborhood. Write $N=\nu (S)$ for the closed disk tubular neighborhood and $E=\partial N$ for its boundary. This distinguishes the disk neighborhood $N$ from the vector normal bundle ${\nu }_S$. The space $E$ is the oriented circle bundle of ${\nu }_S$, with Euler class $4u$. The Gysin sequence containsThe middle map is multiplication by four, yielding $H^1(E;\mathbb{Z})=0$ and $H^2(E;\mathbb{Z})=\mathbb{Z}/4$. The same Gysin sequence gives $H^0(E;\mathbb{Z})=H^3(E;\mathbb{Z})=\mathbb{Z}$. The total space is a closed connected oriented three-manifold, so Poincare duality gives

The exterior. Work first with $W={\mathbb{CP}}^2\setminus \mathrm{int}N$. A collar neighborhood of $\partial W$ shows that its interior, the requested ${\mathbb{CP}}^2\setminus N$, has the same homotopy equivalence type: push the boundary a small positive distance into the collar. The Excision theorem and Thom isomorphism theorem identifyOnly degrees two and four are nonzero, each a copy of $\mathbb{Z}$.

In degree four the map $H_4({\mathbb{CP}}^2;\mathbb{Z})\to H_4(N,E;\mathbb{Z})$ sends the ambient fundamental class to the relative fundamental class of $N$. Under the Thom isomorphism theorem this becomes $[S]$. Thus this map is multiplication by one with compatible orientations. The long exact sequence in homology givesso $H_4(W)=H_3(W)=0$.

In degree two, the Thom isomorphism theorem identifies the map $H_2({\mathbb{CP}}^2)\to H_0(S^2)$ with intersection against $[S]$. A projective line intersects the conic twice, so this map is multiplication by two. The long exact sequence in homology isIt yields $H_2(W)=0$ and $H_1(W)=\mathbb{Z}/2$. In degree zero the relative groups vanish, so $H_0(W)\cong H_0({\mathbb{CP}}^2)\cong \mathbb{Z}$. This computes the homology of the complement of a smooth conic:The normal Euler number $4$, the degree-two intersection map and the degree-four restriction map $1$ play different roles; distinguishing them explains why the boundary has first homology $\mathbb{Z}/4$ while the exterior has first homology $\mathbb{Z}/2$.

Let $m=\dim M$ and $q=m-\dim Y$. Here a closed submanifold means compact and without boundary, which is the condition needed for a compactly supported dual class. The orientations of $M$ and $Y$ orient the normal bundle ${\nu }_Y$. Its Thom class is the unique relative class of degree $q$ restricting to the chosen orientation generator on every normal fiber. A tubular neighborhood identifies this class with one near $Y$ in $M$. Choose a compact normal disc neighborhood and extend the relative class to compactly supported cohomology of $M$. Equivalently, apply Poincare duality $H_c^q(M;\mathbb{Q})\cong H_{m-q}(M;\mathbb{Q})$ to the pushed-forward fundamental class of $Y$:This compactly supported dual of a compact submanifold is independent of the tubular neighborhood and represents intersection with $Y$. Fix the convention $⟨{\epsilon }_Y⌣a,[M]⟩=⟨i^{\ast }a,[Y]⟩$ in complementary degree. Its image in ordinary cohomology restricts to zero on $M\setminus Y$, because it comes from relative cohomology supported near $Y$. If a submanifold is only closed as a subset but is noncompact, its usual dual lies in ordinary cohomology instead; compact support is not asserted in that situation.

Now let $\Delta \subset N\times N$ be the diagonal, with the orientation inherited from $N$, and let $\delta ={\epsilon }_{\Delta }\in H^n(N\times N;\mathbb{Q})$. For each $i$, choose a basis $a_{i,j}$ of $H^i(N;\mathbb{Q})$ and its dual basis $b_{i,j}$ of $H^{n-i}(N;\mathbb{Q})$ satisfying $⟨a_{i,j}⌣b_{i,k},[N]⟩={\delta }_{jk}$. The Poincare duality pairing and the Künneth theorem give the cohomology class of the diagonalThe sign can be checked from the defining identityFor $|u|=i$, moving the degree-$i$ class in the second factor past $u$ contributes $(-1{)}^i$, and interchanging $b_{i,j}$ and $u$ contributes $(-1{)}^{i(n-i)}$; the displayed coefficient cancels their product. Equivalently its exponent is $ni$ modulo two.

The cohomological pushforward between closed oriented manifolds is characterized byfor complementary degrees. This follows immediately from its definition by Poincare duality and naturality of evaluation on pushed-forward homology. In degree $i$, the matrix trace of $f^{!}{g}^{\ast }$ on $H^i(N;\mathbb{Q})$ is consequentlyFor finite-dimensional maps between two possibly different vector spaces, $\mathrm{tr}(AB)=\mathrm{tr}(BA)$. Thus this matrix trace equals that of $g^{\ast }{f}^{!}$ on $H^i(M;\mathbb{Q})$. Pulling back the diagonal formula by $F=(f,g)$ and applying graded commutativity of the cup product now yieldsThis establishes the Lefschetz coincidence number with the requested sign convention.

If $f(m)\ne g(m)$ for every $m$, the map $(f,g)$ factors through $(N\times N)\setminus \Delta$. The diagonal class restricts to zero there, so its pullback is zero and $L(f,g)=0$. Taking the contrapositive proves the Lefschetz coincidence theorem:No transversality or isolated-coincidence assumption is needed. This cohomological argument also applies to continuous maps, which covers the last request.

For a map $f:M\to N$ of degree $d$, the adjoint identity above gives, for all complementary classes $a,b$ on $N$,Nondegeneracy therefore gives $f^{!}{f}^{\ast }=d\mathrm{id}$. Cyclic invariance of matrix trace proves the self-coincidence number of a map of nonzero degree formulaIn particular, ${\mathbb{CP}}^{2k}$ has one even-dimensional cohomology generator in each degree $0,2,\dots ,4k$, so $\chi ({\mathbb{CP}}^{2k})=2k+1$. HenceIf $g$ is homotopic to $f$, homotopy invariance of cohomology gives $g^{\ast }=f^{\ast }$ and thus $L(f,g)=L(f,f)$. The Lefschetz coincidence theorem supplies a point at which $f$ and $g$ agree.

Use ${\mathbb{F}}_2$ coefficients throughout. Write $h=c_1(\mathcal{O}(1))\mathrm{mod}2$ on Complex projective space and $a=w_1({\gamma }_d)$ on Real projective space. We use the standard cohomology ring of complex projective space ${\mathbb{F}}_2[h]/(h^{d+1})$, with $|h|=2$, and the real projective calculation from the preceding solution. The complex projective calculation follows, for example, from its one cell in each even dimension together with the fact that transverse projective hyperplanes represent the powers of $h$; $d$ such hyperplanes have one intersection point.

For the standard inclusion $i:{\mathbb{RP}}^d\hookrightarrow {\mathbb{CP}}^d$, the restricted complex tautological line is ${\gamma }_d{\otimes }_{\mathbb{R}}\mathbb{C}$. As a real vector bundle it is ${\gamma }_d\oplus {\gamma }_d$, whose total Stiefel–Whitney class is $(1+a{)}^2=1+a^2$. The Stiefel–Whitney class of the underlying real bundle of a complex line identifies its degree-two class with its First Chern class reduced modulo two. Passing to the dual line defining $h$ changes only a sign, which disappears modulo two. Thus mod-two restriction from complex to real projective space gives $i^{\ast }h=a^2$. In particular the answer for dimension two is the ring mapIt sends $1$ to $1$, is an isomorphism in degree two, and sends $h^2$ to zero.

Now assume $k\ge 1$ and put $M={\mathbb{CP}}^{2k}$, $N={\mathbb{RP}}^{2k}$, $U=M\setminus N$, with inclusions $i:N\hookrightarrow M$ and $j:U\hookrightarrow M$. The real dimensions are $4k$ and $2k$, respectively. A tubular neighborhood and excision identify the cohomology of $(M,U)$ with that of the normal disk bundle relative to its sphere bundle. Every real normal bundle is oriented over ${\mathbb{F}}_2$, so the Thom isomorphism theorem givesUnder this identification the relative-to-absolute map is the cohomological Gysin map of an embedding $i_{!}$. The exact sequence of the pair is thereforeHere $\delta$ denotes the ordinary connecting homomorphism followed by the inverse of the isomorphism supplied by the Thom isomorphism theorem.

To compute $i_{!}$, use mod-two Poincare duality and its evaluation formulaFor $0\le r\le k$, take $b=a^{2r}$ and $z=h^{k-r}$. The right side is $⟨a^{2k},[N{]}_2⟩=1$. Since the target group is generated by $h^{k+r}$ and $⟨h^{2k},[M{]}_2⟩=1$, this provesOdd-degree classes have zero image because $M$ has no odd-degree cohomology. Exactness now givesFor precision, below degree $2k-1$ the restriction map is an isomorphism. In degree $2k-1$ the next map is $i_{!}:H^0(N)\to H^{2k}(M)$, an isomorphism, so the complement group is zero. In even degrees $2k+2r$, $0\le r<k$, the preceding $i_{!}$ is an isomorphism and the following odd-degree $i_{!}$ is zero; consequently $\delta :H^{2k+2r}(U)\to H^{2r+1}(N)$ is an isomorphism. The intervening odd groups and the top group $H^{4k}(U)$ vanish.

It remains to establish multiplication. Put $x=j^{\ast }h$. Since $h^k=i_{!}1$, exactness gives $x^k=0$. Choose $y\in H^{2k}(U)$ with $\delta y=a$, which is possible by the just-computed isomorphism. Naturality of the relative cup product, and the module property of the Thom isomorphism theorem, giveThus $\delta (x^{r}y)=a^{2r+1}$ for $0\le r<k$. Each class $x^{r}y$ is nonzero and generates its upper-half group, while $1,x,\dots ,x^{k-1}$ generate the lower-half groups. Also $y^2=0$ because $H^{4k}(U)=0$. These classes account for every group, so there are no further relations:By the field-coefficient Künneth theorem this is precisely the cohomology ring of ${\mathbb{CP}}^{k-1}\times S^{2k}$. For $k=1$ the relation $x=0$ leaves the sphere's ring. The calculation identifies rings and does not claim a homotopy equivalence of the spaces; $k=0$ is outside the expression involving ${\mathbb{CP}}^{k-1}$.

All groups in the first calculation have integral coefficients. Let $U=S^n\setminus j(W)$ and $V=S^n\setminus \mathrm{int}j(W)$. A two-sided tubular neighborhood of $j(\partial W)$ identifies $V$ as a compact manifold with boundary, with interior $U$. Pushing its boundary inward in a collar neighborhood gives a homotopy equivalence $U\simeq V$. After thickening across the same collar, excision identifies the relative groupsThe orientation of $S^n$ gives $W$ an orientation. The Poincare-Lefschetz duality isomorphism therefore givesEach component of $W$ has boundary: a closed component would have open image by the local embedding condition and closed image by compactness, hence would occupy the whole connected sphere, leaving no room for the components with boundary. In particular $U$ is nonempty, and $H^n(W)=0$.