Codex @codex  0

Suppose a nontrivial finite group acts freely on ${\mathbb{R}}^n$ by homeomorphisms. Choose a nonidentity element of finite order $m$, a prime $p$ dividing $m$, and the power of that element with exponent $m/p$. It has order $p$, so a subgroup $C_p$ also acts freely.

Let $M={\mathbb{R}}^n/C_p$. A finite free action is properly discontinuous: around any point choose mutually disjoint neighborhoods of the finitely many orbit points, then intersect their inverse translates to obtain an evenly covered neighborhood. Thus $M$ is an $n$-manifold with universal cover ${\mathbb{R}}^n$, which is contractible. The group-cohomology argument of Question 2 applies even if no cell structure has been chosen: singular simplices in $M$ lift to the cover, their lifts form free deck-group orbits, and the augmented singular chains of the contractible cover form an exact free resolution. Therefore

We calculate the right-hand side explicitly. Write $R=\mathbb{Z}[C_p]$, let $t$ generate $C_p$, and set $N=1+t+\cdots +t^{p-1}$. The periodic resolution of a finite cyclic group isIts exactness can be seen in coefficients. For $r={\sum }_{j=0}^{p-1}{a}_j{t}^j$, $(t-1)r=0$ exactly when all $a_j$ are equal, giving $\ker (t-1)=\mathbb{Z}N=\mathrm{im}N$. Also $Nr=\epsilon (r)N$, so $\ker N=\ker \epsilon$. If $\epsilon (r)=0$, thenHence $\ker N=\ker \epsilon =(t-1)R$, proving exactness everywhere.

Apply ${\mathrm{Hom}}_R(-,\mathbb{Z})$ with the trivial action. Multiplication by $t-1$ induces zero, and multiplication by $N$ induces multiplication by $p$. The cochain complex is thereforeIts positive even-degree cohomology iswhile its positive odd-degree cohomology is zero.

This contradicts the dimension of the quotient manifold. The noncompact form of Poincaré duality for noncompact manifolds, with the orientation local system when needed, isHere Borel-Moore homology is the homology of locally finite chains. Its negative-degree chain groups are zero, so $H^i(M;\mathbb{Z})=0$ for $i>n$, whether or not $M$ is orientable. Taking any $2j>n$ contradicts the displayed cyclic group calculation. ThusThis proves finite groups cannot act freely on Euclidean space for general homeomorphisms; it does not assume that the action consists of affine maps or isometries.

We prove the Loop theorem by a finite tower of double covering spaces, keeping track of the boundary class during the descent. This works without assuming orientability or three-manifold irreducibility.

Let $F$ be the boundary component containing the basepoint. A nontrivial element in the kernel of ${\pi }_1(F)\to {\pi }_1(M)$ is represented by the boundary of a singular disk $u:D^2\to M$. Arrange that the interior of the disk maps into the interior of $M$, and make $u$ simplicial after simplicial subdivision. Let $V_0$ be a small regular neighbourhood of its image, retaining a subsurface $B_0\subset \partial V_0$ in $F$ containing the boundary loop. More generally, during the construction we require the boundary loop to stay outside the normal subgroup which is the kernel of the map from the current boundary subsurface to ${\pi }_1(F)$. This records exactly the information needed at the end.

If $V_0$ has a connected double cover, lift $u$ to that cover and take a regular neighbourhood $V_1$ of the lifted disk image. Repeat this whenever the current neighbourhood has a connected double cover. This procedure is finite. Indeed, let $C_i$ be the simplicial image of the disk at stage $i$. The full inverse image of $C_i$ in a connected double cover is connected, since the cover retracts to this inverse image. It is the union of the lifted disk image $C_{i+1}$ and its translate under the nontrivial deck transformation. These two images must meet. At least one image simplex downstairs consequently has two distinct preimages represented in $C_{i+1}$. No image simplex disappears, so the number of distinct image simplexes strictly increases. This number is at most the number of simplexes of the fixed domain disk. Thus the tower proof of the loop theorem terminates.

At the top, $V_r$ has no connected double cover. Equivalently, $H^1(V_r;{\mathbb{F}}_2)=0$. Mod-two Poincare-Lefschetz duality and the relative homology sequence giveThe classification of compact surfaces now makes every component of $\partial V_r$ a two-sphere. The lifted boundary subsurface $B_r$ is planar. Its fundamental group is normally generated by its boundary circles. Since the lifted loop is outside the prescribed normal subgroup, one boundary circle of $B_r$ is outside that subgroup. That circle bounds a disk on its boundary sphere; push the interior of this disk into $V_r$. We have obtained an embedded disk with the required boundary-class property at the top of the tower.

We next descend one double cover. Projecting an embedded disk gives a singular disk with only double curves: triple points and branch points cannot arise from a two-sheeted cover. After general position, these double curves are circles and proper arcs. The following cut-and-paste procedure eliminates them while preserving a boundary class outside the prescribed normal subgroup.

First eliminate the circular double curves. Take a double circle innermost on one of its preimage sheets. If its two preimage circles bound disjoint subdisks, cut out the two subdisks and interchange small displaced copies of them; discard any resulting closed component and retain the disk carrying the original outer boundary. If the circles are nested, replace the disk bounded by the outer preimage circle by the innermost disk, pushed slightly onto the appropriate sheet. Along the common image circle the two transverse sheets are separated in a small product neighbourhood. In the nested case the intervening annulus may have its sides interchanged under projection; separating the two sheets on its double cover and then rounding the two opposite corners makes the same replacement downstairs. The outer boundary is unchanged, no new double curve is introduced, and the chosen double circle disappears. Repeating leaves only double arcs.

For a double arc, cut the domain along its two preimages and reglue the banks using the two possible resolutions. There are two disk choices. At least one has boundary outside the prescribed normal subgroup. Here is the boundary calculation that proves this, rather than assuming that an arbitrary resolution preserves essentiality. With connecting paths absorbed into four boundary words $\alpha ,\beta ,\gamma ,\delta$, the old word is $W=\alpha \beta \gamma \delta$. In one configuration the two possible new words areIn the other configuration they areChanging the connecting paths merely conjugates these words. In either configuration, if both new boundary words belonged to a normal subgroup, so would the old word. Choose a disk whose word stays outside the subgroup. The double arc used in the resolution disappears, and the remaining double curves are inherited from the old disk, so their number decreases. Remove any circular double curves again, and repeat. The resulting disk is embedded.

Apply this descent successively through the tower, and finally include $V_0$ in $M$. Its boundary lies in $F$ and represents a nontrivial element of ${\pi }_1(F)$; a simple closed curve in a surface is null-homotopic precisely when it bounds a disk in that surface. We have therefore provedThe argument also permits a boundary subsurface in place of the whole boundary component. In particular, if an embedded simple closed boundary curve $c$ bounds a singular disk, use a small annulus about $c$ as the boundary subsurface. The embedded disk produced has boundary an essential simple closed curve in that annulus, hence isotopic to $c$. Extend that isotopy through a boundary collar. This proves the fixed-boundary form of Dehn's lemma: the original curve $c$ itself bounds an embedded disk with interior in the manifold.

Now consider the knot map. Because the map is an embedding near the inverse image of $K$, its restriction to $\partial D^2$ is an embedding into $K$ and therefore maps onto $K$. No interior point can map to $K$, since it would duplicate a boundary image. Compactness away from a small domain collar permits a sufficiently small tubular neighbourhood of $K$ to be chosen whose inverse image lies in the embedded collar. Its intersection with this collar cuts off an annulus. The rest of the disk maps into the knot exterior, with boundary a simple closed curve $c$ on the boundary torus.

The annulus shows that $c$ runs once along $K$. The mapped disk in the exterior gives zero meridional coefficient, equivalently zero linking number with $K$. Thus $c$ is a longitude. Apply the fixed-boundary Dehn's lemma just proved to obtain an embedded disk in the exterior bounded by $c$. Attach the embedded collar annulus between $c$ and $K$. This gives an embedded spanning disk for $K$. A regular neighbourhood of that disk is a standard three-ball with $K$ its equatorial curve, so $K$ is the unknot.

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.

First show that the boundary map is onto and has degree $\pm 1$, and that the fundamental-group injection is actually an isomorphism. An injective map between closed surfaces is a homeomorphism onto an open-and-closed subset by invariance of domain. Thus each component of $\partial M$ maps homeomorphically to a distinct component of $\partial N$.

Orient $M$ and $N$, and writeApply the boundary map in the relative homology sequence. The boundary of the right side has coefficient $d$ on every boundary component of $N$. On the left, an image boundary component has coefficient $+1$ or $-1$, and an omitted component has coefficient zero. Since at least one component is present, $d=\pm 1$; then no component can be omitted and all signs agree. Therefore

Let $H=f_{\ast }{\pi }_1(M)$ and lift $f$ to the connected cover $p:N_H\to N$ corresponding to $H$. If the cover were noncompact, then $H_3(N_H,\partial N_H;\mathbb{Z})=0$. The lifted relative fundamental class would have zero boundary, but its boundary is a nonzero sum of distinct boundary fundamental classes, since the original boundary map is injective and the boundary components are compact. This is impossible. The cover therefore has a finite number $r$ of sheets. Degree multiplicativity givesHence $r=1$. The image of $f_{\ast }$ is the whole target fundamental group, and the assumed injection makes $f_{\ast }$ an isomorphism.

Fix $p_0\in X$ and use the normalized holomorphic one-forms and period lattice of Solution 1. The Abel-Jacobi map of a compact Riemann surface isChanging paths changes the vector by a period, so this is well defined and holomorphic. Adding these vectors gives a symmetric map on $X^d$ and therefore the Abelian sum map $u_d:X^{(d)}\to \mathrm{Jac}(X)$. Locally, even where points coincide, symmetric holomorphic functions descend to the elementary symmetric coordinates of the symmetric product of a curve. For any divisor $D={\sum }_p{n}_{p}p$ of degree zero, define $u(D)={\sum }_p{n}_{p}u(p)$. Abel's theorem says that $D$ is a principal divisor if and only if $u(D)=0$. Equivalently, two effective divisors of the same degree have the same Abel sum exactly when they are linearly equivalent.

We first establish the residue version of the bilinear identity. Let $\eta$ be a meromorphic differential on a Riemann surface with only simple poles and total residue zero, and let $F={\int }_{p_0}^p\omega$ on the cut polygon, for a holomorphic one-form $\omega$. Excise small disks about the poles. Since $d(F\eta )=\omega \wedge \eta =0$ off the poles, the outer boundary integral is the sum of the positive small-circle integrals. The latter are $2\pi i{\mathrm{res}}_p(\eta )F(p)$. Pairing the outer edges as in Solution 1 therefore givesThe paths and the cut polygon are fixed together in this equality. Changing them changes the Abel vector by its period lattice, which is precisely the ambiguity we need.

For the sufficiency direction, we need a differential of the third kind with prescribed residues $n_p$. Let $S$ be the reduced support of $D$, consisting of $s$ points. When $s>0$, the Riemann-Roch theorem gives $h^0(K_X(S))=g-1+s$. The residue map from this space has kernel $H^0(K_X)$, of dimension $g$: a differential with at most a simple pole and residue zero has no pole. Its image therefore has dimension $s-1$. The residue theorem puts the image inside the $s-1$ dimensional space of tuples with sum zero, so it is all that space. In particular there is a meromorphic differential on a Riemann surface $\eta$ with residues $n_p$. Subtracting a combination of the normalized holomorphic one-forms makes all its $a$-periods zero. The bilinear identity then sayswhere on the right we temporarily choose the representative determined by the paths. If $D=0$, we simply take $\eta =0$.

Suppose now that $u(D)=0$ in the Jacobian variety, so the chosen vector is $m+Bn$, with $m,n\in {\mathbb{Z}}^g$. Replace $\eta$ byIts $a$-periods are $-2\pi i{n}_j$ and its $b$-periods are $2\pi i{m}_i$. The periods around the punctures are $2\pi i{n}_p$. These loops together generate the homology of the punctured Riemann surface, so every period of ${\eta }^{\prime }$ lies in $2\pi i\mathbb{Z}$. Consequently $f(p)=\exp ({\int }^p{\eta }^{\prime })$ is a single-valued nonzero holomorphic function off the support of $D$. At a support point, ${\eta }^{\prime }=n_{p}dt/t+$ a holomorphic differential, so $f=t^{n_p}$ times a nonvanishing holomorphic function. It extends meromorphically and has divisor exactly $D$. This proves sufficiency, including negative coefficients.

Conversely, if $D=\mathrm{div}(f)$, take $\eta =df/f$. Its residues are the coefficients of $D$, while its $a$- and $b$-periods are integer multiples of $2\pi i$, by the winding numbers of $f$ along these loops. Writing them as $2\pi ia$ and $2\pi ib$, the residue bilinear identity gives $u(D)=b-Ba\in \Lambda$. Thus $u(D)=0$ in the Jacobian variety, proving necessity and henceHere the displayed necessity of degree zero is also the residue theorem applied to $df/f$.

There is a useful geometric version of the conclusion. For an effective divisor $D$, the fiber of $u_d$ through $D$ consists precisely of the effective divisors linearly equivalent to $D$. A nonzero section of ${\mathcal{O}}_X(D)$ is a meromorphic function $f$ with $\mathrm{div}(f)+D\ge 0$, and its zero divisor as a section is $\mathrm{div}(f)+D$. Two such sections give the same effective divisor exactly when their ratio is a nonzero constant. Thus the fiber is the complete linear system of a divisor $|D|=\mathbb{P}{H}^0(X,{\mathcal{O}}_X(D))$.

A finite CW complex can be defined without assuming separation properties in advance. Start with a finite discrete set $X^0$. Inductively form $X^k$ from $X^{k-1}\bigsqcup {\coprod }_{\alpha }{D}_{\alpha }^k$ by identifying each boundary point with its image under a continuous attaching map $a_{\alpha }:S^{k-1}\to X^{k-1}$. Only finitely many disks are attached in total, and $X=X^N$ for some finite $N$. The interiors of the disks are its cells. Its weak topology of a CW complex is the final topology of the characteristic maps: a subset is closed exactly when its inverse image in every characteristic disk is closed. In this finite construction it is the topology obtained by the successive quotients. Closure-finiteness is automatic because there are only finitely many cells.

Here is a direct proof of both requested topological properties, without presupposing that arbitrary attaching maps give a triangulable space. Suppose the compact previous stage $Y$ embeds by $e:Y\hookrightarrow {\mathbb{R}}^m$, and attach one $k$-cell, $k\ge 1$. Extend the boundary coordinate map radially byIt is continuous at zero because the image of the sphere is bounded. Map the two pieces into ${\mathbb{R}}^{m+k+1}$ byAt $|x|=1$ these agree with the attaching identification. They therefore induce a continuous map on the quotient. Its restriction to the old stage is injective. An interior point has positive last coordinate $t=1-|x|$, and the preceding coordinate block recovers $x$ by division by $t$. Hence it cannot coincide with an old point or another interior point. The induced map is injective.

The quotient is compact, being the image of a compact disjoint union. A continuous injection into Euclidean space is Hausdorff, by inverse images of disjoint separating neighborhoods; it is also a homeomorphism onto its image because closed subsets of the compact source have compact, hence closed, images in the Hausdorff target. Finite discrete $X^0$ embeds in $\mathbb{R}$, and induction over the finite list of cells proves $X$ is Hausdorff and its weak topology is exactly an induced Euclidean subspace topology. This is Euclidean embedding by finite cell attachment; no optimal ambient dimension is needed.

For Real projective space, filter by ${\mathbb{RP}}^0\subset \cdots \subset {\mathbb{RP}}^n$. The complement of ${\mathbb{RP}}^{k-1}$ in ${\mathbb{RP}}^k$ is an affine $k$-cell. A characteristic disk is a hemisphere of $S^k$, whose boundary is attached by the antipodal quotient $S^{k-1}\to {\mathbb{RP}}^{k-1}$. Thus there is one cell in every dimension $0,\dots ,n$.

The incidence number is the sum of the two sheet contributions. Fixing the orientation of one sheet gives degree one; the other differs by the antipodal map on $S^{k-1}$, whose degree is $(-1{)}^k$. Consequently the cellular homology of real projective space hasTaking kernels modulo images gives, for $n\ge 1$,Over $\mathbb{R}$, multiplication by two is invertible, soOver ${\mathbb{F}}_2=\mathbb{Z}/2$, every differential vanishes, giving $H_q({\mathbb{RP}}^n;{\mathbb{F}}_2)={\mathbb{F}}_2$ for $0\le q\le n$, and zero otherwise. For $n=0$ the space is a point and only degree-zero homology is present, with the chosen coefficient group.

Interpret the connectivity hypothesis in its usual based sense: $X$ is $(n-2)$-connected, with $n\ge 2$ for the nontrivial suspension range. Form the suspension as two cones $C_{-}X$ and $C_{+}X$ meeting in $X$. Both cones are contractible. Their relative homotopy sequences give ${\pi }_j(C_{\pm }X,X)\cong {\pi }_{j-1}(X)$ in the relevant degrees, so both cone pairs are $(n-1)$-connected.

The homotopy excision theorem says that if two CW pairs meeting in a connected subcomplex are $p$- and $q$-connected, $p,q\ge 1$, the map from the first relative pair to the union relative to the second part is an isomorphism in degrees $i<p+q$ and a surjection in degree $p+q$. Here $p=q=n-1$, givingas an isomorphism for $i<2n-2$ and a surjection for $i=2n-2$. Contractibility identifies the domain with ${\pi }_{i-1}(X)$ and the target with ${\pi }_i(\Sigma X)$. The comparison under these identifications is the suspension map: a sphere representative is coned in each half and the two cones give its suspended sphere representative. Hence the Freudenthal suspension from two cones provesFor connected CW complexes, reduced and unreduced suspension have the same based homotopy type, so either standard suspension convention gives this result. In the low-dimensional range, pointed sets are used where necessary; the displayed positive-degree homomorphism assertion concerns the usual groups.

For $S^r$ with $r\ge 2$, cellular approximation makes every sphere map of degree below $r$ null-homotopic: it can be homotoped into the point skeleton of the CW structure having cells only in degrees zero and $r$. The Hurewicz theorem therefore identifies its first possible nonzero homotopy group with homology. Its cellular chain complex gives $H_r(S^r;\mathbb{Z})=\mathbb{Z}$. For $r=1$, lifting a based loop to the universal cover $\mathbb{R}\to S^1$ classifies it by an integer endpoint displacement, so ${\pi }_1(S^1)=\mathbb{Z}$; the degree-one Hurewicz map is abelianization and is an isomorphism here. ThusAlternatively, after the $r=2$ Hurewicz base case, the suspension isomorphisms above propagate the generator to every higher sphere.

Work in the based CW homotopy category. A homotopy-invariant functor depends on a map only through its based homotopy class. In a sequence of pointed sets, exactness at the middle means that the preceding image equals the inverse image of the distinguished point; it does not imply injectivity or surjectivity at the ends. For groups this is the familiar image-equals-kernel condition. A half-exact functor preserves such middle exactness on the appropriate elementary sequence.

For a contravariant functor $F$, the relevant elementary sequence is a cofibration $A\hookrightarrow X\to X/A$. Half-exactness isexact at $F(X)$. One should not silently turn this into a short exact sequence. In particular, arbitrary group actions and pointed sets at low degrees are not interchangeable with abelian groups.

The basic contravariant example is $F_Y(X)=[X,Y{]}_{\ast }$, often called a cohomotopy-type functor. Suppose a map $f:X\to Y$ becomes null-homotopic on $A$. A chosen null-homotopy extends it over the cone attached to $A$, giving a map from the topological mapping cone to $Y$. Conversely every such extension restricts to a null-homotopic map on $A$. Since the topological mapping cone of a subcomplex inclusion models $X/A$, this proves half-exactness, rather than merely asserting it.

Repeating the topological mapping cone construction gives the Puppe sequenceContravariant mapping into $Y$ givesThis is exact as pointed sets. An H-group structure on $Y$, for example a loop space with path concatenation, provides group structures; repeated looping gives abelian group structures in the usual ranges. The suspension-loop adjunction $[\Sigma X,Y{]}_{\ast }\cong [X,\Omega Y{]}_{\ast }$ explains the shifted terms and their compatibility.

The covariant counterpart is $H_T(Y)=[T,Y{]}_{\ast }$, with ordinary homotopy groups obtained from $T=S^n$. Its natural elementary sequence is a homotopy fiber sequence $F\to E\to B$, not an arbitrary cofibration. A map $a:T\to E$ whose composite to $B$ is null-homotopic, together with that null-homotopy, determines a map into the homotopy fibre. Conversely a fibre map projects to such a null composite. This proves exactness of $[T,F{]}_{\ast }\to [T,E{]}_{\ast }\to [T,B{]}_{\ast }$ at the middle. Adding the loop terms produces the homotopy long exact sequence. For a Serre fibration, its ordinary fibre supplies the same homotopy information.

This variance distinction matters. Include one circle, generated by $a$, in $S^1\vee S^1$, with other generator $b$. The quotient is the second circle. The kernel of ${\pi }_1(S^1\vee S^1)\to {\pi }_1(S^1)$ is the normal closure of $a$, which contains $ba{b}^{-1}$. The image of the included circle is only the cyclic subgroup generated by $a$, and does not contain that reduced word. Thus covariant ${\pi }_1$ is not half-exact on this cofibration. Fibre and cofibre statements cannot be swapped just by reversing notation.

Reduced cohomology gives group-valued contravariant examples. For an abelian coefficient group $G$, the degree-$q$ functor is represented by an Eilenberg–MacLane space:Its cofibration exact sequence continues as $\cdots \to {\tilde{H}}^{q-1}(A)\to {\tilde{H}}^q(X/A)\to {\tilde{H}}^q(X)\to {\tilde{H}}^q(A)\to \cdots$. Ordinary homotopy groups are instead covariant, with decreasing group degree across a fibre connecting map. These are the two complementary representable patterns, maps out of a test object and maps into a representing object.

Brown's representability theorem makes the contravariant pattern structural. A reduced contravariant homotopy functor on based CW complexes is represented by a CW complex if it converts arbitrary wedges to products and satisfies the Mayer-Vietoris weak-pullback gluing axiom. The latter says that compatible classes on two subcomplexes extend to their union. It contains more information than the single kernel statement of half-exactness. Consequently half-exactness alone is not a representability criterion. In a representing functor these axioms are natural: a map on a wedge is a family of based maps, while compatible maps on an intersection can be adjusted by a homotopy and glued across a homotopy pushout.

A reduced generalized cohomology theory with the usual wedge axiom supplies these axioms degree by degree. Its representing spaces $E_q$ have $E_q\simeq \Omega E_{q+1}$ by suspension compatibility, assembling an Omega-spectrum. Thus generalized cohomology can be expressed as $E^q(X)=[{\Sigma }^{\infty }X,{\Sigma }^{q}E{]}_{\mathrm{st}}$, while represented homology is $E_q(X)={\pi }_q(E\wedge {\Sigma }^{\infty }X)$. The appearance of mapping into a spectrum in one case and smashing with it in the other is another form of the same variance distinction.

There is a literal duality after stabilization and a finiteness restriction. Here a finite based CW complex is identified with its suspension spectrum. For a finite spectrum $X$, let $DX=F(X,\mathbb{S})$ be its Spanier-Whitehead dual. Then $D^{2}X\simeq X$ and $[X,Y{]}_{\mathrm{st}}\cong [DY,DX{]}_{\mathrm{st}}$ for finite spectra. Duality reverses a cofibre triangle into a fibre triangle. In particular,This turns a cohomology calculation on a finite object into a homology calculation on its dual, and vice versa. It is Spanier-Whitehead duality, not a claim that the full category of ordinary spaces is self-dual.

Half-exactness is the local image–kernel property; fibres and cofibres, representability, and finite stable duality explain how that property organizes homotopy and cohomology. The distinctions between variance, pointed sets and groups, and finite stable objects and arbitrary spaces are essential to using the duality correctly.

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.

The Wirtinger presentation assigns a meridian generator to each arc between undercrossings. At a crossing the outgoing underarc meridian is the conjugate of the incoming one by the overarc meridian, with the exponent determined by the crossing sign. One crossing relation is redundant for a connected knot diagram. This gives a presentation of the knot group, the fundamental group of its exterior.

For the trefoil knot, choose three arcs x,y,z so that two crossing relations are $z=xy{x}^{-1}$ and $x=yz{y}^{-1}$. The third follows from these. Substituting for z gives $xyx=yxy$, henceWe use the relator $r=xyx{y}^{-1}{x}^{-1}{y}^{-1}$.

Here is the Fox free differential calculus, including its connection with the covering-space chain complex. Let F be the free group on $x_1,\dots ,x_n$, let $\mathbb{Z}F$ be its group ring, and let $\epsilon :\mathbb{Z}F\to \mathbb{Z}$ be augmentation. Define additive maps $D_j:\mathbb{Z}F\to \mathbb{Z}F$ byFor group words u,v the product rule reads $D_j(uv)=D_j(u)+u{D}_j(v)$. Differentiating $x_i{x}_i^{-1}=1$ forces $D_j(x_i^{-1})=-{\delta }_{ij}{x}_i^{-1}$. Thus for a reduced word one sums its prefixes at positive occurrences of $x_j$, and subtracts the prefix including $x_j^{-1}$ at negative occurrences. Inserting or deleting $x_i{x}_i^{-1}$ makes two cancelling contributions, proving that this definition is independent of word reduction. Linear extension proves the group-ring product rule and uniqueness.

The fundamental identity isIt holds on generators and inverses. If it holds for u and v, then $uv-1=(u-1)+u(v-1)$ proves it for their product, so induction proves it for every word. These Fox derivatives form the matrix $(D_j{r}_i{)}_{i,j}$ for any chosen presentation.

Let $\varphi :G\to \mathbb{Z}$ be knot-group abelianization and put $a_j=\varphi (x_j)$. Apply the ring map $\alpha :\mathbb{Z}F\to \Lambda$ sending $x_j$ to $t^{a_j}$, and set $J_{ij}=\alpha (D_j{r}_i)$. Build the presentation two-complex P with one vertex, n oriented edges, and m two-cells attached along the relators. Its infinite cyclic cover has cellular chain complexTo verify the second boundary, lift an attaching word starting at deck level zero. An occurrence $x_j$ after prefix u traverses the jth edge at level $\varphi (u)$, contributing $t^{\varphi (u)}{e}_j$. An occurrence $x_j^{-1}$ traverses it backwards starting at level $\varphi (u{x}_j^{-1})$, contributing $-t^{\varphi (u{x}_j^{-1})}{e}_j$. Their sum is exactly the evaluated Fox derivative. The fundamental identity, with $\alpha (r_i)=1$, gives $d_1{J}^T=0$.

The cover of P and the cover of the knot exterior both have fundamental group $\ker \varphi$. Their first homology is the abelianization of that group, with the same deck action induced by conjugation. Thus no assumption that the presentation complex is aspherical is needed, and the complete relation with the Alexander module of a knot isFor meridian generators all $a_j=1$. Then $\ker d_1$ consists of vectors with coefficient sum zero, with basis $e_j-e_n$, $j<n$. Each Fox row has coefficient sum zero, so deleting its nth entry gives its coordinates in this basis. Deleting one column of J therefore presents the Alexander module; if a Wirtinger relator is redundant it can also be deleted. This proves the matrix procedure used for the displayed knot above. For arbitrary generators, the exponent vector is primitive because the generators map onto $\mathbb{Z}$; elementary changes of free generators implement the Euclidean algorithm and reduce it to $(1,0,\dots ,0)$. In that basis $\ker d_1$ is free on the last $n-1$ edges, and their relator coefficients give a presentation. This explains the general case rather than treating the full Fox cokernel as the Alexander module.

For the trefoil relator, with $\alpha (x)=\alpha (y)=t$, direct differentiation givesThe first boundary is $(t-1,t-1)$, whose kernel is generated by $(1,-1)$. The relator boundary is $f(t)(1,-1)$, whenceFor a connected sum of knots, join Seifert surfaces in separated balls. Their Seifert matrix is block diagonal, so the Alexander module of a connected sum is a direct sum. Thus ${\mathcal{A}}_{T\#T}\cong (\Lambda /(f){)}^2$.

This module is not cyclic. The polynomial f is irreducible modulo two, since it has no root in ${\mathbb{F}}_2$, so $\mathfrak{m}=(2,f)$ is a maximal ideal of $\Lambda$ with residue field ${\mathbb{F}}_4$. The quotient ${\mathcal{A}}_{T\#T}/\mathfrak{m}{\mathcal{A}}_{T\#T}$ has dimension two over that field, whereas the corresponding quotient of a cyclic module has dimension at most one.

Every two-generator knot group, however, has a cyclic Alexander module, regardless of its number of relators. Indeed its primitive abelianization vector can be changed to $(1,0)$ as above. The covering boundary is then $(t-1,0)$; its kernel is the free rank-one module on the second edge, and any quotient by lifted relators is cyclic. This proves the two-generator knot groups have cyclic Alexander modules obstruction. Consequently the group of the connected sum of two trefoils cannot even be generated by two elements, and in particular it has no two-generator, one-relator presentation.

The standard CW complex structure on Complex projective space ${\mathbb{CP}}^3$ has one cell in each dimension $0,2,4,6$ and no other cells. Its cellular chain complex therefore has zero differentials: each boundary has an odd-dimensional target group, which is zero. The product of the usual two-cell structures on $S^2$ and $S^4$ likewise has exactly one cell in dimensions $0,2,4,6$. Thus cellular homology gives, for both spaces,The cohomology rings detect more than these additive groups. Let $c\in H^2({\mathbb{CP}}^3;\mathbb{Z})$ be the hyperplane class. The cohomology ring of complex projective space isFor the normalization, a hyperplane is Poincare dual to $c$; intersecting three generic hyperplanes gives one point with positive complex orientation, so $⟨c^3,[{\mathbb{CP}}^3]⟩=1$. Together with the computed additive groups this makes $c,c^2,c^3$ generators in their respective degrees.