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For an $n$-dimensional singular simplex $\sigma$ and a degree-$p$ singular cochain $\alpha$, use the front-face/back-face conventionand set it to zero for $p>n$. Extend linearly over $R$. Passing to homology classes of cycles and cohomology classes of cocycles defines the cap product $H_n(X;R)\times H^p(X;R)\to H_{n-p}(X;R)$.

For a continuous map $f$, the cochain pullback satisfies $(f^{\ast }\alpha )(\tau )=\alpha (f\circ \tau )$, and the induced chain map sends $\sigma$ to $f\circ \sigma$. Since composition commutes with restrictions to faces,The identity holds on chains, so it descends directly to classes and proves the naturality of the cap product: $f_{\ast }(x⌢f^{\ast }\alpha )=f_{\ast }(x)⌢\alpha$. No additional well-definedness verification is required here.

For the triangle quotient, the universal coefficient theorem for homology gives $H_j(X;{\mathbb{F}}_2)=0$ for every $j>0$, because its only positive integral homology is $\mathbb{Z}/3$. A connected positive-dimensional closed manifold has a nonzero top mod-two fundamental class, independently of orientability. Thus this quotient cannot be homotopy equivalent to such a manifold. A connected zero-dimensional closed manifold is a point, which its nonzero $H_1(X;\mathbb{Z})$ also excludes.

For the attached-band space, the same universal coefficient theorem for homology gives nonzero $H_2(X;{\mathbb{F}}_2)$ and zero homology above degree two. A connected closed manifold homotopy equivalent to it would consequently have dimension two. Its Euler characteristic would be one. The classification theorem for surfaces says that a closed connected surface of Euler characteristic one is the real projective plane. But its integral $H_1$ is $\mathbb{Z}/2$, differing from both possibilities computed in part 2. Thus this space is never homotopy equivalent to a closed manifold.

The antipodal mapping torus is already a closed manifold: gluing the two boundary components of $S^n\times [0,1]$ by a diffeomorphism gives a smooth $(n+1)$-manifold without boundary. For $n\ge 1$, it is orientable exactly when $(-1{)}^{n+1}=1$, namely when $n$ is odd. In that case $-I$ on ${\mathbb{R}}^{n+1}$ is isotopic to the identity through rotations, so the mapping torus is diffeomorphic to $S^n\times S^1$. The $n=0$ case is a circle as above. Only the third family can be homotopy equivalent to a closed manifold.

Take the Hopf map $\eta :S^3\to {\mathbb{CP}}^1\cong S^2$, given by $\eta (z_0,z_1)=[z_0:z_1]$. The domain has positive-dimensional homology only in degree three, while $H_3(S^2;\mathbb{Z})=0$. In degree two the source group is zero, and all remaining positive-degree source groups are zero too. Hence the induced map on homology is zero in every positive degree.

For nontriviality, use the Hopf fibration $S^1\to S^3\stackrel{\eta }{\to }{S}^2$. It is a fiber bundle: phase multiplication gives each fibre, and choosing a normalized representative on either standard projective chart gives a local trivialization. The theorem invoked is the long exact sequence of homotopy groups of a fibration. Since the universal cover of $S^1$ is contractible, ${\pi }_2(S^1)={\pi }_3(S^1)=0$. Its exact sequence consequently gives an isomorphismThe identity class of $S^3$ is nonzero, since an identity map cannot be null-homotopic while inducing the identity on $H_3(S^3;\mathbb{Z})$. Its image under ${\eta }_{\ast }$ is the class of $\eta$, so that class is nonzero. The Hopf map is not homotopic to a constant despite inducing zero on all positive-degree homology.

There are no one-handles, so attaching the three two-handles to a four-ball cannot create a fundamental group generator: ${\pi }_1(W)=1$. The handle chain complex has $C_2={\mathbb{Z}}^3$, $C_1=0$ and $C_0=\mathbb{Z}$, givingTo calculate the boundary, orient the two long components down their central strands, and orient the small component clockwise. The six mutual crossings of the long components have negative sign, giving linking number $-3$. Each long component has two positive crossings with the small component, giving linking number $1$. The surgery linking matrix, in the order of framings $4,3,2$, is consequentlyChanging a component orientation changes the corresponding row and column signs, and does not change the resulting homology. The long exact sequence of a pair and Poincare-Lefschetz duality identify $H_2(\partial W)=\ker Q$ and $H_1(\partial W)=\mathrm{coker}Q$. The determinant is nonzero, so the kernel vanishes. The gcd of matrix entries is one; the two-by-two minors include $-6$ and $7$, so their gcd is also one. The Smith normal form therefore has diagonal $1,1,7$, andIn particular the sign of the braid crossings matters: replacing $-3$ by $+3$ while leaving both meridian linkings positive would represent a different link and would give the wrong determinant.

Let $x,y$ be the generators associated with the two pairs of one-handle feet. Following the attaching circle across the identifications marked $a,b$ and $c,d$ gives the word $xy{x}^{-1}{y}^{-1}$, up to reversal and changing generator orientations. Thus the van Kampen theorem givesThe handle decomposition has cellular groups $C_0=\mathbb{Z}$, $C_1={\mathbb{Z}}^2$, $C_2=\mathbb{Z}$, with no higher groups. The boundary of the two-cell is the abelianized attaching word, which is zero. HenceThe zero framing is the product framing of the commutator handlebody of the torus, so the represented manifold is $T^2\times D^2$ and its intersection form is $(0)$. In particular its boundary is $T^2\times S^1=T^3$.

The boundary homology can also be obtained without using the product identification. By Poincare-Lefschetz duality, $H_3(W,\partial W)=H^1(W)={\mathbb{Z}}^2$ and $H_2(W,\partial W)=H^2(W)=\mathbb{Z}$. The map $H_2(W)\to H_2(W,\partial W)$ is its zero intersection form. The long exact sequence of a pair then gives split sequencesThe boundary is connected and oriented, so its zeroth and third homology are both $\mathbb{Z}$. Therefore

Use the boundary-orientation convention in which $-p/q$ surgery on the unknot gives $L(p,q)$. The negative continued fraction isAttach three two-handles to a zero-handle along a chain of unknots, with adjacent linking numbers $+1$ and framings $-3,-2,-2$. Equivalently, take the plumbing of oriented disk bundles over three spheres with Euler numbers $-3,-2,-2$. Successive slam-dunk moves change the boundary surgery coefficient to $-3-1/(-2-1/(-2))=-7/3$, proving that its oriented boundary is $L(7,3)$ in this convention.

The zero sections are a basis of second homology, and the intersection form isIts successive leading principal minors are $-3,5,-7$. Equivalently those of $-Q$ are $3,5,7>0$, so Sylvester's criterion proves that $Q$ is negative definite. This explicitly constructs the required filling.

If lens spaces are instead labeled by positive surgery with its boundary orientation, use the chain with framings $-2,-4$, whose coefficient is $-7/4$; its boundary is orientation-preservingly the positive-surgery $L(7,3)$, since orientation reversal changes $q=4$ to $q=3$. Its negative-definite form is $\left(\begin{array}{cc}-2& 1\\ 1& -4\end{array}\right)$. This is only a convention change in the named boundary, and specifies a valid filling in either labeling system.

For the shaded bigon, a holomorphic strip is a conformal parametrization of its interior. Send its two corners to the two ends of $\mathbb{R}\times [0,1]$. The Riemann map is unique up to translation, so quotienting the moduli space by translations leaves a single point. The embedded strip is regular, and a coherent orientation assigns it sign $+1$ or $-1$. Thus the signed count is $\pm 1$. The other three small bigons have the same argument.

For the shaded rectangle, a disk in ${\mathrm{Sym}}^2(\Sigma )$ is represented by a two-sheeted surface over the strip, with its projection to $\Sigma$ covering the rectangle once. There is one simple branch point. The four alternating boundary arcs and the conformal modulus of the rectangular region determine the branched cover up to strip translation: cutting along an arc from the branch point produces two simply connected strips, whose conformal maps are fixed by their marked corners, and gluing fixes the relative translation. The index calculation in part (ii) gives a zero-dimensional quotient; regularity of this embedded empty rectangle makes its one solution an isolated regular point. Hence its signed count is also $\pm 1$. The same construction applies to the lower rectangle. This is the empty bigon and rectangle counts in Heegaard Floer homology calculation, including the geometric reason for the counts.

The listed bigons and rectangles give all positive index-one domains in these diagrams. One can check this by following the alpha and beta boundary arcs: the four two-corner regions in Figure 2a and the two empty four-corner regions in Figure 2b are the possibilities with convex corners; a domain reaching around a handle has an additional corner or a negative region coefficient. There is no nonzero basepoint-normalized periodic domain, because the displayed homology presentations have nonzero determinant. Adding the whole surface increases the Maslov index by two, so does not give another index-one representative. Thus no extra chain differential term is hidden in the calculations below.

For Figure 2a with the basepoint outside the four small bigons, coherent signs can be absorbed into the choices of generator orientations, yielding a chain complex of the formThe first summand has homology $R⟨[1]⟩$, with $[3]=-[1]$. The singleton $4$ contributes $R$. In the last summand, $6$ is a boundary and the cycles are $R⟨5-7⟩$. Consequentlyone free tower in each Spin-c structure. With the normalizations in part (i), their top relative Heegaard Floer gradings are $0,0,1$.

The absent visible basepoint in Figure 2a does not change this homology answer. For the multiplicities of part (i), replace the three nonzero chain differential formulas by $\partial 2=\pm U^{a}1\pm U^{b}3$, $\partial 5=\pm U^{c}6$, $\partial 7=\pm U^{d}6$. At most one of $a,b,c,d$ equals one, and the others are zero, so each two-term relation contains a unit. Elementary elimination again leaves one free $R$ tower in each class; its relative Heegaard Floer grading is read from the multiplicity-dependent grading formulas.

For Figure 2b, write $A=(1,a)$, $B=(2,b)$, $C=(3,c)$. The two rectangles give, after choosing orientations,and both singleton generators have zero chain differential. The first summand has cycles $RA\oplus RC$ modulo the primitive relation $A+C$, hence is $R$. Thereforeagain with one free tower in each Spin-c structure. All three can have top relative Heegaard Floer grading zero in the independent normalizations of part (ii).

Remove two disjoint four-balls from $M$, giving $W:S^3\to S^3$. An Admissible cut for a Heegaard Floer mixed map is a separating three-manifold $N$ withThe last condition is the Mayer-Vietoris connecting-map condition. It makes a Spin-c structure on $W$ uniquely determined by its restrictions to the two pieces, avoiding an uncontrolled sum of extensions in the composition law. Such cuts exist when $b_2^{+}(M)>1$: represent an integral positive-square class by an embedded surface whose tubular neighborhood carries one positive direction and whose complement retains another. One can construct an embedded representative by resolving intersections of a representative cycle; the genus may increase without changing its class or square. The nonzero normal Euler number makes the restriction $H^1(\nu T)\to H^1(\partial \nu T)$ surjective, so the connecting-map condition holds. Puncture on either side of the cut.

Write $\mathfrak{t}=\mathfrak{s}{|}_N$ and ${\mathfrak{s}}_i=\mathfrak{s}{|}_{W_i}$. Both infinity Heegaard Floer cobordism maps vanish by positivity. Naturality with respect to the exact sequencegives $\mathrm{im}{F}_{W_1,{\mathfrak{s}}_1}^{-}\subset \ker \iota$ and $F_{W_2,{\mathfrak{s}}_2}^{+}\pi =0$. Exactness identifies $\delta$ with a degree-$-1$ isomorphism from $H{F}^{+}(N)/\mathrm{im}\pi$ onto $\ker \iota$, namely the two versions of Reduced Heegaard Floer homology. Hence the inverse on these reduced groups is defined, and one formsThis is the construction of the Mixed Heegaard Floer invariant, rather than the zero ordinary composite.

Let ${\Theta }^{-}$ be the top minus generator, of degree $-2$, and ${\Theta }^{+}$ the bottom plus generator, of degree zero. For $\zeta \in {\Lambda }^{\ast }(H_1(M)/\mathrm{torsion})$, insert its degree-one homology actions in the Heegaard Floer cobordism maps and defineThis is an integer up to the overall orientation sign. cobordism composition and associativity show independence of the admissible cut and handle choices. Equivalently, fixing the relevant orientation data removes the overall sign ambiguity.

The grading condition can be checked explicitly. Since $\chi (W)=\chi (M)-2$ and $\sigma (W)=\sigma (M)$, the sum of the ordinary cobordism shifts is $d(\mathfrak{s})+1$, where $d(\mathfrak{s})=(c_1(\mathfrak{s}{)}^2-2\chi (M)-3\sigma (M))/4$. The inverse connecting map contributes another $+1$. Therefore the output degree on $U^r{\Theta }^{-}\otimes \zeta$ is $d(\mathfrak{s})-2r-\mathrm{deg}\zeta$, and the displayed invariant vanishes unless $2r+\mathrm{deg}\zeta =d(\mathfrak{s})$.

Finally, two positive pieces are needed, so that both infinity maps vanish and the reduced-group factorization is available. For an admissible cut the connecting-map condition and signature additivity imply that the two positive subspaces inject as orthogonal subspaces of the closed intersection form; hence $b_2^{+}(M)\ge 2$. With only one positive direction no such admissible cut exists, and a chamber-dependent construction requires extra choices. With no positive direction, the ordinary infinity maps need not vanish at all. This explains the stated bound for the invariant defined here.

The map is a homogeneous Laurent polynomial ring homomorphism between the two rank-one infinity towers. Choose homogeneous generators $e_i$ with absolute degrees $q_i$. Its degree isConsequently the general graded-module description isIf the grading does not allow an integer exponent, the map is zero. Negative exponents are allowed in the infinity theory. The notation $F$ here denotes the map called ${\Phi }^{\infty }$ in the question.

The standard topological statements for Heegaard Floer cobordism maps, which require no proof in this part, areIn the second case $a=\pm 1$ and the map is therefore $\pm U^r$ with the exponent determined by the displayed degree formula. Since both boundary components are rational homology spheres, the rational intersection form is nondegenerate, so $b_2^{+}=0$ means that it is negative definite, allowing also the rank-zero form.

The printed hypothesis $b_1(Y_i)=0$ does not imply $b_1(W)=0$. Thus one must not replace the second implication by an unconditional isomorphism assertion for every negative-definite cobordism. A concrete check is $W=(S^1\times S^3)\setminus (B^4\bigsqcup B^4):S^3\to S^3$. It has $b_2^{+}=0$ and $b_1(W)=1$. A one-handle followed by a three-handle gives $e\mapsto e\otimes {\theta }^{+}$ and then kills the ${\theta }^{+}$ summand; hence its infinity map is zero. Its grading shift is one, also incompatible with a nonzero map between the even-graded $S^3$ towers. With only the stated boundary assumption, the general homogeneous Laurent-module description and the two topological implications above are the valid conclusions; determining any remaining coefficient requires the actual cobordism, not just its boundary Betti numbers.

For a rational homology three-sphere and any Spin-c structure, the infinity version of Heegaard Floer homology isIt is one Laurent tower, with one integral generator in each degree of a fixed parity and none in the opposite parity. Its absolute rational grading has an overall offset determined by $(Y,\mathfrak{s})$.

To compute the Euler characteristic of the hat group, tensor with $\mathbb{Q}$, which preserves integral ranks. Put $V=H{F}^{-}(Y,\mathfrak{s};\mathbb{Q})$. It is a finitely generated graded $\mathbb{Q}[U]$ module. Localizing at $U$ gives $H{F}^{\infty }$, so the structure theorem for finitely generated modules over a principal ideal domain gives one free summand and finitely many $U$-primary torsion summands. The torsion is $U$-primary because it vanishes after localization, or directly because the reduced graded group is supported in finitely many degrees.

The exact sequence of Heegaard Floer chain complexes given by multiplication by $U$ and passage to the hat quotient identifies the hat group, as a parity-graded vector space, with an extension of $V/UV$ by $\ker (U:V\to V)$ with the latter parity reversed. The extra absolute shift by two between $C{F}^{-}/UC{F}^{-}$ and $\widehat{CF}$ is even, so it does not affect this argument. ThusThe free summand contributes one vector to $V/UV$ and none to $\ker U$. A torsion summand $\mathbb{Q}[U]/U^r$, whose top generator has parity $ϵ$, contributes one vector of parity $ϵ$ to the quotient and one of the same parity to the kernel, since $U^{r-1}$ changes degree by an even number. Their contributions cancel. Only the single free summand remains, provingInteger torsion does not enter an Euler characteristic defined by ranks.

Here the usual homological algebra convention is that the right-exact functor $F$ is additive. Module categories have enough projective modules, because any module is a quotient of a free module. Choose a projective resolutionApply $F$ to the unaugmented chain complex and define the left derived functors byAdditivity ensures that $F$ preserves zero maps, sums of maps and homotopy identities, so the transformed differentials still compose to zero.

For a module homomorphism $u:M\to N$, lift it to a chain map between chosen projective resolutions. In degree zero, lift through $Q_0\twoheadrightarrow N$ using projectivity of $P_0$. In each higher degree, the already constructed map lands in the next kernel; exactness of $Q_{\bullet }$ and projectivity of $P_i$ supply the next lift. Any two lifts are chain homotopic: their degree-zero difference lands in $\ker (Q_0\to N)=\mathrm{im}{d}_1^Q$, so it lifts to $h_0:P_0\to Q_1$. Inductively lift the remaining difference after subtracting the preceding homotopy term. This givesApplying $F$ preserves this identity, so the lifts induce the same homology map. Identity lifts and composable lifts show that $L_{i}F$ is a functor. Comparison lifts between two resolutions over the identity of $M$ are inverse up to chain homotopy; hence the construction is independent of the chosen resolution up to its canonical natural isomorphism.

Right exactness identifies $L_{0}F(M)=\mathrm{coker}(F(P_1)\to F(P_0))$ naturally with $F(M)$. For a projective module $P$, the length-zero resolution gives $L_{i}F(P)=0$ for $i>0$. For a short exact sequence $0\to M^{\prime }\to M\to M^{\prime \prime }\to 0$, the horseshoe lemma builds a degreewise split short exact sequence of projective resolutions. Concretely, lift the projective generators covering $M^{\prime \prime }$ into $M$ and combine them with generators covering $M^{\prime }$; repeat on the kernels. The middle projective term in each degree is the direct sum of the two outer ones. An additive functor preserves these split sequences, and lifting cycles, taking their differentials and returning to the subcomplex produces the connecting maps inThis is the natural long exact derived-functor sequence. In particular, if $0\to K\to P\to M\to 0$ has $P$ projective, then $L_{i}F(M)\cong L_{i-1}F(K)$ for $i\ge 2$, and $L_{1}F(M)=\ker (F(K)\to F(P))$. These are dimension-shifting formulas. Positive degrees can be killed by the surjection $P\to M$. If $F$ is exact, it preserves every resolution's exactness and all its positive left derived functors vanish. Conversely, if $L_{1}F$ vanishes on all modules, the displayed long exact sequence makes $F$ exact. A natural transformation between additive right-exact functors induces transformations on the left derived functors by applying it termwise.

For extension of scalars $F(M)=B{\otimes }_{A}M$, these derived functors are the Tor functors ${\mathrm{Tor}}_i^A(B,M)$. When $B=A/(a)$ and $a$ is a non-zero-divisor, there is the length-one free resolutionAlthough the construction above resolves $M$, one may instead use this resolution of $B$. To justify that calculation, tensor a free resolution of $M$ with the two-term resolution of $B$ to form a double complex. Taking homology first along the resolution of $B$ leaves $B{\otimes }_A{P}_{\bullet }$; taking it first along the resolution of $M$ leaves $M\stackrel{a}{\to }M$. The free terms preserve exactness under tensoring. The double complex is first-quadrant with finitely many terms on each total diagonal, so both filtrations compute the same total homology. This proves the required balancing directly.

The kernel and cokernel of multiplication by $a$ now give Tor for a principal non-zero-divisor quotient:These are naturally $B$-modules. For $M=A$, the degree-one group is zero; for $M=B$, multiplication by $a$ is zero and both degrees zero and one are $B$. Thus, when $B\ne 0$, degree one explicitly detects the failure of this base-change functor to be exact.

Label the six vertices $A,B,C,D,E,F$ clockwise, starting at the upper-left vertex. The arrows identify $AB$ with $ED$, $BC$ with $FE$, and $FA$ with $DC$, preserving the indicated directions. Consequently the quotient space has exactly two vertices,Orient its three topological cells by $a:A\to B$, $b:B\to C$, and $c:F\to A$. The clockwise boundary of the hexagonal topological cell traversesEvery topological cell of dimension one therefore occurs once with each sign. The cellular chain complex, with integral coefficients, isHere the rows correspond to $P,Q$. The image of ${\partial }_1$ is the primitive subgroup generated by $Q-P$, and its kernel is freely generated by $a+b$ and $a+c$. Thus the homology groups areThe two vertex classes are important: identifying all six vertices would give an incorrect cellular chain complex. The result agrees with the homology of a torus.

Write a point of the larger sphere as $(u,v)\in {\mathbb{R}}^{m+1}\times {\mathbb{R}}^{n-m}$. In the complement, $v\ne 0$. There is a homeomorphismwhose inverse sends $(u,w)$ to $(u,\sqrt{1-\Vert u{\Vert }^2}w)$. In particular the complement has a deformation retraction onto its transverse sphere. Explicitly,The denominator stays positive, the homotopy stays in the complement, and points with $u=0$ remain fixed. This establishes the homology of an equatorial sphere complement. By homotopy invariance of homology, its homology is therefore the homology of a sphere of dimension $d=n-m-1$:In the codimension-one case, the complement has two contractible components; this is why degree-zero homology has rank two. Equivalently, in every case its only nonzero reduced homology is ${\tilde{H}}_d(X;\mathbb{Z})=\mathbb{Z}$.

The torus map acts on $H_1(T^2;\mathbb{Z})$ byIts determinant is one, so it is an orientation-preserving homeomorphism and acts as the identity on both $H_0(T^2)$ and $H_2(T^2)$.

To calculate the homology of its mapping torus, cover the time circle by two arcs. Each inverse image has the homotopy type of $T^2$, while their intersection has two components of that homotopy type. The map in the Mayer–Vietoris sequence can be writtenAdding the output coordinates and using the input coordinate $a+b$ separates an identity map from $f_{\ast }-I$. Canceling the identity summands gives the Wang sequence, and henceIn degree one,Its image and kernel are both generated by $(1,-1)$; its cokernel is $\mathbb{Z}$ since this image is primitive. Its Smith normal form is $\mathrm{diag}(1,0)$, so its torsion subgroup is zero. On the other two nonzero homology groups of the torus, $f_{\ast }-I=0$. The displayed short exact sequences split because their right-hand groups are free. ThereforeIn particular, the mapping torus contributes an additional degree-one class from the time circle, as well as the invariant degree-one class in the fiber.

The Snake lemma applies to a commutative diagram of modules with exact rows $0\to A^{\prime }\stackrel{i^{\prime }}{\to }{B}^{\prime }\stackrel{j^{\prime }}{\to }{C}^{\prime }\to 0$ and $0\to A\stackrel{i}{\to }B\stackrel{j}{\to }C\to 0$, and vertical maps $f,g,h$. It supplies the exact sequenceFor $c^{\prime }\in \ker h$, choose $b^{\prime }$ with $j^{\prime }{b}^{\prime }=c^{\prime }$. Commutativity gives $jg{b}^{\prime }=0$, so $g{b}^{\prime }=i(a)$ for a unique $a\in A$. Define $\delta (c^{\prime })=[a]$ modulo $f(A^{\prime })$. Changing $b^{\prime }$ by $i^{\prime }(a^{\prime })$ changes $a$ by $f(a^{\prime })$, proving well-definedness. This construction is additive. Exactness follows by the same lift-and-chase: $\delta (c^{\prime })=0$ exactly when the lift can be adjusted into $\ker g$; a class in $\mathrm{coker}f$ maps to zero in $\mathrm{coker}g$ exactly when it is obtained from such a lift. The remaining kernel and cokernel positions follow directly from row exactness.

For an open cover $X=U\cup V$, use the short exact sequence of singular chain complexeswhere the last complex consists of chains whose simplices lie in one member of the cover. This is exact because a simplex present in both summands lies in their intersection. Barycentric subdivision and its chain homotopy to the identity show that this small-chain inclusion induces an isomorphism on homology: compactness of each simplex and a Lebesgue-number argument place sufficiently subdivided simplices inside the cover.

Applying the Snake lemma to a short exact sequence of chain complexes gives the connecting homomorphism $[c]\mapsto [da]$: lift a quotient cycle $c$ to the middle complex, take its differential, and identify that differential with an element of the subcomplex. Its independence of the lift and representative follows by the same chase. Repeating this degree by degree, or chasing the cycles and boundaries directly, givesThis is the Mayer–Vietoris sequence. Specifically, split a small cycle as $a+b$; then $da=-db$ lies in the intersection, and $\delta [a+b]=[da]$ with the sign convention above.

For the final truncation, use homological grading $d:C_j\to C_{j-1}$. The span $B$ in degrees below $n$ is a subcomplex. The span called $A$ in degrees at least $n$ need not be a subcomplex with the inherited differential: $d{C}_n$ may lie in $C_{n-1}$. Give $A$ the quotient differential, with $d:A_n\to A_{n-1}=0$ zero and all higher differentials inherited. Thenis a short exact sequence of chain complexes. Its only potentially nonzero connecting map isIndeed $H_n(A)=C_n/d{C}_{n+1}$; $da$ is a cycle in $B$, and changing $a$ by $db$ changes $da$ by zero. In other degrees either the source or the relevant target of the connecting map vanishes.

The three handle attachments give $H_2(W;\mathbb{Z})\cong {\mathbb{Z}}^3$: there are no one-handles or three-handles. Use the capped core disks of handles as the homology basis, ordered by the framing labels $-2,-1,+1$. The surgery linking matrix gives the intersection form. Orient the three components so that the outer components have linking number $3$ and the middle component has linking number $1$ with each outer component. There are six crossings between the outer components, and two crossings between the middle component and each outer component; counting their signs gives these linking numbers. Reversing a component changes the corresponding row and column signs and does not change the invariants.

Thus the intersection form isAn exact matrix congruence diagonalizes this intersection form toConsequently its signature is $\boxed{\sigma (W)=-1}$, and $\det Q_W=18$.

The long exact sequence in relative homology and Poincare-Lefschetz duality identifywith the intersection form matrix, and identify $H_1(\partial W)$ with its cokernel, since $H_1(W)=0$. The Smith normal form has first invariant factor $1$, because the entries have greatest common divisor $1$, and second invariant factor $1$, because the upper-left two-by-two matrix minor has determinant $1$. The last invariant factor is $18$. Hence

All homology and cohomology in this solution use ${\mathbb{F}}_2$ coefficients, so no orientation hypothesis is necessary. The Poincare-Lefschetz duality degrees are complementary:The starred notation in the question must be understood with this degree reversal.

Turn the handle decomposition upside down. An absolute $k$-handle becomes a relative $(n-k)$-handle based on $\partial W$. The absolute cellular boundary coefficient counts intersections of an attaching sphere with a belt sphere. In the reversed handle decomposition, the same intersections are counted in the opposite order. Over ${\mathbb{F}}_2$ the two incidence matrices are transposes, with no sign ambiguity. Thus the absolute chain complex is the complementary-degree dual of the relative chain complex. Taking homology and using the universal coefficient theorem for cohomology over a field proves the displayed isomorphism. This is mod-two handle duality.

To obtain a perfect pairing from the assumed closed case, form the double of a manifold $D(W)$ from two copies of $W$. The fold map $D(W)\to W$ is a retraction, so the inclusion of either copy induces an injection on homology. If $a\ne 0$ in $H_k(W)$, its image in $H_k(D(W))$ is nonzero. Closed-manifold mod-two Poincare duality supplies a class $z\in H_{n-k}(D(W))$ with intersection $1$ against that image. Cut a representative of $z$ along the boundary and retain its part in the first copy; it defines $b\in H_{n-k}(W,\partial W)$. Equivalently, apply the quotient map $D(W)\to D(W)/W_{-}\cong W/\partial W$ and excision. Intersections with a representative of $a$ in the interior are unchanged, so $a\cdot b=1$. This proves nondegeneracy in the first variable. Mod-two handle duality gives equal finite dimensions for the two spaces, hence nondegeneracy in the second variable as well.

For the three-dimensional conclusion, setThe long exact sequence in relative homology says $L=\mathrm{im}\partial$, where $\partial :H_2(W,\partial W)\to V$. The closed-surface intersection pairing on $V$ is nondegenerate. A collar calculation gives the adjoint identityIndeed, push $x$ slightly into the collar: intersections with the relative surface representing $b$ correspond to its boundary intersections with $x$. By the absolute-relative perfect pairing already proved, $x$ is orthogonal to every $\partial b$ precisely when $i_{\ast }x=0$. Therefore $L^{\perp }=L$.

For any finite-dimensional space with a perfect pairing, $\dim L+\dim L^{\perp }=\dim V$. Hence the half-lives-half-dies theorem givesBut $H_1({\mathbb{RP}}^2;{\mathbb{F}}_2)\cong {\mathbb{F}}_2$ has dimension one. It cannot be the entire boundary of a compact three-manifold, because the displayed dimension would be $1/2$. This excludes nonorientable three-manifolds as well as orientable ones.

Choose an oriented Seifert surface $F$ of genus $g$ and an integral basis $a_1,\dots ,a_{2g}$ of $H_1(F;\mathbb{Z})$. Push $a_i$ in the positive normal direction to $F$. The Seifert matrix isIt represents the Seifert form. Its skew-symmetrization $V-V^T$ is the integral intersection form of $F$. The relation to the Alexander polynomial of a knot isThis follows from the presentation of the Alexander module of a knot obtained by cutting the knot exterior along $F$ and stacking the resulting copies; a proof is not needed here.

The printed genus inequality is false. The useful consequence of the determinant relation is the Alexander breadth bound on Seifert genus:Here the breadth of a Laurent polynomial is its largest exponent minus its smallest exponent, so it is unchanged by multiplying by $\pm t^m$. Indeed, for a minimal-genus Seifert surface, the matrix has size $2g_s(K)$ and each entry of $tV-V^T$ has degree at most one. Its nonzero determinant is an ordinary polynomial of degree at most $2g_s(K)$; its breadth is no larger than that degree. Equivalently, the highest exponent of a symmetrically normalized Alexander polynomial of a knot is at most $g_s(K)$. An unnormalized degree is not invariant under Laurent units.

For a counterexample to the printed inequality and the requested example, take the untwisted Whitehead double of a trefoil knot. The picture specifies the two-strand tangle in the zero Seifert framing; the two exterior caps form a Whitehead double clasp. The inset identifies the trefoil knot used as companion.

Its usual genus-one Seifert surface is a zero-framed annulus following the trefoil knot, joined by the clasp band. Choose the annulus core and a curve traversing the clasp band as the basis of its first homology. Zero annulus twisting gives the first self-linking number $0$. The clasp contributes a Hopf band with self-linking number $-1$ in the positive-clasp convention used here. Plumbing the bands contributes one linking in one push-off direction and none in the other. Orienting the basis suitably givesTying the annulus into the trefoil knot changes neither these local linking counts nor its prescribed zero Seifert framing. ThereforeThis Whitehead double is not the unknot, as allowed without proof in the question. Its Seifert genus is consequently at least one and at most one, hence exactly one. Thus $g_s(K)=1$ while every normalized constant Alexander polynomial of a knot has degree zero: a direct counterexample to the printed inequality.

Choose a generator $[S^n]$ of ${\tilde{H}}_n(S^n;\mathbb{Z})\cong \mathbb{Z}$. The mapping degree of a continuous map is the integer determined byFor $n\ge 1$ this is ordinary top-dimensional homology; using reduced homology also gives a convention for $n=0$. Domain and target use the same orientation, so reversing the chosen generator in both leaves the integer unchanged. Homotopy invariance of homology makes the mapping degree invariant under homotopy.

We interpret the manifold criterion for a suspension in the category of nonempty topological manifolds without boundary, and “sphere” as homeomorphic to the standard sphere. Let $X$ have dimension $d$. If $X\cong S^d$, an explicit homeomorphism from its suspension to $S^{d+1}$ isIt is a continuous map that is a bijection: the two collapsed ends become the poles, and every other point determines $t$ and $x$ uniquely. The compact-to-Hausdorff continuous bijection theorem makes it a homeomorphism.

Conversely suppose $SX$ is a topological manifold. Its open part $X\times (0,1)$ shows its dimension is $d+1$. At a suspension vertex $v$ there is an open cone neighbourhood $U$ on $X$. It is contractible, while $U\setminus \{v\}$ admits a deformation retraction onto $X$. The long exact sequence in relative homology and excision give the local homology from a link calculationA point of a $(d+1)$-dimensional topological manifold without boundary has local homology $\mathbb{Z}$ in degree $d+1$ and zero in all other degrees. ConsequentlyThus $X$ is an integral homology sphere. This conclusion alone does not identify its homeomorphism type.

If $d\ge 2$, it is also simply connected. Choose a cone neighbourhood $U$ of the vertex and an open coordinate ball $B$ around the vertex inside $U$. By compactness of $X$, sufficiently short entire cone neighbourhoods fit inside any neighbourhood of $v$: the inverse image of that neighbourhood contains $X\times \{0\}$, and a finite subcover supplies one uniform collar length. Choose such a shorter cone $V\subset B$. The inclusions of punctured neighbourhoods factor asThe composite induces an isomorphism of fundamental groups, since both cone punctures are products of $X$ with intervals and their inclusion is a homotopy equivalence. The middle punctured ball has the homotopy type of $S^d$ and is simply connected for $d\ge 2$. Hence the composite is also zero on the fundamental group, forcing ${\pi }_1(X)=0$. This uses the nested neighbourhoods, not an unsupported claim that an arbitrary punctured neighbourhood is a sphere.

The Hurewicz theorem, applied successively to the vanishing lower homology groups, now gives ${\pi }_j(X)=0$ for $j<d$ and ${\pi }_d(X)\cong H_d(X)\cong \mathbb{Z}$. A representative of its generator is a continuous map $S^d\to X$ inducing an isomorphism in every integral homology degree. Topological manifolds have the homotopy type of CW complexes, so the homological Whitehead theorem makes it a homotopy equivalence. We have proved that $X$ is a homotopy sphere.

At this point the topological classification theorem is essential: the topological generalized Poincare theorem says that a closed homotopy sphere is homeomorphic to the standard sphere. Applying it completes the converse for $d\ge 2$. For $d=1$, the local calculation makes $X$ a connected compact one-dimensional manifold without boundary, hence a circle. For $d=0$, a compact zero-dimensional manifold is a finite set, and ${\tilde{H}}_0(X)\cong \mathbb{Z}$ forces exactly two points. Thus it is $S^0$. Altogether,The last classification step is not a consequence of homology alone, and we make no assertion of diffeomorphism with a prescribed smooth structure. If the word manifold allows boundary on both sides, the printed assertion needs this correction: the suspension of the interval is a closed disk. For instance $[x,t]\mapsto ((2x-1)(1-|2t-1|),2t-1)$ identifies it with the filled diamond $|u|+|v|\le 1$, although the interval is not a sphere.

To construct sphere maps of arbitrary integer degree, first use $h_m:S^1\to S^1$, $h_m(z)=z^m$ for any integer $m$, including the constant map when $m=0$. Lifting its argument to the real line gives $\theta \mapsto m\theta$, so the fundamental one-cycle is sent to $m$ times itself. Therefore $\mathrm{deg}{h}_m=m$.

Two cone neighbourhoods cover a suspension; their intersection admits a deformation retraction onto the original space. The reduced Mayer–Vietoris theorem gives a natural suspension isomorphismNaturality means that the suspension of a continuous map induces the same integer multiplier. This proves degree under suspension. Using the standard identification $S(S^k)\cong S^{k+1}$, setThus every integer occurs in every positive sphere dimension, with negative degrees handled just as well as positive ones.

If “every $n$” includes zero, that final printed assertion is false. The generator of ${\tilde{H}}_0(S^0)$ is the difference of its two points. A map of a two-point set is the identity, the transposition, or one of two constant maps, giving preciselyThese degrees of maps of the zero-sphere are the explicit exception to the arbitrary-degree construction.

For a CW complex $X$, let $X^k$ be its $k$-skeleton. The cellular chain complex isThe isomorphism uses excision and $H_k(D^k,S^{k-1})\cong \mathbb{Z}$, after choosing an orientation for each cell. In degree zero, take $X^{-1}=\varnothing$ and the free group on the vertices. The differential is the compositeThe long exact sequences of skeleton pairs imply $d_{k-1}{d}_k=0$. The cellular boundary formula describes its coefficient at a $(k-1)$-cell as the mapping degree of the attaching sphere after all other $(k-1)$-cells are collapsed. In degree one this is the signed difference of the two endpoints. The cellular homology theorem identifies the homology of this complex with singular homology: relative homology of successive skeletons is concentrated in their cell dimension, and the exact skeleton sequences leave precisely these kernels modulo images. This is the construction, rather than merely a count of cells.