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An augmented chain complex includes an augmentation $\epsilon :C_0\to \mathbb{Z}$ with $\epsilon d_1=0$. For a nonempty simplicial complex, the augmentation sends every vertex to $1$ and the augmented homology in degree zero is its reduced homology.

The boundary operator of a chain complex is its degree-minus-one differential $\partial :C_i\to C_{i-1}$. The identity ${\partial }^2=0$ says that every chain boundary is a chain cycle.

The chain group $C_i$ is the group in degree $i$ of a chain complex. Its elements are $i$-chains.

A contracting homotopy on a chain complex is a degree-one map $h$ whose displayed identity makes the identity chain-homotopic to zero. It implies that all homology vanishes. The converse is not valid for arbitrary complexes of modules.

The chain differential of a homological chain complex is a family of group or module homomorphisms ${\partial }_n:C_n\to C_{n-1}$ satisfying ${\partial }_{n-1}{\partial }_n=0$. The identity implies $\mathrm{im}{\partial }_{n+1}\subset \ker {\partial }_n$, so the quotient $H_n=\ker {\partial }_n/\mathrm{im}{\partial }_{n+1}$ is defined. In a graded complex over $\mathbb{Z}[U]$, the chain differential can have degree $-1$ while $U$ has degree $-2$, with $\partial (Ux)=U\partial x$.

For $n\ge 1$, the chain complex with copies of $R$ in degrees $n,n-1$ and identity differential. It is acyclic; a map from it chooses an arbitrary degree-$n$ element and its boundary.

The normalized-chain functor gives an equivalence between simplicial abelian groups and nonnegatively graded chain complexes of abelian groups. Its inverse is the Eilenberg–MacLane denormalization functor $K$.

A double complex has modules $C_{p,q}$ and differentials lowering either index, each squaring to zero and anticommuting with the other. The total chain complex has term ${⨁}_{p+q=i}{C}_{p,q}$ and differential the sum of the two differentials. For the tensor product of two chain complexes, a sign $(-1{)}^p$ on the second differential ensures anticommutation. If the double complex is in the first quadrant and one direction has homology only in degree zero, its total homology is computed by the surviving degree-zero complex. Applying this twice to two free resolutions proves the balanced calculation of the Tor functor.

A connected Eilenberg–MacLane space $K(G,n)$, $n\ge 1$, has ${\pi }_n\cong G$ and all other positive-degree homotopy groups trivial. The group $G$ must be abelian when $n\ge 2$, but may be nonabelian when $n=1$. For an abelian group, the Dold–Kan correspondence constructs a simplicial model from the chain complex concentrated in degree $n$. A classifying space of a discrete group gives the general $K(G,1)$ case.

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.

For an admissible pointed Heegaard diagram, $C{F}^{-}$ is a free $\mathbb{Z}[U]$ chain complex with one generator for each point of ${\mathbb{T}}_{\alpha }\cap {\mathbb{T}}_{\beta }$. Each generator is an unordered selection containing exactly one intersection on each alpha circle and exactly one on each beta circle. After choosing coherent orientations,Here $\hat{\mathcal{M}}$ is the moduli space modulo strip translation, $\mu$ is the Maslov index, and $n_z$ is multiplicity at the basepoint. Holomorphic representatives have nonnegative domain multiplicities. The differential lowers grading by one and $U$ lowers it by two. Boundary strata of one-dimensional moduli spaces give the cancellation proving ${\partial }^2=0$. Its homology is $H{F}^{-}$.

Heegaard Floer homology assigns groups $H{F}^{\circ }(Y,\mathfrak{s})$ to a closed oriented three-manifold with a Spin-c structure. A pointed Heegaard diagram of genus $g$ gives product Lagrangian tori ${\mathbb{T}}_{\alpha },{\mathbb{T}}_{\beta }\subset {\mathrm{Sym}}^g(\Sigma )$; the chain complex is generated by their intersection points and its differential counts Whitney disk classes represented by holomorphic disks. The minus, infinity, plus and hat versions use different powers or quotients of the variable $U$. The Heegaard Floer cobordism maps and the Mixed Heegaard Floer invariant relate these groups to smooth four-dimensional topology.

Homological algebra studies chain complexes, homology, cohomology, and the derived constructions that measure failure of exactness.

A homology class is an equivalence class of chain cycles modulo chain boundaries in a chain complex.

For an endomorphism of a finite-dimensional bounded chain complex over a field, the alternating chain trace equals the alternating trace on homology. Split each chain space into boundaries, representatives for homology, and a complement mapping isomorphically to the preceding boundaries. The chain-map relation makes the traces on the first and third summands cancel in successive degrees, leaving the homology traces. Over a field the induced cohomology maps are dual and have the same traces. This identity turns the absence of diagonal simplex contributions into a vanishing Lefschetz number.

The group $I{S}_i^{\bar{p}}(X)$ consists of finite singular chains whose nonzero simplices are allowable singular simplices and whose boundary, after cancellation, also consists of allowable simplices. Since ${\partial }^2=0$, these groups form a chain complex. Its homology is intersection homology.

Intersection homology is the homology of a chain complex in which chains and their boundaries satisfy codimension-dependent perversity restrictions at singular strata. Unless specified otherwise coefficients here are $\mathbb{Q}$. Ordinary homology is recovered on manifolds.

Reverse a handle decomposition of a compact $n$-dimensional manifold. Absolute $k$-handles become relative $(n-k)$-handles based on the boundary. The absolute boundary matrices count attaching sphere-belt sphere intersections; the corresponding relative matrices are their transposes over ${\mathbb{F}}_2$. The absolute chain complex is thus the complementary-degree dual of the relative chain complex. The universal coefficient theorem for cohomology proves Poincare-Lefschetz duality without an orientation hypothesis.

Use integral singular chains. The inclusion $Y\subseteq X$ identifies $C_k(Y)$ with the subgroup of $C_k(X)$ generated by the singular simplices whose images lie in $Y$. It is a chain subcomplex. DefineThis boundary is well defined because $\partial C_k(Y)\subseteq C_{k-1}(Y)$, and its square is zero because ${\partial }^2=0$. Its homology is, by definition, $H_k(X,Y)$. A relative cycle has a representative $c$ with $\partial c\in C_{k-1}(Y)$; representatives of the same relative homology class differ by $\partial b+y$, where $b\in C_{k+1}(X)$ and $y\in C_k(Y)$.

Inclusion and quotient induce homomorphisms $i_{\ast }:H_k(Y)\to H_k(X)$ and $j_{\ast }:H_k(X)\to H_k(X,Y)$. Define the relative homology connecting homomorphism byHere $\partial c$ is a cycle in $Y$ since ${\partial }^{2}c=0$. If $c$ is changed to $c+\partial b+y$, its boundary changes by $\partial y$, a boundary in $Y$, proving independence of representatives. The formula is additive, so $\delta$ is a homomorphism.

The long exact sequence in relative homology isand ends inWe prove all three types of exactness directly, so no unproved abstract lemma about chain complexes is required.

At $H_k(X)$, $j_{\ast }{i}_{\ast }=0$ because a chain in $Y$ is zero in the relative chain group. Conversely, let an absolute cycle $c$ have $j_{\ast }[c]=0$. The definition of a relative boundary gives $c=\partial b+y$ with $y\in C_k(Y)$. Since $c$ is a cycle, $\partial y=0$. Thus $[c]=i_{\ast }[y]$, proving $\ker j_{\ast }=\mathrm{im}{i}_{\ast }$.

At $H_k(X,Y)$, $\delta j_{\ast }=0$ because an absolute cycle has zero boundary. Conversely, suppose $\delta [c]=0$. Then $\partial c=\partial y$ for some $y\in C_k(Y)$. Therefore $c-y$ is an absolute cycle and $j_{\ast }[c-y]=[c]$. Hence $\ker \delta =\mathrm{im}{j}_{\ast }$.

At $H_{k-1}(Y)$, $i_{\ast }\delta =0$ because $\partial c$ is a boundary in $X$. Conversely, if a cycle $y\in C_{k-1}(Y)$ becomes zero in $H_{k-1}(X)$, then $y=\partial c$ for some $c\in C_k(X)$. That $c$ is a relative cycle, and $\delta [c]=[y]$. Thus $\ker i_{\ast }=\mathrm{im}\delta$.

These arguments also prove the assertions at shifted indices. Finally, every zero-dimensional relative cycle lifts to a chain in $C_0(X)$, and every zero-chain is an absolute cycle; this proves surjectivity onto $H_0(X,Y)$. All maps, representative changes, compositions and exactness assertions have consequently been justified from the chain definitions.

For $t$ tetrahedra, a normal surface has a vector $x\in {\mathbb{Z}}_{\ge 0}^{7t}$ counting four triangle types and three quadrilateral types in each tetrahedron. On every glued face, the numbers of arcs of each of its three types must agree on the two sides. These give homogeneous integer linear equationsFor closed surfaces the coordinates of disks meeting an unglued boundary face must also vanish. The nonnegativity and matching equations are not the whole admissibility condition: in each tetrahedron at most one of its three quadrilateral coordinates can be nonzero. Conversely, for an admissible vector, arrange the prescribed parallel disks and glue their face arcs in order. This reconstructs an embedded normal surface, unique up to normal isotopy.

A fundamental normal surface corresponds to a nonzero admissible vector which cannot be written as $y+z$ with both $y$ and $z$ nonzero admissible vectors. These indecomposable vectors form a fundamental set: every admissible vector is their finite sum. To find them, fix which quadrilateral type is allowed in each tetrahedron. This leaves the integer points of a rational polyhedral cone $Ax=0,x\ge 0$, with some coordinates set to zero. Compute its Hilbert basis of a rational cone, for example by subdividing into rational simplicial cones and enumerating the lattice points in their fundamental parallelepipeds along with their integral ray generators, then removing decomposable generators. Repeat the finite list of quadrilateral choices, remove duplicates, and remove vectors decomposable in their compatible sector. Finiteness follows because each of these pointed rational cones has a finite Hilbert basis.

Let $E$ be the link exterior. By Alexander duality, for a two-component linkAn embedded connected closed surface separates $S^3$; its mod-two class is nonzero exactly when the two link components lie on different sides. The connectedness qualification matters: two parallel splitting spheres form a disconnected surface with zero total class. We use the criterion only for connected surfaces.

Normalize a splitting sphere by isotopy, disk compressions and removal of components lying inside tetrahedra. A compression replaces a sphere by two spheres whose mod-two classes add to the old class; at least one is still a splitting sphere. The discarded tetrahedral spheres have zero class. Thus there is a normal splitting sphere. Choose one, $S$, of least normal surface weight.

Suppose its vector is decomposable. Realize $S=A+B$ as a Haken sum, choosing a decomposition with the least possible number of intersection curves. Two elementary exchange observations are useful. First, $A$ and $B$ can be taken connected. If their total union has more than two components, perform regular exchanges connecting components until exactly two embedded groups remain, resolving any intersections within each group. Since the final sum is connected, this can be done before the last joining exchange; it uses up at least one of the old intersection curves. The resulting two normal surfaces still have total vector $x(S)$, contradicting minimality. Second, no intersection curve separates both $A$ and $B$. If it did, resolve that curve first: the two separated sides give two groups. Resolve any remaining self-intersections within each group and continue exchanges between groups only while the result remains disconnected. Just before the exchange making the final connected surface, there are again exactly two embedded normal groups with fewer intersection curves. Both assertions follow from the local four-bank resolution at an intersection circle; internal crossings are resolved before regarding a group as one embedded summand.

Every closed embedded surface in $S^3$ is orientable. Additivity of Euler characteristic in the Haken sum givesFor connected orientable closed surfaces the positive Euler characteristic is $2$, and the only zero Euler characteristic is that of a torus. Thus one summand, say $A$, is a sphere and the other $B$ is a torus. Both have smaller weight than $S$. The sphere $A$ cannot split the link by minimality, soEach intersection circle separates the sphere $A$, so the second exchange observation makes it nonseparating on $B$. In particular, these circles are essential parallel curves on the torus.

Cut $A$ along the intersection circles. There are at least two innermost disk patches. Choose one of smallest weight, $D$; then $2w(D)\le w(A)$. Its interior misses $B$, and its boundary is essential on $B$. Compress the torus along two slightly separated copies of $D$. The result is a sphere $C$, and the compression takes place in the exterior, soThus $C$ is a splitting sphere.

This alone is not quite a contradiction if the last weight inequality is equality: $C$ need not yet be normal. At the compression seam it uses the alternative, rather than regular, exchange between the disk patch and the torus. An intersection circle of normal disk sheets must cross a tetrahedral face; it cannot be a closed curve entirely in two disk interiors in one tetrahedron, which can be put in standard position with only intersection arcs. In a face, parallel arcs of the same type do not cross. At a crossing of different normal arc types, the alternative exchange joins two endpoints on the same edge, producing a returning arc. Pushing across its edge bigon strictly reduces weight. Normalize the resulting surface, retaining a nonzero-class sphere at each compression as before. We obtain a normal splitting sphere of weight strictly less than $w(S)$, a contradiction. Therefore

The algorithm is now finite. Triangulate the exterior from a link diagram, enumerate its fundamental admissible vectors with the closed-surface boundary condition, reconstruct their surfaces, and inspect the connected components and Euler characteristics. A fundamental vector yields a connected surface, since otherwise its component vectors decompose it. A connected closed surface in $S^3$ with Euler characteristic two is a sphere. Compute its mod-two class in the finite simplicial chain complex, or equivalently check which link boundary tori lie on either side. Answer yes exactly when such a sphere has nonzero class. The argument proves that every split link is detected, and every detected sphere witnesses splitting. For more than two components, $H_2(E;{\mathbb{F}}_2)\cong {\mathbb{F}}_2^{\ell -1}$ and a nonzero sphere class separates a nonempty proper subset of the components; the same least-weight argument and finite enumeration apply.