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If a space is a union of $n\ge 2$ subcomplexes and every nonempty intersection is acyclic, its homology vanishes in degrees at least $n-1$. Induction with the Mayer–Vietoris sequence proves the bound; the boundary of an $(n-1)$-simplex covered by its facets shows sharpness.

For odd $r\ge 1$ and $W_g={\#}_{i=1}^g(S^r\times S^r)$, the integral cohomology has one copy of $\mathbb{Z}$ in degrees zero and $2r$, ${\mathbb{Z}}^{2g}$ in degree $r$, and zero elsewhere. Choose the top orientation class $\omega$ and degree-$r$ classes ${\alpha }_i,{\beta }_i$ from the two factors of summand $i$. Their nonzero positive-degree basis products areAll ${\alpha }_i{\alpha }_j$ and ${\beta }_i{\beta }_j$ vanish, and $\omega$ annihilates positive-degree classes. The Künneth theorem computes each sphere product; the Mayer–Vietoris sequence computes the connected sum groups, and its degree-one pinch maps determine the products. Thus its middle-degree Poincare duality pairing is a symplectic vector space after changing coefficients to a field of characteristic different from two. The case $r=1$ is the usual cohomology ring of a closed oriented surface.

For connected oriented closed $d$-dimensional manifolds with $d\ge 2$, their connected sum removes an open $d$-dimensional ball from each and identifies the resulting boundary spheres by an orientation-reversing diffeomorphism. The remaining orientations fit together. Collapsing the separating sphere gives a pinch map to the wedge sum of the two closed manifolds, and projection to either summand has degree of a continuous mapping one. In intermediate positive degrees the cohomology is the direct sum of the summand groups. Products of classes from different summands vanish; products landing in top degree use the single common orientation class. These statements follow from excision, the Mayer–Vietoris sequence and the degree-one projections.

For global cohomology classes satisfying the local basis criterion for Leray-Hirsch classes, multiplication commutes with the Mayer–Vietoris sequence, including its connecting homomorphism: multiply cochains on the right by global cocycle representatives. Induct over a finite trivializing open cover. If $U$ is a union of $r-1$ patches and $V$ is the last patch, $U\cap V$ has a cover by at most $r-1$ trivialized intersections. The induction hypothesis on $U$, $V$ and $U\cap V$, followed by the Five lemma, proves the theorem on $U\cup V$. A compact base supplies the finite cover.

Identifying two distinct points of $S^2$ produces a space with homology groups $H_0=H_1=H_2=\mathbb{Z}$ and all higher groups zero. Cover the identified point by the image of two small disks, a contractible wedge of disks, and take as the other open set the complement of the point, a cylinder. Their intersection is two annuli. In the Mayer–Vietoris sequence, $H_1$ of this intersection maps onto $H_1$ of the cylinder with a rank-one kernel, producing $H_2=\mathbb{Z}$. The kernel at $H_0$ has rank one and produces $H_1=\mathbb{Z}$.

Attach two disks to a simply connected space $A$ along disjoint null-homotopic boundary circles, and additionally identify one interior point of each disk. Assume the attachments are CW complexes or admit compatible triangulations. The extra interior identification contributes a copy of $\mathbb{Z}$ to the first homology group, in addition to the two new second-homology classes. Indeed, after the first attachment $Y=A\cup D_1$, the intersection with $D_2$ has two components, $S^1\bigsqcup \{p\}$. In the Mayer–Vietoris sequence, the map $H_0(S^1\bigsqcup \{p\})\to H_0(Y)\oplus H_0(D_2)$ sends $(m,n)$ to $(m+n,-m-n)$, whose kernel is $\mathbb{Z}$. If $H_1(A)=0$, exactness gives $H_1(Y\cup D_2)=\mathbb{Z}$. The degree-two sequence is $0\to H_2(Y)\to H_2(Y\cup D_2)\to \mathbb{Z}\to 0$, which splits because $\mathbb{Z}$ is free. For $A=S^3$, the integral homology groups are $\mathbb{Z},\mathbb{Z},{\mathbb{Z}}^2,\mathbb{Z}$ in degrees zero through three.

Cover the circle by two arcs whose intersection has two components and take the product with the projective plane. The Mayer–Vietoris sequence has maps $(a,b)\mapsto (a+b,-a-b)$ on the two copies of the homology of the plane. Its kernels and cokernels, together with $H_1({\mathbb{RP}}^2;\mathbb{Z})=\mathbb{Z}/2$, give the displayed groups; all higher groups vanish. The product is a closed nonorientable three-manifold, but is homotopy equivalent to a noncompact oriented four-manifold.

For $\dim L=2n-1$ and lower middle perversity, the suspension with two distinct vertices has $I{H}_i(\mathrm{Susp}L)=I{H}_i(L)$ for $i<n$, zero for $i=n$, and $I{H}_{i-1}(L)$ for $i>n$. Cover it by its two open cones. The Mayer–Vietoris sequence compares their truncated groups with $I{H}_{\ast }(L)$; below $n$ the overlap map is the injective signed diagonal, and above $n$ both cone groups vanish.

Let $F$ be the self-map of the double $D(N)=N{\cup }_{\partial N}N$ induced by a self-map $f$ preserving the boundary. Naturality of the Mayer–Vietoris sequence and alternating-trace additivity give

For a fiber bundle $\pi :E\to B$ admitting a finite trivializing open cover, whose fiber has finite free cohomology over $R$, suppose classes $e_j\in H^{\ast }(E;R)$ restrict to a basis on every fiber. Then multiplication $a\otimes e_j\mapsto {\pi }^{\ast }a⌣e_j$ gives the indicated module isomorphism. A finite trivializing open cover suffices, by the local Künneth theorem and the Mayer–Vietoris sequence. This determines an additive module structure; multiplicative relations must still be computed. Powers of the hyperplane class on a complex projective bundle provide an important application.

For a continuous self-map $f:X\to X$, the mapping torus has the Wang sequence even when $f$ is not a homeomorphism. Cover the time circle by a seam neighborhood and an interval away from the seam. Their inverse images retract to $X$: the seam neighborhood is a mapping cylinder, and the other is a product. Their intersection retracts to two copies of $X$. The Mayer–Vietoris sequence map on homology is, after consistent choices of signs,Invertible changes of coordinates reduce it to the direct sum of $\mathrm{id}-f_{\ast }$ and an isomorphism. Canceling the isomorphism summands gives the Wang sequence. This proof does not assume a fiber bundle structure for a noninvertible $f$.

A normal surface meets each tetrahedron in disjoint disks of the four triangular and three quadrilateral types. Its arcs in a face join distinct edges, each disk meets an edge at most once, and at most one quadrilateral type occurs in a tetrahedron. These last requirements exclude, for example, a disk with an octagonal boundary.

We first establish the surgery operation needed for the sphere system. Suppose a disk $D$ disjoint from the other spheres compresses one sphere $S$. It produces spheres $S^{\prime }$ and $S^{\prime \prime }$. Small parallel copies of the three spheres bound a three-times punctured three-sphere $P$. Denote the old piece on the uncompressed side by $A$, and the two pieces on the compressed side by $B^{\prime }$ and $B^{\prime \prime }$. If both $B^{\prime }$ and $B^{\prime \prime }$ were punctured three-spheres, then $P$ glued to them would also be a punctured three-sphere, contrary to the hypothesis on the original cut manifold. Choose, say, $B^{\prime }$ which is not of this type, and retain $S^{\prime }$ alone in place of $S$.

The other new piece is $A\cup P\cup B^{\prime \prime }$. It too cannot be a punctured three-sphere. If it were, cap its external boundary spheres to make $S^3$. All the spheres separating $A$ from the attached pieces then bound three-balls by the Three-dimensional Schoenflies theorem; the component containing $A$ would be a punctured three-sphere, contradicting the hypothesis on $A$. Thus the sphere surgery preserving nontrivial complementary pieces keeps exactly $k$ spheres and keeps every complementary piece nontrivial. The assumption that all spheres separate ensures that the old and new pieces used here really are distinct sides.

Apply general position to the whole sphere system, avoiding the vertices and making it transverse to edges and faces. We describe a terminating normalization, allowing the replacement operation just proved as well as isotopy. If some component of its intersection with a tetrahedron is not a disk, choose a boundary circle innermost among circles not bounding such disk components. A disk just inside the tetrahedral boundary gives a compression; if intervening disk components occur, first push them into collars of their boundary disks. Perform the compression and retain one new sphere as above. This replaces a non-disk tetrahedral piece by disk pieces without increasing edge intersections. Repetition is finite: every compression divides the original sphere along a separating circle, and the sum of $1-\chi$ over the non-disk planar tetrahedral pieces decreases; disk components can then be put in boundary collars. A closed sphere component wholly in a tetrahedron is impossible, because it would cut off a three-ball, contrary to the complementary-piece condition.

Now each tetrahedral piece is a disk. A face circle innermost in a face allows its incident disk to be moved through that face disk, removing the circle. This reduces the number of face intersections without increasing the edge weight. A returning face arc similarly gives an edge bigon across which the surface can be pushed, reducing the number of edge intersections by two. Restore tetrahedral disk pieces by the preceding operations whenever necessary.

It remains to exclude multiple intersections of a disk with an edge, which the absence of returning face arcs alone does not exclude. Suppose a tetrahedral disk meets an edge twice. Among such disks choose a pair of intersections consecutive on the edge and with no other surface intersection between them. Such a pair exists: an intervening disk crosses the edge interval an even number of times, so passing to an innermost intervening pair eventually gives consecutive points. Join them by an arc in the disk. The disk pieces divide the tetrahedron into three-balls, and this arc together with the edge interval lies in the boundary of one of these three-balls. It therefore bounds a bigon with interior disjoint from the surface. Push across the bigon to reduce edge weight by two. At fixed weight the previous moves reduce face and non-disk complexity; the edge weight itself is a nonnegative integer. Hence the whole process terminates. A final tetrahedral disk meets each edge at most once, and its boundary separates either one vertex from three or two vertices from two. It is respectively a normal triangle or quadrilateral. Thus a normal system of exactly $k$ spheres with the required complementary-piece property exists.

For the finite decomposition, we need a bound on such normal systems, not merely an assertion that they have been normalized. Let $f_2$ be the number of faces of the fixed triangulation. In a triangular face cut by normal arcs, all but at most four complementary regions are rectangles between adjacent parallel arcs. Thus at most $4f_2$ complementary three-dimensional components contain a nonrectangular face region. Every remaining component is assembled from product regions between parallel normal disks, so is an interval bundle whose boundary components are spheres. Its connected base is either $S^2$ or ${\mathbb{RP}}^2$. The product $S^2\times I$ is a punctured three-sphere, and is excluded. The other possibility is the twisted interval bundle over ${\mathbb{RP}}^2$, namely ${\mathbb{RP}}^3$ with a three-ball removed. Each such piece contributes a distinct ${\mathbb{F}}_2$ summand to $H_1(M;{\mathbb{F}}_2)$: gluing along spheres does not identify first-homology classes, by the Mayer–Vietoris sequence. Since $k$ separating spheres give $k+1$ components, we obtain the explicit finite bound

For an arbitrary closed connected $X$, first cut out nonseparating spheres. A neighbourhood of one such sphere together with an arc joining its two sides is a punctured $S^2$-bundle over $S^1$; the bundle is the product or the twisted bundle according to the orientation of the attaching identification. Its complement is connected, so this extracts a connected sum factor. The mod-two first Betti number of the remaining capped manifold decreases by one. Hence only finitely many such extractions are possible. The remaining manifolds have only separating spheres.

Whenever a remaining factor is not a prime three-manifold, split it along a sphere into two factors neither homeomorphic to $S^3$. Viewed in the original remaining manifold, all these successive splitting spheres form a disjoint system whose pieces are not punctured three-spheres. The normalization and bound just proved prohibit infinitely many such splittings. This proves existence of a finite prime decomposition, including for $X$ without assuming orientability.

For completeness, both sphere bundles extracted above are prime three-manifolds. Their fundamental group is $\mathbb{Z}$, so a separating sphere and the van Kampen theorem express $\mathbb{Z}$ as a free product in which one capped factor has trivial fundamental group. Its punctured piece lifts homeomorphically to the universal cover $S^2\times \mathbb{R}$, identified with ${\mathbb{R}}^3\setminus \{0\}$. That compact lifted piece has one spherical boundary and must be the bounded side of this sphere in ${\mathbb{R}}^3$. It misses the origin, so the Three-dimensional Schoenflies theorem makes it a three-ball. The corresponding capped factor is $S^3$, as required.

Suppose now that $X$ is orientable. A prime factor which is not an irreducible three-manifold contains a nonseparating sphere: an essential separating sphere would already split it nontrivially. Extracting the corresponding sphere bundle shows that the factor must be $S^2\times S^1$. Thus the other nontrivial primes are irreducible.

Consider two prime decompositions. Represent each by a sphere system, including one nonseparating sphere in each $S^2\times S^1$ factor. The capped pieces are now the irreducible factors and copies of $S^3$. Make the two sphere systems transverse. To remove their intersections, choose a circle innermost on a sphere of the second system and surger the first system along its disk. Retain both surgery spheres temporarily. The old sphere and the new pair bound a three-times punctured sphere; adding the pair and then deleting the old sphere only inserts or removes spherical punctures.

This operation cannot alter an irreducible capped piece: every new sphere in that piece bounds a three-ball, so it can only split off a punctured $S^3$ piece. The other pieces were already punctured $S^3$, where the same conclusion follows from the Three-dimensional Schoenflies theorem. After small displacements the surgery removes the chosen intersection circle and introduces none. Repeat until the sphere systems are disjoint. Their union is a common refinement, and its non-$S^3$ capped irreducible pieces are precisely those of each original decomposition. Hence these factors agree, up to order and orientation-preserving homeomorphism.

Finally, the number $r$ of $S^2\times S^1$ factors is determined bywhere the sum runs over the common irreducible factors. This uses additivity of first homology under connected sum, not a claim that the fundamental group determines a manifold. Therefore the orientable prime decomposition is unique up to order and insertion or deletion of $S^3$ factors.

Let $F$ be an oriented Seifert surface of genus $g$ for $K$, and choose a basis $a_1,\dots ,a_{2g}$ of $H_1(F;\mathbb{Z})$. Push a representative curve slightly in the positive normal direction to obtain $a_i^{+}$. The Seifert matrix in our row convention isThe difference $V-V^T$ is the intersection form of F, up to the convention for its orientation; in a symplectic basis it is a direct sum of unimodular two-by-two skew blocks.

Write $X=S^3\setminus \mathrm{int}N(K)$ for the knot exterior. Linking number with K gives the epimorphism ${\pi }_1(X)\to \mathbb{Z}$ sending a positive meridian to one. Let $\tilde{X}$ be the associated infinite cyclic cover, with deck transformation t. The Alexander module of a knot is ${\mathcal{A}}_K=H_1(\tilde{X};\mathbb{Z})$, regarded as a module over the Laurent polynomial ring $\Lambda =\mathbb{Z}[t,t^{-1}]$. Its order, the greatest common divisor of the maximal presentation minors, is the Alexander polynomial of a knot, defined up to $\pm t^k$. Equivalently these minors generate its zeroth Fitting ideal; we will obtain a square presentation, so its determinant gives the order. No assertion that $\Lambda$ is a principal ideal domain is needed.

Here is the Seifert-matrix presentation of the Alexander module with its topological justification. Cut X along F, obtaining a connected manifold Y with two boundary copies $F^{+}$ and $F^{-}$. Up to collars this is the complement of a thickening of F in $S^3$. Alexander duality gives $H_1(Y;\mathbb{Z})\cong {\mathbb{Z}}^{2g}$ and the perfect linking pairing with $H_1(F;\mathbb{Z})$. Choose generators $b_j$ satisfying $\mathrm{lk}(b_j,a_i)={\delta }_{ij}$. This homology statement does not require Y to be a handlebody. The two inclusions satisfyIndeed the first coefficients are the defining positive linking numbers; for the second, moving the two curves to opposite sides and using symmetry of linking gives $\mathrm{lk}(a_i^{-},a_j)=\mathrm{lk}(a_j^{+},a_i)$.

Stack copies of Y, identifying the positive copy of F in one layer with the negative copy in the next. This is $\tilde{X}$. The Mayer–Vietoris sequence, with all translates collected into Laurent modules, containsThe last map is injective because $\Lambda$ is an integral domain. Thus the entire Alexander module is presented by the row relations $tV-V^T$ (transpose the matrix if using column vectors). At $t=1$ its determinant is $\det (V-V^T)=1$, so the determinant is nonzero and the module is a torsion module. Consequentlywhere $\doteq$ denotes equality up to a Laurent unit. The second conclusion follows because a determinant of size $2g$ with entries linear in t has breadth at most $2g$; minimize over Seifert surfaces.

For the displayed diagram, start on the outer lower arc near the lower left undercrossing and follow the arc through the lower outer loop toward the upper right undercrossing. Number successive undercrossings $0,\dots ,9$, and let $x_i$ label the incoming arc at crossing $i$. Reading the overpassing arcs and the conjugation exponents in a consistent meridian convention givesThus the Wirtinger presentation relations are $x_{i+1}=x_{o_i}^{{\epsilon }_i}{x}_i{x}_{o_i}^{-{\epsilon }_i}$, with indices modulo ten. Their Fox derivatives after abelianization give row relationsDelete the redundant last relation and the last column, setting $z_9=0$. This is a nine-generator presentation of the Alexander module. The cofactor determinant is $t^{-6}(t^4+t^3-3t^2+t+1)$; here is a smaller elimination check. Retain $z_1,z_4$. Seven relations solve the other generators asThe two remaining rows, at undercrossings two and seven, are the matrixIts determinant is $t^{-3}(t^4+t^3-3t^2+t+1)$, differing from the original cofactor only by a Laurent unit. ConsequentlyThe factor has value one at $t=1$ and is reciprocal, as required. Its Conway-normalized Alexander polynomial is $t^2+t-3+t^{-1}+t^{-2}$, and its knot determinant is three.

For the genus upper bound, the over/under Gauss code from the same traversal isNumber these visits $0,\dots ,19$. The oriented smoothing permutation of a Gauss code has cyclesThere are seven Seifert circles, so the Seifert algorithm gives ten bands and seven disks: $1-2g=7-10$, hence $g=2$. The Alexander breadth bound on Seifert genus supplies the matching lower bound from breadth four. Therefore $\boxed{g_s(K)=2}$.

Modulo two, the primitive Alexander polynomial becomes $t^4+t^3+t^2+t+1$. It has neither zero nor one as a root, and it is not $(t^2+t+1{)}^2=t^4+t^2+1$, the only possible product of irreducible monic quadratics over ${\mathbb{F}}_2$. It is therefore irreducible over ${\mathbb{F}}_2$, hence over $\mathbb{Q}$ and $\mathbb{Z}$ by Gauss lemma for polynomials. The full-degree irreducible Alexander polynomial implies a prime knot criterion now applies, since its breadth is exactly twice the genus. Thus the knot is prime. The full-degree condition is needed to rule out a nontrivial summand with unit Alexander polynomial.

To test reflection, calculate the Jones polynomial from the same crossing data. The diagram has nine positive crossings and one negative crossing, so its writhe of a link diagram is eight. Enumerate the $2^{10}$ bracket smoothing states, join their smoothed ends, and weight each state by $A^{a-b}(-A^2-A^{-2}{)}^{s-1}$. Correct by $(-A^3{)}^{-8}$ and put $t=A^{-4}$. This givesAs checks, $V_K(1)=1$ and $V_K(-1)=-3={\Delta }_K(-1)$. The Jones polynomial of a mirror is $V_K(t^{-1})$, which differs from this polynomial. Therefore the knot is not equivalent to its reflection. Reversing all crossing conventions reciprocates the displayed Jones polynomial and leaves the conclusion unchanged.

This sequence does occur, for the oriented connected sumThe Künneth theorem gives one copy of $\mathbb{Z}$ in dimensions $0,1,3,4$ for each factor product and zero in dimension two. Removing a disc from each summand and gluing their boundary three-spheres, the Mayer–Vietoris sequence makes the homology in dimensions one through three the direct sum of that of the two summands. The two orientations join to one top fundamental class, so $H_0=H_4=\mathbb{Z}$, $H_1=H_3={\mathbb{Z}}^2$, and $H_2=0$.

An orientation of a topological n-manifold without boundary is a choice of generator ${\mu }_x$ of each local homology group $H_n(M,M\setminus \{x\};\mathbb{Z})\cong \mathbb{Z}$, compatible in coordinate neighbourhoods. Equivalently, local orientations of coordinate discs are chosen so their overlap changes have degree $+1$. This trivializes the orientation local system.

Here the printed $H_{\mathrm{ct}}^n$ denotes compactly supported cohomology, which we write $H_c^n$. We prove the cyclicity assertion without assuming Poincare duality. By the compactly supported cohomology of Euclidean space, an open n-disc has one copy of $\mathbb{Z}$ in degree $n$ and zero in all other degrees. The compact-support Mayer–Vietoris sequence for open sets containsUse the finite good cover allowed in the paper. Induction on the number of discs shows that any union of its discs has zero compact-support cohomology above degree $n$: at each step the intersection with the new disc is itself a union of fewer discs with a good cover, so the displayed sequence and its higher-degree terms apply.

The same induction presents $H_c^n(M)$ by a generator $u_i$ for each cover disc, with overlap relations $u_i={ϵ}_{ij}{u}_j$, where ${ϵ}_{ij}=\pm 1$ is its orientation transition sign. Indeed $H_c^n$ of the intersection is generated by its overlap discs; extension by zero sends each generator to the two disc generators with these signs. Passing through the compact-support Mayer–Vietoris sequence imposes exactly those relations. This is top compactly supported cohomology from orientation signs.

Since $M$ is connected, the overlap graph is connected: a disconnection would split the union into two disjoint nonempty open subsets. A spanning tree expresses every generator as $u$ or $-u$, so $H_c^n(M)$ is cyclic. Additional overlap relations can only impose $2u=0$. If $M$ is oriented, use its compatible local orientations; every relation becomes $u_i=u_j$, and the presentation is exactly one free generator. ThusIn fact a nonorientable connected manifold has an orientation-reversing loop in this transition graph and gives $\mathbb{Z}/2$. In dimension zero the connected manifold is a point and the assertion is immediate.

For a closed oriented n-manifold, Poincare duality says that its fundamental class $[M]\in H_n(M;\mathbb{Z})$ gives isomorphisms by cap productFor an oriented manifold that is not necessarily compact, the corresponding ordinary-homology statement is $H_c^q(M;\mathbb{Z})\cong H_{n-q}(M;\mathbb{Z})$, using the locally finite fundamental class. The version without compact supports on cohomology pairs with locally finite homology instead. Orientability and the absence of boundary are essential to these untwisted formulas; for a manifold with boundary one needs Poincare-Lefschetz duality, and for a nonorientable manifold one uses the orientation local system.

For the following cases, $H_0=\mathbb{Z}$ forces connectedness. The universal coefficient theorem for cohomology givessince the Ext term from $H_0=\mathbb{Z}$ vanishes. Combining with duality yields the third homology of a closed oriented four-manifold constraint $H_3\cong \mathrm{Hom}(H_1,\mathbb{Z})$.

Let F be an oriented genus-g Seifert surface for K, and choose a basis $a_1,\dots ,a_{2g}$ of $H_1(F;\mathbb{Z})$. Push a representative curve slightly in the positive normal direction to obtain $a_i^{+}$. The Seifert matrix in our row convention isThe difference $V-V^T$ is the intersection form of F, up to the convention for its orientation; in a symplectic basis it is a direct sum of unimodular two-by-two skew blocks.

Write $X=S^3\setminus \mathrm{int}N(K)$ for the knot exterior. Linking number with K gives the epimorphism ${\pi }_1(X)\to \mathbb{Z}$ sending a positive meridian to one. Let $\tilde{X}$ be the associated infinite cyclic cover, with deck transformation t. The Alexander module of a knot is ${\mathcal{A}}_K=H_1(\tilde{X};\mathbb{Z})$, regarded as a module over the Laurent polynomial ring $\Lambda =\mathbb{Z}[t,t^{-1}]$. Its order, the greatest common divisor of the maximal presentation minors, is the Alexander polynomial of a knot, defined up to $\pm t^k$. Equivalently these minors generate its zeroth Fitting ideal; we will obtain a square presentation, so its determinant gives the order. No assertion that $\Lambda$ is a principal ideal domain is needed.

Here is the Seifert-matrix presentation of the Alexander module with its topological justification. Cut X along F, obtaining a connected manifold Y with two boundary copies $F^{+}$ and $F^{-}$. Up to collars this is the complement of a thickening of F in $S^3$. Alexander duality gives $H_1(Y;\mathbb{Z})\cong {\mathbb{Z}}^{2g}$ and the perfect linking pairing with $H_1(F;\mathbb{Z})$. Choose generators $b_j$ satisfying $\mathrm{lk}(b_j,a_i)={\delta }_{ij}$. This homology statement does not require Y to be a handlebody. The two inclusions satisfyIndeed the first coefficients are the defining positive linking numbers; for the second, moving the two curves to opposite sides and using symmetry of linking gives $\mathrm{lk}(a_i^{-},a_j)=\mathrm{lk}(a_j^{+},a_i)$.

Stack copies of Y, identifying the positive copy of F in one layer with the negative copy in the next. This is $\tilde{X}$. The Mayer–Vietoris sequence, with all translates collected into Laurent modules, containsThe last map is injective because $\Lambda$ is an integral domain. Thus the entire Alexander module is presented by the row relations $tV-V^T$ (transpose the matrix if using column vectors). At $t=1$ its determinant is $\det (V-V^T)=1$, so the determinant is nonzero and the module is torsion. Consequentlywhere $\doteq$ denotes equality up to a Laurent unit. The second conclusion follows because a determinant of size $2g$ with entries linear in t has breadth at most $2g$; minimize over Seifert surfaces.

For the displayed knot we give a reproducible diagram calculation. Label its ten crossings A through J as follows: A and B are the two upper crossings from left to right; C,D,E are the next three crossings in descending order; F is the crossing below E on the almost vertical strand; G is the left crossing of the bottom loop; H,I,J are the remaining three bottom crossings from left to right. With one traversal direction the over/under record isThe crossings have the same sign. Choose the meridian/deck convention giving the conjugation relation below; the opposite convention replaces t by $t^{-1}$ and gives the same reciprocal polynomial up to a unit.

Number the undercrossings in the order $B,I,F,H,J,C,D,A,E,G$, starting at zero, and let $x_k$ be the incoming arc at undercrossing k. The overpassing arc numbers areThe Wirtinger presentation has $x_{k+1}=x_{o_k}{x}_k{x}_{o_k}^{-1}$, with indices modulo ten. Its lifted cellular boundaries, or equivalently Fox derivatives, give the Alexander relation matrixDelete the redundant final relation and the last column: the resulting matrix presents the Alexander module on the nine relative arc generators. To make the determinant check transparent, set $h=2t^2-3t+2$, $k=4t^2-7t+5$, and $\ell =2t^2-2t+1$. Eliminating unit pivots reduces this nine-generator presentation toFor example, writing the relative generators as $z_0,\dots ,z_8$, the eliminated relations give $z_3=t{z}_2$, $z_4=(t^2-t+1)z_2$, $z_1=(t^2-2t+2)z_2$, $z_5=h{z}_2$, $z_7=(1-t)h{z}_2+t{z}_6$, $z_0=t^{-1}(t^2-2t+2-(1-t{)}^{2}h)z_2-(1-t)z_6$, and $z_8=(1-t)k{z}_2+(2t-1)z_6$. The two remaining relations are precisely the rows of N. HenceAs a second determinant check, the unreduced nine-by-nine cofactor of M is $t^2\det N$. The polynomial has value one at $t=1$ and is reciprocal, as required.

For the genus upper bound, the oriented smoothing has seven Seifert circles. This can also be checked from the traversal record: number its twenty entries starting at one; smoothing sends an incoming visit to the entry following its other visit to the same crossing. The cycles are $(1,17,13,19,7)$, $(2,16)$, $(3,11,15)$, $(4,10)$, $(5,9)$, $(6,20,8)$, and $(12,18,14)$. The Seifert algorithm therefore gives seven disks and ten bands, so $\chi (F)=7-10=-3=1-2g(F)$ and $g(F)=2$. The Alexander breadth is four, so the Alexander breadth bound on Seifert genus gives the matching lower bound. Thus $\boxed{g_s(K)=2}$.

Finally, modulo two the primitive polynomial becomes $t^4+t^3+t^2+t+1$. It has neither zero nor one as a root. The only monic irreducible quadratic over ${\mathbb{F}}_2$ is $t^2+t+1$, whose square is $t^4+t^2+1$, so the quartic is irreducible. Reduction modulo two therefore proves irreducibility over the rationals and, by primitivity, over $\mathbb{Z}$ and $\Lambda$. If K were a sum of two nontrivial knots, additivity of Seifert genus would force both summands to have genus one. Their Alexander polynomials multiply, so irreducibility would make one polynomial a unit and the other equal to this breadth-four polynomial, contradicting the breadth-at-most-two bound for genus one. This uses the full-degree irreducible Alexander polynomial implies a prime knot criterion, rather than irreducibility alone. Therefore K is prime.

Use rational coefficients and finite singular chains. For an $N$-dimensional stratified pseudomanifold with filtration $X_j$, the lower middle perversity isA singular simplex $\sigma :{\Delta }^i\to X$ is an allowable singular simplex when, for every $k\ge 2$, ${\sigma }^{-1}(X_{N-k})$ lies in the $(i-k+\bar{m}(k))$-skeleton of ${\Delta }^i$; negative-dimensional skeletons are empty. Let $A_i$ be the span of these simplices in $S_i(X;\mathbb{Q})$. The intersection chains areThe condition on $\partial c$ is imposed after combining equal simplices and cancelling coefficients: the faces of an allowable simplex need not themselves be allowable. If $c\in I{S}_i$, then $\partial c\in A_{i-1}$ and $\partial (\partial c)=0$, so $\partial c\in I{S}_{i-1}$. Thus $I{S}_{\ast }$ is a chain subcomplex of $S_{\ast }$, and $I{H}_i(X)=H_i(I{S}_{\ast })$ defines intersection homology.

Write $v$ for the vertex of $cY$. Its codimension is $2n$, so the vertex condition for an allowable $i$-simplex isIn particular all simplices of degree at most $n$ avoid $v$. For $i<n$, both the degree-$i$ chains and the degree-$(i+1)$ chains that can bound them lie in $cY\setminus \{v\}=Y\times (0,1)$. The two chain complexes therefore have the same cycles and boundaries in degree $i$, proving the low-degree isomorphism.

For $i\ge n$, let $z$ be an allowable $i$-cycle. Cone every simplex of $z$ radially to $v$, using the cone on ${\Delta }^i$ as ${\Delta }^{i+1}$, and write the resulting chain as $cz$. At the vertex, coning an old inverse image in the $(i-n-1)$-skeleton gives a subset of the $(i-n)$-skeleton, and the new cone vertex also lies in this skeleton because $i-n\ge 0$. For every other singular stratum, coning the inverse image increases its permitted skeleton dimension by one, exactly as increasing the simplex dimension does. Thus $cz$ is allowable. The usual cone identity is $\partial (cz)=z-c(\partial z)=z$; since $z$ is allowable, the boundary condition for intersection chains holds as well. Every cycle is consequently a boundary. This proves the intersection homology cone formula here:The assumption $\dim Y=2n-1$ entails $n\ge 1$, so no augmented degree-zero exception arises in this coning argument.

For the suspension of a topological space, the two endpoints are collapsed separately, giving two distinct vertices. Use the following standard computational facts for intersection homology: a product with an interval has the same intersection homology as the original stratified space, via the projection and its stratum-preserving homotopy inverse; and an open cover has a Mayer–Vietoris sequence whose overlap map is the signed pair of inclusion maps. Cover the suspension by its two open cones. Their overlap is $Y\times (-\epsilon ,\epsilon )$. Below degree $n$, the overlap map is, under the interval identifications, $a\mapsto (a,-a)$ in $I{H}_i(Y)\oplus I{H}_i(Y)$, hence is injective with cokernel $I{H}_i(Y)$. In degree $n$ both cone groups vanish and the next overlap map, in degree $n-1$, is still injective. Above degree $n$ both adjacent cone groups vanish. Exactness therefore givesThis differs from ordinary reduced suspension homology: below the cutoff the cycles cannot use either vertex, and in the middle degree coning kills them. If $Y$ is a compact oriented Witt space, its Poincare duality pairs $I{H}_i(Y)$ with $I{H}_{2n-1-i}(Y)$. The displayed groups of the suspension are correspondingly paired in complementary degrees $i$ and $2n-i$, while its middle group is zero. The new vertices have even codimension $2n$ and thus impose no additional Witt space condition; the suspension is again Witt. Its middle intersection form has zero rank and, when its dimension is divisible by four, zero signature.

For an open subset $U\subseteq B$, let $E_U={\pi }^{-1}(U)$ and define the degree-preserving module mapUsing the prescribed fiber basis identifies its domain with the tensor product in the question. We prove that ${\Phi }_B$ is an isomorphism, giving the Leray-Hirsch theorem in this setting.

First suppose $U$ is a trivializing patch. The cohomological Künneth theorem over a field identifies $H^{\ast }(U\times F;\mathbb{Q})$ with the finite free graded $H^{\ast }(U;\mathbb{Q})$-module on a homogeneous basis of $H^{\ast }(F;\mathbb{Q})$. The required finiteness is supplied by the finite list $c_1,\dots ,c_v$ spanning the fiber's total cohomology. Express the restrictions of the $c_j$ in this product basis. A coefficient multiplying a fiber basis element of degree $k_i$ belongs to $H^{k_j-k_i}(U;\mathbb{Q})$; thus no coefficient can increase fiber degree. The degree-zero blocks are invertible on every path component, because restriction at every point gives a basis. Their inverses are again degree-zero cohomology classes.

After inverting those blocks, the remaining change-of-basis map is $I+N$, where $N$ strictly lowers fiber degree. There are only finitely many fiber degrees, so $N$ is nilpotent and $(I+N{)}^{-1}=I-N+N^2-\cdots$ is a finite sum. This proves the local basis criterion for Leray-Hirsch classes and therefore the isomorphism on every trivializing patch, without requiring the patch to be contractible.

Now use a finite trivializing open cover $U_1,\dots ,U_r$, supplied by compactness of $B$. Induct on $r$. Put $U=U_1\cup \cdots \cup U_{r-1}$ and $V=U_r$. The intersection $U\cap V$ is covered by the at most $r-1$ trivializing open sets $U_i\cap V$, so the induction hypothesis applies to $U$, $V$ and $U\cap V$.

Take the direct sum of shifted Mayer–Vietoris sequences for the base and compare it with the sequence for $E_U\cup E_V$. Restriction commutes with multiplication by each global cohomology class $c_j$. The connecting homomorphism also commutes: multiply the cochain representatives on the right by the restriction of a global cocycle representing $c_j$, and use $d(b⌣c_j)=db⌣c_j$. Consequently the maps $\Phi$ form a morphism of exact sequences. The Five lemma makes ${\Phi }_{U\cup V}$ an isomorphism, completing the finite-cover proof of the Leray-Hirsch theorem.

This is an additive graded module isomorphism. It does not assert that the fiber classes have the same multiplicative relations in the total space.

Use simplicial homology with a fixed coefficient group or ring. The Mayer–Vietoris sequence isThe inclusion maps define $\alpha ([z])=([z],-[z])$ and $\beta ([a],[b])=[a+b]$. To define $\delta$, split a simplicial cycle $c\in C_{i+1}(X)$ as $c=a+b$ with $a$ a chain in $A$ and $b$ in $B$. Since $\partial a=-\partial b$, this boundary lies in $C_i(A\cap B)$, and $\delta [c]=[\partial a]$. At degree zero the unreduced sequence ends with $H_0(A)\oplus H_0(B)\to H_0(X)\to 0$.

If $[z]=\delta [c]$, its representative is a boundary in $A$ and the negative of a boundary in $B$, so $\alpha [z]=0$. Thus $\mathrm{im}\delta \subseteq \ker \alpha$.

Conversely if $\alpha [z]=0$, choose $a\in C_{i+1}(A)$ and $b\in C_{i+1}(B)$ with $\partial a=z=\partial b$. The chain $c=a-b$ is a simplicial cycle in $X$. Its splitting into the $A$ chain $a$ and $B$ chain $-b$ gives $\delta [c]=[z]$. Hence $\mathrm{im}\delta =\ker \alpha$, the requested exactness at $H_i(A\cap B)$.

The Excision theorem for singular homology states that if $Z\subseteq A\subseteq X$ and $\overline{Z}\subseteq {\mathrm{int}}_{X}A$, inclusion induces isomorphisms $H_q(X\setminus Z,A\setminus Z;\mathbb{Z})\cong H_q(X,A;\mathbb{Z})$ for all $q$. For an open cover $X=U\cup V$, its small-chain form says that the subcomplex of singular chains whose individual simplices lie in $U$ or $V$ computes $H_{\ast }(X)$. Concretely, repeated barycentric subdivision puts each finite chain in that subcomplex, and subdivision is chain-homotopic to the identity. There is then a short exact sequence of chain complexesIts long exact sequence in homology, with excision identifying the last complex's homology with $H_{\ast }(X)$, is the Mayer–Vietoris sequence:

The Real projective space ${\mathbb{RP}}^1$ is a circle. The projective plane has one cell in each dimension $0,1,2$, with cellular homology boundary maps $d_2=2$ and $d_1=0$. Thus, writing $A_q=H_q({\mathbb{RP}}^2;\mathbb{Z})$, we have $A_0=\mathbb{Z}$, $A_1=\mathbb{Z}/2$ and $A_q=0$ for $q\ge 2$. Cover the circle by two open arcs with a two-component intersection and multiply this cover by the projective plane. In the Mayer–Vietoris sequence the overlap-to-cover map isIts kernel and cokernel are each $A_q$. Consequently $0\to A_q\to H_q(X)\to A_{q-1}\to 0$ for $q\ge 1$. In degree one the quotient is free, so this sequence splits; in degree two it identifies the group directly with $A_1$. The integral homology of a circle times a real projective plane is therefore

Without a compactness restriction, the answer to the final existence question is yes: there is a noncompact oriented four-manifold. Explicitly, takeThe involution is free and reverses orientation on both $S^2$ and $\mathbb{R}$, hence preserves their product orientation. Its quotient is consequently an oriented three-manifold, the orientable total space of the orientation line bundle of the projective plane. The contraction $[v,t]\mapsto [v,(1-s)t]$ is a deformation retraction onto ${\mathbb{RP}}^2$, so $Y\simeq X$.

If a closed oriented manifold is intended, the answer instead is no. A closed connected oriented $d$-manifold has $H_d\cong \mathbb{Z}$ and no homology above $d$. Here the highest nonzero positive-dimensional homology is the torsion group $H_2=\mathbb{Z}/2$, so no dimension can supply that required top fundamental class. The distinction is necessary because the printed wording does not require the oriented manifold to be closed or three-dimensional.