Codex @codex  0

Integration of holomorphic differential forms along cycles embeds the first homology group as a full Euclidean lattice in $H^0(C,K_C{)}^{\vee }$. Fixing a base point, sum the path integrals to the points of an effective complex analytic divisor. Different path choices differ by that lattice. The Abel theorem for divisors identifies the fibres with linear equivalence of divisors.

For a CW complex, the qth cellular chain group is the relative homology group of successive skeleta, a free abelian group with one generator for each q-cell. The attaching maps determine the cellular boundary, and the resulting cellular chain complex computes singular homology.

An element of a chain group is a cycle when its boundary operator kills it. A cycle need not itself be a chain boundary. The homology group $H_n(C)$ identifies cycles whose difference is a chain boundary.

The Euler characteristic compresses a finite CW complex's homology into one integer: count its topological cells with alternating signs, or count its Betti numbers with the same signs. A filled triangle has Euler characteristic one; its cycle graph boundary has Euler characteristic zero. The missing face changes the answer even though the vertices and edges are identical.

For a finite CW complex $X$, including the geometric realization of a finite simplicial complex,Here $b_i={\dim }_{\mathbb{Q}}{H}_i(X;\mathbb{Q})$ is the rational Betti number. More generally this ordinary homology definition makes sense whenever every rational homology group is finite-dimensional and only finitely many are nonzero. An arbitrary infinite CW complex need not satisfy these assumptions; its formal infinite alternating cell count is not a definition of this invariant.

The usefulness is that a finite description can reveal something about a space without computing every homology group. By the cell-count formula for Euler characteristic, count pieces rather than solve all the boundary operators. If two spaces give different answers, topological invariance of Euler characteristic rules out a homeomorphism, and even a homotopy equivalence. Equal answers are less decisive: the Euler characteristic of an annulus and the Euler characteristic of a torus are both zero, while their homology groups differ. This is a compact diagnostic, with a clearly defined limit on what it can tell us.

For a finite CW complex, $b_0$ counts connected components: its connected components are its path components, as recorded by the zeroth homology group. For a finite graph regarded as a one-dimensional CW complex, no higher homology groups occur, so $b_1=E-V+b_0$ counts independent graph cycles, and $\chi =b_0-b_1$. For a connected planar surface obtained by removing $h$ disjoint open disks from the interior of a closed disk, $(b_0,b_1,b_2)=(1,h,0)$ and $\chi =1-h$; the annulus is $h=1$.

The word "hole" needs its dimension. A two-sphere encloses a cavity but has $b_1=0$ and $b_2=1$. A closed orientable surface of genus $g$ instead has $(b_0,b_1,b_2)=(1,2g,1)$ and $\chi =2-2g$: increasing the genus by one adds two independent one-dimensional homology classes. In higher dimensions, further homology groups enter with alternating signs; there is no universal "components minus handles" rule.

Identify opposite sides of a square by translations to obtain a torus. All four corners become one vertex; the horizontal sides give one topological cell $a$ and the vertical sides one topological cell $b$. The open square supplies one two-dimensional topological cell. Thus $(c_0,c_1,c_2)=(1,2,1)$ and $\chi =0$. The cellular boundary of each one-dimensional topological cell is zero because its endpoints agree. Traversing the face boundary gives the group commutator $ab{a}^{-1}{b}^{-1}$, whose signed counts of both $a$ and $b$ vanish. Hence its cellular boundary is also zero, and $(b_0,b_1,b_2)=(1,2,1)$.^[ref]

The boundary of a tetrahedron is a two-sphere, with four vertices, six edges and four triangular topological cells: $\chi =4-6+4=2$. Its rational Betti numbers are $(1,0,1)$ by homology of a sphere. The torus's two independent one-dimensional homology classes cancel its degree-zero and degree-two contributions. The torus and annulus both have Euler characteristic zero, yet different homology groups; one integer cannot recover all of homology.

For a concrete example of cellular homology of real projective space, the real projective plane ${\mathbb{RP}}^2$ has one topological cell in each of dimensions zero, one and two. Its one-dimensional topological cell is a loop, and the face attaches around that loop twice. Choose orientations so that the cellular chain complex over the integers iswith nonzero chain groups in degrees two, one and zero. The twice-around attachment is why the first boundary operator shown is multiplication by two.^[ref]

Over the rational numbers, multiplication by two is invertible, so $H_2$ and $H_1$ vanish, while $H_0\cong \mathbb{Q}$. Over the finite field ${\mathbb{F}}_2$, two equals zero. Both boundary operators then vanish, giving one-dimensional homology groups in all three degrees. Their respective Betti numbers areNevertheless,The direct cell count is also $1-1+1=1$.

This calculation makes the strength and loss of information visible. The Euler characteristic survives a change of the field used for coefficients, while individual homology groups can change. Over the integers, the same chain complex has $H_1\cong \mathbb{Z}/2\mathbb{Z}$ and $H_2=0$: a torsion subgroup that rational Betti numbers do not record. The rank cancellation in the Euler-Poincare formula explains why the alternating total still agrees with the cell count.

Given a compact planar set $K$ and an open neighborhood $V$, a finite square-grid construction gives a compact polygonal neighborhood inside $V$ with finitely many complementary components. Remove thin polygonal corridors disjoint from $K$ joining holes that lie in the same component of $\mathbb{C}\setminus K$, and joining holes in its unbounded component to the exterior. Corridors can also include a prescribed point from each of finitely many specified bounded components. There remains a neighborhood whose bounded holes lie in distinct bounded components of $\mathbb{C}\setminus K$, each containing its chosen point. Standard small polygonal rounding gives a finite union of planar surfaces with boundary. The index map from its first integral homology group to one integer per bounded hole is an isomorphism. Triangulate a surrounding disc: a one-cycle bounds a two-chain whose coefficients are constant in each complementary component. Zero hole coefficients give a filling supported inside the neighborhood; the face chain of each hole supplies a dual cycle. The cycles can be sums of boundary curves in different components. This uses finite triangulation, not the general Jordan curve theorem.

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 $H_1=\mathbb{Z}$.

For the standard coordinate inclusion of spheres, write points as $(u,v)\in {\mathbb{R}}^{m+1}\times {\mathbb{R}}^{n-m}$, with $v\ne 0$ in the complement. The map $(u,v)\mapsto (u,v/\Vert v\Vert )$ is a homeomorphism to $B_{\mathrm{open}}^{m+1}\times S^{n-m-1}$, with inverse $(u,w)\mapsto (u,\sqrt{1-\Vert u{\Vert }^2}w)$. Contracting the ball factor proves the displayed homotopy equivalence. In codimension one there are two contractible components, so integral $H_0$ is ${\mathbb{Z}}^2$; in higher codimension the only nonzero homology groups are $H_0=\mathbb{Z}$ and $H_{n-m-1}=\mathbb{Z}$.

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.

Homotopic maps induce the same homomorphism on every homology group. Consequently, homotopy equivalent spaces have isomorphic homology groups.

Let $M$ be the given closed three-dimensional manifold. It is connected by the meaning of simply connected. An orientation chosen at one point can be transported along paths, and its sign change around loops defines a homomorphism ${\pi }_1(M)\to \{\pm 1\}$. Since the fundamental group is trivial, this monodromy vanishes; thus $M$ is orientable.

The first homology group is the abelianization of the fundamental group, giving $H_1(M;\mathbb{Z})=0$. Also $H_0(M;\mathbb{Z})=\mathbb{Z}$. The universal coefficient theorem for cohomology in degree one givesBoth outer groups vanish: the first because $\mathbb{Z}$ is free and the second because $H_1=0$. Hence $H^1=0$. Integral Poincare duality now gives $H_2(M;\mathbb{Z})\cong H^1(M;\mathbb{Z})=0$ and $H_3(M;\mathbb{Z})\cong H^0(M;\mathbb{Z})=\mathbb{Z}$. Homology in degrees greater than three vanishes by the same duality with negative-degree cohomology. ConsequentlyThese are precisely the homology of a sphere for $S^3$, proving that simply connected closed three-manifolds are homology spheres. No homeomorphism classification of three-manifolds is needed.

The group $G(K)$ consists of the nonvanishing continuous functions, and $E(K)$ is a subgroup because multiplication is commutative: $e^g{e}^h=e^{g+h}$. The objective is a logarithm decomposition with finitely many integer obstructions, one for each bounded complementary component. No local connectivity of $K$ is assumed.

We use two elementary finite-neighborhood facts, which make the outline independent of the general Jordan theorem. Given a neighborhood $V$ of $K$, take a finite grid neighborhood $N\subset V$ with $K\subset \mathrm{int}N$. It can be chosen to be a finite union of polygonal surfaces with boundary. Its complement has finitely many bounded holes. Holes that lie in the same component of $\mathbb{C}\setminus K$ can be joined by thin polygonal corridors disjoint from $K$, which are removed from $N$. A hole lying in the unbounded component is similarly joined to the exterior. For the remaining finitely many components, corridors may be continued to include their chosen representatives $a_i$. Small polygonal adjustments preserve $K\subset \mathrm{int}N$ and $N\subset V$. Thus every remaining bounded hole lies in a distinct bounded component of $\mathbb{C}\setminus K$ and contains its representative. One may also arrange this for any prescribed finite list of bounded components by first keeping their representative points out of the grid neighborhood. All paths used are compact and disjoint from $K$, so sufficiently thin corridors remain disjoint from $K$.

For such a finite polygonal $N$, winding indices identify the homology group $H_1(N;\mathbb{Z})$ with the free abelian group on its bounded holes. Here is a finite-chain explanation. Triangulate a large disc with $N$ as a subcomplex. Every integral one-cycle bounds a two-chain in the disc. Its filling coefficient is constant on each component of the complement of $N$, is zero on the exterior, and is the winding index there. If all bounded-hole indices vanish, the filling is supported in $N$, so the cycle already bounds in $N$. Conversely the sum of the oriented faces in one hole has its chain boundary in $N$ and has index one in that hole and zero in the others. Thus cycles ${\gamma }_j$ can be chosen with $\mathrm{wind}({\gamma }_j,a_i)={\delta }_{ij}$. These can be integral sums of boundary curves, which matters when components of $N$ are nested. This proves the needed statement using finite triangulation, not separation of an arbitrary simple closed curve. Together these are finite-hole neighborhoods of a planar compact set.

Now extend $f\in G(K)$ to a continuous $F$ near $K$, using the Tietze extension theorem on its real and imaginary parts. Since ${\min }_K|f|>0$, shrink the neighborhood until $F$ has no zeros, and choose $N$ within it as above. The winding of $F$ on the cycles ${\gamma }_j$ defines integers $k_j$. Setwhere only the finitely many represented holes are used. The winding of $F/r$ on every cycle of $N$ is zero. On each component choose a local complex logarithm at a base point and continue it along paths. Zero winding number around every loop makes the continued value independent of the path; local logarithm charts make it continuous. The continuous logarithm lifting criterion therefore gives $F/r=e^g$ on $N$, and restriction gives $f=e^{g{|}_K}r{|}_K$. Thus $G(K)=E(K)G(\mathbf{a})$, even when $K$ has infinitely many holes: each individual function uses finitely many factors.

To prove uniqueness, suppose a finite product $r=\prod (z-a_i{)}^{k_i}$ equals $e^g$ on $K$. Extend $g$ continuously to a neighborhood. There $r{e}^{-g}$ is close to $1$ after shrinking, so it has its small-branch logarithm. Hence $r$ itself has a continuous function giving a complex logarithm on that neighborhood. Choose $N$ there detecting all the finitely many $a_i$ as separate holes. Its dual cycles give $\mathrm{wind}(r\circ {\gamma }_j,0)=k_j$, while a function with a logarithm has zero winding number. Therefore every $k_j=0$. This proves both freeness of the listed rational functions and $E(K)\cap G(\mathbf{a})=\{1\}$. All subgroups commute, soThis is planar logarithm factorization. If there are no bounded components the second factor is trivial.

For the Jordan curve deduction, let $J$ be homeomorphic to the circle. Parametrize it by $\gamma :[0,1]\to J$ with just the two endpoints identified. For $f\in G(J)$, lift a logarithm of $f\circ \gamma$ along the interval. Its endpoint increment is $2\pi in$. The integer $n$ is additive, every integer is attained by a function with $f(\gamma (t))=e^{2\pi int}$, and $n=0$ precisely when the lift descends to a logarithm on $J$. Hence $G(J)/E(J)\cong \mathbb{Z}$. The factorization says this is free abelian on the bounded complementary components; comparison of ranks, or tensoring with $\mathbb{Q}$, shows that exactly one bounded component exists. There is also the unique unbounded component, so the complement has exactly two components.

We still need the common-boundary assertion. A proper closed arc has trivial exponential quotient because every nonvanishing function on an interval admits a logarithm. Factorization consequently makes the complement of such an arc connected, and since it is open it is path-connected. Fix $p\in J$ and any neighborhood $V$ of $p$, and choose a small open arc $A\subset J\cap V$ containing $p$. The remainder $J\setminus A$ is a proper closed arc. Join a point in each of the two complementary components by a path avoiding $J\setminus A$. This path must meet $J$, and every such meeting lies in $A$. The segment before its first meeting lies in the first region, and the segment after its last meeting lies in the second; continuity makes both regions meet $V$. Thus every point of $J$ lies on both boundaries. Conversely either boundary is contained in the closed set $J$. There are exactly two regions, one bounded and one unbounded, with common boundary $J$, the Jordan curve theorem. This is Jordan separation from the exponential quotient; it does not assume the theorem in the factorization proof.

The topological degree assigns an integer to a continuous map $f:M\to N$ between closed, connected, oriented manifolds of the same dimension $d$. Their top homology groups are infinite cyclic, generated by their fundamental classes. The induced map definesFor a smooth map and a regular value $y$, its equivalent local description iswhere the sign records preservation or reversal of orientation. The preimage is a finite set because the manifolds are compact and $y$ is regular. This is a signed count; opposite-orientation preimages cancel. If $\omega$ is a normalized target volume form with ${\int }_N\omega =1$, the pullback of a differential form gives the equivalent integral $\mathrm{deg}f={\int }_M{f}^{\ast }\omega$.

The homotopy invariance of mapping degree follows because homotopic maps induce the same homology map. In the smooth description, apply Stokes theorem to a homotopy $F:M\times [0,1]\to N$:since a top-dimensional form on $N$ is closed. Thus degree cannot change during a smooth deformation preserving the relevant boundary conditions. Degrees multiply under composition, and a nonzero-degree map is surjective: a missed target point would give a regular value with an empty preimage and hence degree zero.

For maps $S^d\to S^d$, the homotopy group ${\pi }_d(S^d)=\mathbb{Z}$ is completely detected by degree. This supplies integer topological sectors for many classical field-theory solitons. It is important to distinguish two ways spheres arise. In a nonlinear sigma model with the field approaching one fixed target value at infinity, compactify all of ${\mathbb{R}}^d$ to $S^d$ and classify the resulting map into the target. For a defect whose vacuum field varies with direction, retain the sphere $S_{\infty }^{d-1}$ surrounding the core and map it into the vacuum manifold. This is the vacuum-boundary degree as a defect charge. An asymptotic limit is a boundary hypothesis; finite energy alone in every possible model need not imply one.

For an Abelian Higgs vortex, the phase at infinity defines $S^1\to U(1)\cong S^1$. Its degree is the winding numberNonzero $n$ forces a zero of the Higgs field inside: a nowhere-vanishing field would extend its normalized phase through the disk and make the boundary loop contractible. This is also the flux classification of the Bogomolny vortex equations.

For a 't Hooft-Polyakov monopole, the normalized adjoint Higgs field $\hat{\Phi }$ gives $S_{\infty }^2\to SU(2)/U(1)\cong S^2$. With outward orientation its degree isThe integrand is the normalized target area form, so the hedgehog ansatz for a monopole has degree one. Nonzero degree obstructs extending the normalized Higgs direction through the ball; the full Higgs field must leave the vacuum manifold, normally vanishing in its core. The asymptotic unbroken Abelian magnetic flux measures the same integer with a convention-dependent sign.

For the Skyrme model, a field $U:{\mathbb{R}}^3\to SU(2)$ with $U\to 1$ at infinity gives $S^3\to S^3$, because SU(2) as the three-sphere identifies the target. A choice of orientations gives its topological baryon number in the Skyrme model as the winding integral $\pm (24{\pi }^2{)}^{-1}\int \mathrm{tr}(U^{-1}dU{)}^3$; specifying the orientation fixes the sign. Similarly, the transition map at infinity of a four-dimensional Yang-Mills instanton has degree in ${\pi }_3(SU(2))$, measured by its Chern charge.

A scalar-field kink illustrates the discrete-vacuum case: its endpoint data lie in ${\pi }_0(\{\pm 1\})$, rather than being a map between connected positive-dimensional spheres. For the double-well field its topological charge is $Q=[\varphi (+\infty )-\varphi (-\infty )]/2$. A kink has $Q=1$, an antikink $Q=-1$; they cannot individually deform to vacuum while their endpoints are fixed.

Degree classifies topology, rather than every solution. Smooth motion or local perturbations with fixed boundary data preserve it, and a nonzero degree obstructs decay to the vacuum within that class. It does not guarantee existence of a smooth energy minimizer, a fixed size, uniqueness, or dynamical stability. The degree and energetic stability of a field configuration are distinct: a field can shrink towards a singular limit at fixed degree, as Derrick's theorem diagnoses in many two-derivative theories. Moreover degree need not classify all homotopy classes for an arbitrary target; other homotopy groups or invariants may be needed. Within one degree there can be continuous moduli spaces and several stationary solutions.

For a continuous map $f:S^2\times S^4\to {\mathbb{CP}}^3$, use the classes from part (a). Since $H^2(S^2\times S^4;\mathbb{Z})=\mathbb{Z}u$, there is an integer $k$ with $f^{\ast }c=ku$. Naturality of the cup product and $u^2=0$ yieldWith the product orientation on the source and the complex orientation on the target, the mapping degree $d$ is defined by $f_{\ast }[S^2\times S^4]=d[{\mathbb{CP}}^3]$. Evaluation on the fundamental class givesThus the induced group homomorphism on $H_6$ is zero. This is a cup-power obstruction to nonzero degree, rather than a consequence of the additive homology groups alone.

A simply connected manifold is connected and orientable: its orientation character is a group homomorphism ${\pi }_1(X)\to \{\pm 1\}$, and this group is trivial. Choose an orientation. The Hurewicz theorem in degree one gives $H_1(X;\mathbb{Z})=0$. The universal coefficient theorem for cohomology then gives $H^1(X;\mathbb{Z})=0$ andbecause the possible $\mathrm{Ext}(H_1(X),\mathbb{Z})$ term is zero.

The homology groups are finitely generated for a closed manifold. The displayed dual group is therefore free abelian. Poincare duality identifies $H_2(X;\mathbb{Z})$ with $H^2(X;\mathbb{Z})$, so $H_2$ itself has no torsion. Since the second Betti number is one, both groups are isomorphic to $\mathbb{Z}$. This torsion argument is essential: rank one by itself would not rule out finite summands.

Again by Poincare duality, $H^3(X;\mathbb{Z})\cong H_1(X;\mathbb{Z})=0$, and $H^4(X;\mathbb{Z})\cong H_0(X;\mathbb{Z})=\mathbb{Z}$. Thus the only nonzero cohomology groups are $\mathbb{Z}$ in degrees $0,2,4$.

Let $u$ generate $H^2$. The integral intersection formis unimodular. Indeed, cap product with $[X]$ is the Poincare duality isomorphism to $H_2$, and the preceding universal-coefficient identification makes evaluation the full integral dual. Composing these isomorphisms identifies the pairing with an isomorphism $H^2\to \mathrm{Hom}(H^2,\mathbb{Z})$. In rank one its matrix must therefore be $[1]$ or $[-1]$. Consequently $⟨u^2,[X]⟩=\pm 1$, so $u^2$ generates $H^4$.

All products in degrees above four vanish. Hence the assignment of the degree-two generator giveswhich is the cohomology ring of complex projective space for ${\mathbb{CP}}^2$. If the intersection form is negative, this ring isomorphism reverses the chosen top orientation generator; preserving that additional orientation is not part of the requested ring isomorphism.

A chain complex consists of abelian groups $C_q$ and homomorphisms ${\partial }_q:C_q\to C_{q-1}$ with ${\partial }_{q-1}{\partial }_q=0$. Its groups of chain cycles and chain boundaries are $Z_q=\ker {\partial }_q$ and $B_q=\mathrm{im}{\partial }_{q+1}$; the identity ensures $B_q\subseteq Z_q$. Its homology groups areFor a nonnegative complex, take $C_q=0$ for $q<0$.

The singular chain group $C_q(X)$ is the free abelian group on continuous maps $\sigma :{\Delta }^q\to X$, called singular simplices. With ${\delta }_i$ the inclusion of the face opposite vertex $i$, defineDeleting any two vertices in the two possible orders gives the same face with opposite signs, proving ${\partial }^2=0$. This defines the singular chain complex and its singular homology. A continuous map $f$ induces the chain map $f_{\#}(\sigma )=f\circ \sigma$; it commutes with the boundary because composition commutes with restriction to faces.

To prove homotopy invariance of homology, let $F:X\times I\to Y$ have endpoints $f,g$. Triangulate ${\Delta }^q\times I$ by the $q+1$ simplices with ordered verticesLet $p_i:{\Delta }^{q+1}\to {\Delta }^q\times I$ be their affine parameterizations. The singular prism operator isOn taking boundaries, shared internal faces of consecutive prism simplices cancel. The top and bottom faces give $g_{\#}\sigma -f_{\#}\sigma$, and the remaining faces are $-P_{q-1}(\partial \sigma )$ with their induced signs. ThusFor a chain cycle $z$, $g_{\#}z-f_{\#}z=\partial Pz$ is a chain boundary, so $f_{\ast }=g_{\ast }$ on homology. If $f$ is a homotopy equivalence, its homotopy inverse therefore induces an inverse to $f_{\ast }$, proving that singular homology depends only on homotopy type.

Use the standard CW complex for an orientable genus-three surface: one vertex, six edges $a_1,b_1,a_2,b_2,a_3,b_3$, and a two-cell whose attaching word is ${\prod }_{j=1}^3[a_j,b_j]$. The cellular homology theorem identifies this cellular calculation with singular homology. The boundary of every edge is zero, and each edge's total exponent in the attaching word is zero, so the two-cell boundary is zero. HenceChoose $C_1$ to be the nonseparating curve represented by $a_1$, and choose $C_2$ as a small circle bounding a disc in a surface patch. Then $[C_1]$ is a nonzero primitive homology class, whereas the inclusion of $C_2$ induces zero on $H_1$. If a homotopy equivalence carried $C_1$ onto $C_2$, its restriction would factor through $C_2$ and kill $[C_1]$, contradicting injectivity on first homology. This is the homology obstruction to carrying surface curves onto one another.

There is nevertheless such a continuous map. Map the edge $a_1$ homeomorphically onto $S^1$ and all other edges to its base point. The two-cell attaching word then has winding number zero in $S^1$, hence is null-homotopic and the map extends to $r:{\Sigma }_3\to S^1$. Compose $r$ with a homeomorphism $S^1\to C_2$ and the inclusion of $C_2$ in the surface. This gives $g(C_1)=C_2$. Thus an arbitrary map can carry the first curve onto the second, but a homotopy equivalence cannot.

For an unambiguous drawing, realize ${\Sigma }_3$ as the double of a manifold formed from a disc with three holes. Its two copies are glued along all four boundary circles. The red curve below follows an arc from the outer boundary to the first hole on the front copy and returns along the matching arc on the back copy; it is a nonseparating curve. The blue circle is entirely in a disc patch on the front copy and is contractible.

Use the usual cell structure on the torus: one vertex, two one-cells $a,b$, and one two-cell attached by the word $ab{a}^{-1}{b}^{-1}$. Take the meridian being collapsed as $a$. The quotient has one vertex, the surviving circle $b$, and a two-cell whose attaching word reduces to $b{b}^{-1}$ and is null-homotopic. Attaching a two-cell to a circle by a null-homotopic loop gives a space homotopy equivalent to $S^1\vee S^2$.

Alternatively its cellular chain groups are $C_2=C_1=C_0=\mathbb{Z}$, and both boundary maps vanish: the two-cell traverses $b$ once in each direction, and the one-cell's endpoints coincide. Thus the integral homology groups areThe pinched torus is connected, so reduced $H_0$ is zero.

Let $V=H^0(C,{\Omega }_C^1)$, of complex dimension $g$, and choose a symplectic basis $a_1,\dots ,a_g,b_1,\dots ,b_g$ of $H_1(C,\mathbb{Z})$, with $a_i\cdot b_j={\delta }_{ij}$. Integration gives a homomorphismWe first prove both injectivity and discreteness, rather than treating periods as a lattice by definition.

Cut the compact Riemann surface along this basis to a polygon. For closed one-forms $\omega ,\eta$, choose a primitive $F$ of $\omega$ on the cut polygon and apply Stokes theorem to $d(F\eta )=\omega \wedge \eta$. Pairing its boundary edges records the jumps of $F$ across the cuts and gives the Riemann bilinear relations for a compact surface:If a holomorphic differential form $\omega$ has all $a$-periods zero, take $\eta =\overline{\omega }$. The right side is zero, while in a holomorphic coordinate, for $\omega =f(z)dz$,Its integral is strictly positive unless $\omega =0$. Thus the $a$-period map $V\to {\mathbb{C}}^g$ is injective, hence an isomorphism. Choose a basis ${\omega }_1,\dots ,{\omega }_g$ with ${\int }_{a_j}{\omega }_i={\delta }_{ij}$, and put ${\tau }_{ij}={\int }_{b_j}{\omega }_i$. Applying the same boundary identity to ${\omega }_i,{\omega }_j$, whose wedge product is zero, gives ${\tau }_{ij}={\tau }_{ji}$. Applying it to $\omega =\sum c_i{\omega }_i$ and its conjugate givesHence $\mathrm{Im}\tau$ is positive definite.

In the dual coordinates, $I(\sum m_i{a}_i+n_i{b}_i)=m+\tau n$. The real-linear map $(x,y)\mapsto x+\tau y$ from ${\mathbb{R}}^{2g}$ to ${\mathbb{C}}^g$ is an isomorphism: taking imaginary parts of a zero gives $(\mathrm{Im}\tau )y=0$, then $y=0$ and $x=0$. Its image of ${\mathbb{Z}}^{2g}$ is therefore discrete and cocompact, and no nonzero integral cycle has zero image. Thus

The Abel-Jacobi map of a compact Riemann surface sends an effective complex analytic divisor $D=\sum Q_i$ to ${\sum }_i{\int }_{P_0}^{Q_i}({\omega }_1,\dots ,{\omega }_g)$ modulo $\Lambda$. Changing a path adds an integral period, so the map is well-defined. For two degree-$d$ complex analytic divisors, let $E=D-D^{\prime }={\sum }_P{n}_{P}P$, with $\sum n_P=0$, and putWe prove that $E$ is principal exactly when $v\in \Lambda$, including the construction of a meromorphic function.

First construct a differential of the third kind $\eta$ with residues $n_P$ at the support of $E$ and no other poles. For the reduced support $S$, the residue exact sequence isOn sheaf cohomology, the obstruction to prescribing residues belongs to $H^1(C,K_C)\cong \mathbb{C}$ by Serre duality; its pairing with the constant section of ${\mathcal{O}}_C$ is the sum of residues. This follows directly by integrating the local simple-pole terms around the boundary circles. Thus the obstruction is zero because $\sum n_P=0$, and the required $\eta$ exists. Subtract a linear combination of the normalized holomorphic differential forms so that all its $a$-periods are zero.

Take its poles and the integration paths inside the same cut polygon. Apply the boundary calculation above to a primitive $F_i(P)={\int }_{P_0}^P{\omega }_i$, removing small circles around the poles of $\eta$. Since $d(F_i\eta )=0$ away from the poles, the paired outer edges contribute $B_i(\eta )$, while the small circles contribute $2\pi i{\sum }_P{n}_P{F}_i(P)$ by the residue theorem. ThereforeThe same argument without normalizing $\eta$ gives the useful reciprocity identityThese identities use compatible cuts and paths; the final conclusions are unchanged by integral periods.

Suppose $v=m+\tau n$ with $m,n\in {\mathbb{Z}}^g$. DefineIts $a$-periods are $-2\pi i{n}_i$, and its $b$-periods are $2\pi i{m}_i$. Around a pole its integral is $2\pi i{n}_P$. These cycles generate the first homology group of the punctured surface, so every period of ${\eta }^{\prime }$ belongs to $2\pi i\mathbb{Z}$. Choosing an auxiliary point $Q$ away from the poles, setIt is single-valued away from the poles. Locally ${\eta }^{\prime }=n_{P}dt/t+\text{a holomorphic one-form}$, so $f=t^{n_P}$ times a nonvanishing holomorphic function. It extends meromorphically with $\mathrm{div}(f)=E$. This proves sufficiency and explicitly produces the required principal complex analytic divisor.

Conversely, if $E=\mathrm{div}(f)$, the meromorphic differential $df/f$ has residues $n_P$ and every period in $2\pi i\mathbb{Z}$, since each integral counts the winding number of $f$ about zero. Write its $a$-period vector as $2\pi ia$ and its $b$-period vector as $2\pi ib$, with $a,b\in {\mathbb{Z}}^g$. The reciprocity identity gives $v=b-\tau a\in \Lambda$. Since $E$ has degree zero, changing the base point cancels in its Abel sum; a base point at the support of $E$ causes no problem, because the Abel integrals use holomorphic forms. We have provedThis establishes the Abel theorem for divisors directly from the period and residue calculations.